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For this Question: Given a matrix of N*N dimensions. Take the input and display the matrix as an output.First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Print the given matrix.Input:
2
3 4
7 6
Output:
3 4
7 6
Input:
3
1 2 3
4 5 6
7 8 9
Output:
1 2 3
4 5 6
7 8 9, I have written this Solution Code: n = int(input())
for _ in range(n):
print(input()), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions. Take the input and display the matrix as an output.First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Print the given matrix.Input:
2
3 4
7 6
Output:
3 4
7 6
Input:
3
1 2 3
4 5 6
7 8 9
Output:
1 2 3
4 5 6
7 8 9, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin>>n;
int arr[n][n];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>arr[i][j];
}}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cout<<arr[i][j]<<" ";
}
cout<<endl;}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Arun has gone shopping for his new restaurant along with his 8-year-old daughter, Shruthi. They have bought N items. The items are numbered from 1 to N, and item i weighs W<sub>i</sub> grams. Shruthi is eager about helping carry the groceries for her father. She asks her father to give her some items to carry. Arun doesn't want to burden his daughter. But unless he offers her a few things to carry, she won't stop nagging him. Arun decides to give her a few things as a result. Arun obviously wants to give the child fewer things to carry. His daughter, though, is a smart kid.
She advises that the goods be divided into two groups, one of which has exactly K things, in order to prevent being given the absolute least weight to carry. The larger group will then be carried by Arun and the lighter group by his daughter. Help Arun decide which items the daughter should carry. You'll have an easy job to do. Give Arun the maximum possible difference between the weight carried by him and the weight carried by his daughter.The first line of input contains an integer T, denoting the number of test cases. Then T test cases follow. The first line of each test contains two space-separated integers N and K. The next line contains N space- separated integers W<sub>1</sub>, W<sub>2</sub>,...., W<sub>N</sub>.
<b>Constraints</b>
1 ≤ T ≤ 100
1 ≤ K ≤ N ≤ 100
1 ≤ W<sub>i</sub> ≤ 100,000 (10<sup>5</sup>)For each test case, output the maximum possible difference between the weights carried by both in grams.Sample Input :
2
5 2
8 4 5 2 10
8 3
1 1 1 1 1 1 1 1
Sample Output :
17
2
Explanation :
<ul><li>he optimal way is that Arun gives his daughter K = 2 items with weights 2 and 4. Arun carries the rest of the items himself. Thus the difference is (8+5+10)-(4+2) = 23-6 = 17. </li><li>Arun gives his daughter 3 items and he carries 5 items himself.</li></ul>, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int tt;
cin >> tt;
while (tt--) {
int n, k;
cin >> n >> k;
std::vector<int> v(n);
int sum = 0;
for (int i = 0; i < n; i++) {
cin >> v[i];
sum += v[i];
}
sort(v.begin(), v.end());
int start = 0, end = 0;
for (int i = 0; i < k; i++) {
start += v[i];
end += v[n - i - 1];
}
if (sum - 2 * start > 2 * end - sum) {
cout << sum - 2 * start << "\n";
} else {
cout << 2 * end - sum << "\n";
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Arun has gone shopping for his new restaurant along with his 8-year-old daughter, Shruthi. They have bought N items. The items are numbered from 1 to N, and item i weighs W<sub>i</sub> grams. Shruthi is eager about helping carry the groceries for her father. She asks her father to give her some items to carry. Arun doesn't want to burden his daughter. But unless he offers her a few things to carry, she won't stop nagging him. Arun decides to give her a few things as a result. Arun obviously wants to give the child fewer things to carry. His daughter, though, is a smart kid.
She advises that the goods be divided into two groups, one of which has exactly K things, in order to prevent being given the absolute least weight to carry. The larger group will then be carried by Arun and the lighter group by his daughter. Help Arun decide which items the daughter should carry. You'll have an easy job to do. Give Arun the maximum possible difference between the weight carried by him and the weight carried by his daughter.The first line of input contains an integer T, denoting the number of test cases. Then T test cases follow. The first line of each test contains two space-separated integers N and K. The next line contains N space- separated integers W<sub>1</sub>, W<sub>2</sub>,...., W<sub>N</sub>.
<b>Constraints</b>
1 ≤ T ≤ 100
1 ≤ K ≤ N ≤ 100
1 ≤ W<sub>i</sub> ≤ 100,000 (10<sup>5</sup>)For each test case, output the maximum possible difference between the weights carried by both in grams.Sample Input :
2
5 2
8 4 5 2 10
8 3
1 1 1 1 1 1 1 1
Sample Output :
17
2
Explanation :
<ul><li>he optimal way is that Arun gives his daughter K = 2 items with weights 2 and 4. Arun carries the rest of the items himself. Thus the difference is (8+5+10)-(4+2) = 23-6 = 17. </li><li>Arun gives his daughter 3 items and he carries 5 items himself.</li></ul>, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner in=new Scanner(System.in);
int test=in.nextInt();
while(test-->0)
{
int n=in.nextInt();
int k=in.nextInt();
int arr[]=new int[n];
for(int i=0; i<n; i++){
arr[i]=in.nextInt();
}
Arrays.sort(arr);
if(k>n/2)
{
k=n-k;
}
int sum=0;
for(int i=0; i<k; i++){
sum=sum+arr[i];
}
int sum1=0;
for(int i=k; i<n; i++){
sum1=sum1+arr[i];
}
if(sum>sum1){
int total=sum-sum1;
System.out.println(total);
}else{
int total=sum1-sum;
System.out.println(total);
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S (containing only letters of the English alphabet), find all the capital letters in it (in exactly same order as the string)
Note:- It is guaranteed that the given string will contain at least one capital.Input contains a single line containing the string S.
Constraints:-
1 <= |S| <= 100Print all capital letters present in the string separated by space, in the order of their appearance in the main string.Sample input:-
AbcdEF
Sample Output:-
A E F
Sample Input:-
NewtonSchool
Sample Output:-
N S, I have written this Solution Code: n=input()
for element in n:
if element.isupper():
print(element,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S (containing only letters of the English alphabet), find all the capital letters in it (in exactly same order as the string)
Note:- It is guaranteed that the given string will contain at least one capital.Input contains a single line containing the string S.
Constraints:-
1 <= |S| <= 100Print all capital letters present in the string separated by space, in the order of their appearance in the main string.Sample input:-
AbcdEF
Sample Output:-
A E F
Sample Input:-
NewtonSchool
Sample Output:-
N S, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1000001
#define MOD 1000000000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int cnt[max1];
signed main(){
string s;
cin>>s;
int n=s.length();
//out(s);
FOR(i,n){
if(s[i]>='A' && s[i]<='Z'){
out1(s[i]);
}
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S (containing only letters of the English alphabet), find all the capital letters in it (in exactly same order as the string)
Note:- It is guaranteed that the given string will contain at least one capital.Input contains a single line containing the string S.
Constraints:-
1 <= |S| <= 100Print all capital letters present in the string separated by space, in the order of their appearance in the main string.Sample input:-
AbcdEF
Sample Output:-
A E F
Sample Input:-
NewtonSchool
Sample Output:-
N S, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
String s = sc.next();
int n = s.length();
for(int i=0;i<n;i++){
if(s.charAt(i)>='A' && s.charAt(i)<='Z'){
System.out.print(s.charAt(i)+" ");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono loves to do shopping. Today, she went to the market where there are N different types of products. She wants to buy exactly K of them at the minimum cost. Although she is super smart, she wants to check your smartness as well. Can you tell her the minimum cost required to buy exactly K products if she has already decided to buy product J?
<b>Note:</b> Tono does not buy the same product twice, and Tono will definitely buy product J (J is the <b>index</b> of the item).The first line of the input contains three integers, N, K, and J, denoting the number of products in the market, the number of products Tono needs to buy, and the product that Tono will definitely buy.
The next line contains N singly spaced integers, the cost of the N products C[1], C[2], ..., C[N].
<b>Constraints:</b>
1 <= N <= 200000
1 <= K <= N
1 <= J <= N
1 <= C[i] <= 1000
Output a single integer, the minimum amount Tono needs to pay.Sample Input 1:
5 3 4
1 2 3 4 5
Sample Output 1:
7
Sample Input 2:
5 1 3
2 4 3 1 1
Sample Output 2:
3
<b>Explanation:</b>
Tono needs to buy exactly 3 products, and she will definitely buy the 4th product. Thus, she will buy the 1st, 2nd, and the 4th product. The total cost she pays is 1+2+4=7.
, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main {
public static void main(String[] args) throws IOException {
int n = io.nextInt(), k = io.nextInt(), j = io.nextInt() - 1;
int[] arr = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = io.nextInt();
}
int cost = arr[j];
arr[j] = Integer.MAX_VALUE;
Arrays.sort(arr);
for(int i = 0; i < k - 1; i++) {
cost += arr[i];
}
io.println(cost);
io.close();
}
static IO io = new IO();
static class IO {
private byte[] buf;
private InputStream in;
private PrintWriter pw;
private int total, index;
public IO() {
buf = new byte[1024];
in = System.in;
pw = new PrintWriter(System.out);
}
public int next() throws IOException {
if(total < 0)
throw new InputMismatchException();
if(index >= total) {
index = 0;
total = in.read(buf);
if(total <= 0)
return -1;
}
return buf[index++];
}
public int nextInt() throws IOException {
int n = next(), integer = 0;
while(isWhiteSpace(n))
n = next();
int neg = 1;
if(n == '-') {
neg = -1;
n = next();
}
while(!isWhiteSpace(n)) {
if(n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = next();
}
else
throw new InputMismatchException();
}
return neg * integer;
}
public int[] nextIntArray(int n) throws IOException {
int[] arr = new int[n];
for(int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
public long nextLong() throws IOException {
long integer = 0l;
int n = next();
while(isWhiteSpace(n))
n = next();
int neg = 1;
if(n == '-') {
neg = -1;
n = next();
}
while(!isWhiteSpace(n)) {
if(n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = next();
}
else
throw new InputMismatchException();
}
return neg * integer;
}
public double nextDouble() throws IOException {
double doub = 0;
int n = next();
while(isWhiteSpace(n))
n = next();
int neg = 1;
if(n == '-') {
neg = -1;
n = next();
}
while(!isWhiteSpace(n) && n != '.') {
if(n >= '0' && n <= '9') {
doub *= 10;
doub += n - '0';
n = next();
}
else
throw new InputMismatchException();
}
if(n == '.') {
n = next();
double temp = 1;
while(!isWhiteSpace(n)) {
if(n >= '0' && n <= '9') {
temp /= 10;
doub += (n - '0') * temp;
n = next();
}
else
throw new InputMismatchException();
}
}
return doub * neg;
}
public String nextString() throws IOException {
StringBuilder sb = new StringBuilder();
int n = next();
while(isWhiteSpace(n))
n = next();
while(!isWhiteSpace(n)) {
sb.append((char)n);
n = next();
}
return sb.toString();
}
public String nextLine() throws IOException {
int n = next();
while(isWhiteSpace(n))
n = next();
StringBuilder sb = new StringBuilder();
while(!isEndOfLine(n)) {
sb.append((char)n);
n = next();
}
return sb.toString();
}
private boolean isWhiteSpace(int n) {
return n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1;
}
private boolean isEndOfLine(int n) {
return n == '\n' || n == '\r' || n == -1;
}
public void print(Object obj) {
pw.print(obj);
}
public void println(Object... obj) {
if(obj.length == 1)
pw.println(obj[0]);
else {
for(Object o: obj)
pw.print(o + " ");
pw.println();
}
}
public void flush() throws IOException {
pw.flush();
}
public void close() throws IOException {
pw.close();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono loves to do shopping. Today, she went to the market where there are N different types of products. She wants to buy exactly K of them at the minimum cost. Although she is super smart, she wants to check your smartness as well. Can you tell her the minimum cost required to buy exactly K products if she has already decided to buy product J?
<b>Note:</b> Tono does not buy the same product twice, and Tono will definitely buy product J (J is the <b>index</b> of the item).The first line of the input contains three integers, N, K, and J, denoting the number of products in the market, the number of products Tono needs to buy, and the product that Tono will definitely buy.
The next line contains N singly spaced integers, the cost of the N products C[1], C[2], ..., C[N].
<b>Constraints:</b>
1 <= N <= 200000
1 <= K <= N
1 <= J <= N
1 <= C[i] <= 1000
Output a single integer, the minimum amount Tono needs to pay.Sample Input 1:
5 3 4
1 2 3 4 5
Sample Output 1:
7
Sample Input 2:
5 1 3
2 4 3 1 1
Sample Output 2:
3
<b>Explanation:</b>
Tono needs to buy exactly 3 products, and she will definitely buy the 4th product. Thus, she will buy the 1st, 2nd, and the 4th product. The total cost she pays is 1+2+4=7.
, I have written this Solution Code:
a=input().split()
b=input().split()
for j in [a,b]:
for i in range(0,len(j)):
j[i]=int(j[i])
n,k,j=a[0],a[1],a[2]
c_j=b[j-1]
b.sort()
if b[k-1]<=c_j:
b[k-1]=c_j
sum=0
for i in range(0,k):
sum+=b[i]
print(sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono loves to do shopping. Today, she went to the market where there are N different types of products. She wants to buy exactly K of them at the minimum cost. Although she is super smart, she wants to check your smartness as well. Can you tell her the minimum cost required to buy exactly K products if she has already decided to buy product J?
<b>Note:</b> Tono does not buy the same product twice, and Tono will definitely buy product J (J is the <b>index</b> of the item).The first line of the input contains three integers, N, K, and J, denoting the number of products in the market, the number of products Tono needs to buy, and the product that Tono will definitely buy.
The next line contains N singly spaced integers, the cost of the N products C[1], C[2], ..., C[N].
<b>Constraints:</b>
1 <= N <= 200000
1 <= K <= N
1 <= J <= N
1 <= C[i] <= 1000
Output a single integer, the minimum amount Tono needs to pay.Sample Input 1:
5 3 4
1 2 3 4 5
Sample Output 1:
7
Sample Input 2:
5 1 3
2 4 3 1 1
Sample Output 2:
3
<b>Explanation:</b>
Tono needs to buy exactly 3 products, and she will definitely buy the 4th product. Thus, she will buy the 1st, 2nd, and the 4th product. The total cost she pays is 1+2+4=7.
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n, k, j; cin>>n>>k>>j;
vector<int> vect;
int ans = 0;
For(i, 1, n+1){
int a; cin>>a;
if(i!=j)
vect.pb(a);
else
ans += a;
}
sort(all(vect));
for(int i=0; i<k-1; i++){
ans += vect[i];
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Modify the <code>takeMultipleNumbersAndAdd </code> such that it can take any number of arguments
and return its sum.
This is JS only question.Function should be able to take any number of argsSum of the numberstakeMultipleNumbersAndAdd(1, 2, 2) should return 5 because 1 + 2 + 2
takeMultipleNumbersAndAdd(-1, 2, -1, 5) should return 5, I have written this Solution Code:
function takeMultipleNumbersAndAdd (...nums){
// write your code here
return nums.reduce((prev,cur)=>prev+cur,0)
// return the output using return keyword
// do not console.log it
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Myra has x red colour balls and y blue colour balls. She wants to distribute these balls into identical groups without any balls left. What is the largest number of groups she can make?
Identical groups mean there are equal number of same coloured balls in each group.A single line containing two integers x and y.
Constraints:
1<=x, y<=10000000Return the largest number of groups that can be made.Sample Input
3 9
Sample Output
3
Explanation:-
1 red ball and 3 blue ball will be in each group.
Sample Input:
4 9
Sample Output:
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int gcd(int x,int y)
{
if(x%y==0)
return y;
if(y%x==0)
return x;
int gc=1;
int min=x;
if(min>y)
min=y;
for(int i=2;i<=Math.sqrt(min);i++)
{
if(x%i==0 )
{
if(x%i==0 && y%i==0)
{
if(i>gc)
gc=i;
}
int a=x/i;
if(x%a==0 && y%a==0)
{
gc=a;
break;
}
}
}
return gc;
}
public static void main (String[] args) throws IOException {
BufferedReader bf =new BufferedReader(new InputStreamReader(System.in));
String str[]=bf.readLine().split(" ");
int x=Integer.parseInt(str[0]);
int y=Integer.parseInt(str[1]);
System.out.println(gcd(x,y));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Myra has x red colour balls and y blue colour balls. She wants to distribute these balls into identical groups without any balls left. What is the largest number of groups she can make?
Identical groups mean there are equal number of same coloured balls in each group.A single line containing two integers x and y.
Constraints:
1<=x, y<=10000000Return the largest number of groups that can be made.Sample Input
3 9
Sample Output
3
Explanation:-
1 red ball and 3 blue ball will be in each group.
Sample Input:
4 9
Sample Output:
1, I have written this Solution Code: def gcd(a, b):
result = min(a, b)
while result:
if a % result == 0 and b % result == 0:
break
result -= 1
return result
one,two = map(int,input().split())
print(gcd(one,two)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Myra has x red colour balls and y blue colour balls. She wants to distribute these balls into identical groups without any balls left. What is the largest number of groups she can make?
Identical groups mean there are equal number of same coloured balls in each group.A single line containing two integers x and y.
Constraints:
1<=x, y<=10000000Return the largest number of groups that can be made.Sample Input
3 9
Sample Output
3
Explanation:-
1 red ball and 3 blue ball will be in each group.
Sample Input:
4 9
Sample Output:
1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long x,y;
cin>>x>>y;
cout<<__gcd(x,y);
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings a and b consisting of lowercase characters. The task is to check whether two given strings are an anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “act” and “tac” are an anagram of each other.Input consists of two strings in lowercase english characters.
Constraints:
1 ≤ |s1|, |s2| ≤ 10^5Print "YES" without quotes if the two strings are anagram else print "NO".Sample Input
naman
manan
Sample Output
YES
Explanation: Both String contain 2 'a's, 2 'n's and 1 'm'., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s1 = br.readLine();
String s2 = br.readLine();
boolean flag = true;
int[] arr1 = new int[26];
int[] arr2 = new int[26];
for(int i=0; i<s1.length(); i++){
arr1[s1.charAt(i)-97]++;
}
for(int i=0; i<s2.length(); i++){
arr2[s2.charAt(i)-97]++;
}
for(int i=0; i<25; i++){
if(arr1[i]!=arr2[i]){
flag = false;
break;
}
}
if(flag==true)
System.out.print("YES");
else System.out.print("NO");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings a and b consisting of lowercase characters. The task is to check whether two given strings are an anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “act” and “tac” are an anagram of each other.Input consists of two strings in lowercase english characters.
Constraints:
1 ≤ |s1|, |s2| ≤ 10^5Print "YES" without quotes if the two strings are anagram else print "NO".Sample Input
naman
manan
Sample Output
YES
Explanation: Both String contain 2 'a's, 2 'n's and 1 'm'., I have written this Solution Code: s1 = input().strip()
s2 = input().strip()
dict1 = dict()
dict2 = dict()
for i in s1:
dict1[i] = dict1.get(i, 0) + 1
for j in s2:
dict2[j] = dict2.get(j, 0) + 1
print(("NO", "YES")[dict1 == dict2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings a and b consisting of lowercase characters. The task is to check whether two given strings are an anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “act” and “tac” are an anagram of each other.Input consists of two strings in lowercase english characters.
Constraints:
1 ≤ |s1|, |s2| ≤ 10^5Print "YES" without quotes if the two strings are anagram else print "NO".Sample Input
naman
manan
Sample Output
YES
Explanation: Both String contain 2 'a's, 2 'n's and 1 'm'., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
int A[26],B[26];
signed main()
{
string s,p;
cin>>s>>p;
for(int i=0;i<s.size();i++)
{
int y=s[i]-'a';
A[y]++;
}
for(int i=0;i<p.size();i++)
{
int y=p[i]-'a';
B[y]++;
}int ch=1;
for(int i=0;i<26;i++)
{
if(B[i]!=A[i])ch=0;
}
if(ch==1) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings a and b consisting of lowercase characters. The task is to check whether two given strings are an anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “act” and “tac” are an anagram of each other.Input consists of two strings in lowercase english characters.
Constraints:
1 ≤ |s1|, |s2| ≤ 10^5Print "YES" without quotes if the two strings are anagram else print "NO".Sample Input
naman
manan
Sample Output
YES
Explanation: Both String contain 2 'a's, 2 'n's and 1 'm'., I have written this Solution Code: // str1 and str2 are the two input strings
function isAnagram(str1,str2){
// Get lengths of both strings
let n1 = str1.length;
let n2 = str2.length;
// If length of both strings is not same,
// then they cannot be anagram
if (n1 != n2)
return "NO";
str1 = str1.split('')
str2 = str2.split('')
// Sort both strings
str1.sort();
str2.sort()
// Compare sorted strings
for (let i = 0; i < n1; i++)
if (str1[i] != str2[i])
return "NO";
return "YES";
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a postfix expression, your task is to evaluate given expression.
Infix expression: The expression of the form a operator b. When an operator is in-between every pair of operands.
Postfix expression: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.The first line denotes the size of the string(which contains number or operand in form of string) i.e. N.
The next line contains string. The string contains an integer called operand or any of the four operators (+ - * /)
Constraints:-
1 <= n <= 40
1<=number<=500
Output the value of arithmetic expression formed using reverse Polish Notation.Input 1:
5
2 1 + 3 *
Output 1:
9
Explaination 1:
starting from backside:
*: ( )*( )
3: ()*(3)
+: ( () + () )*(3)
1: ( () + (1) )*(3)
2: ( (2) + (1) )*(3)
((2)+(1))*(3) = 9
Input 2:
5
4 13 5 / +
Output 2:
6
Explanation 2:
+: ()+()
/: ()+(() / ())
5: ()+(() / (5))
1: ()+((13) / (5))
4: (4)+((13) / (5))
(4)+((13) / (5)) = 6, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
String str[] = br.readLine().split(" ");
Stack <Integer> st = new Stack<>();
for(int i=0;i<t;i++){
char ch = str[i].charAt(0);
if(Character.isDigit(ch)){
st.push(Integer.parseInt(str[i]));
}
else{
int str1 = st.pop();
int str2 = st.pop();
switch(ch) {
case '+':
st.push(str2+str1);
break;
case '-':
st.push(str2- str1);
break;
case '/':
st.push(str2/str1);
break;
case '*':
st.push(str2*str1);
break;
}
}
}
System.out.println(st.peek());
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a postfix expression, your task is to evaluate given expression.
Infix expression: The expression of the form a operator b. When an operator is in-between every pair of operands.
Postfix expression: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.The first line denotes the size of the string(which contains number or operand in form of string) i.e. N.
The next line contains string. The string contains an integer called operand or any of the four operators (+ - * /)
Constraints:-
1 <= n <= 40
1<=number<=500
Output the value of arithmetic expression formed using reverse Polish Notation.Input 1:
5
2 1 + 3 *
Output 1:
9
Explaination 1:
starting from backside:
*: ( )*( )
3: ()*(3)
+: ( () + () )*(3)
1: ( () + (1) )*(3)
2: ( (2) + (1) )*(3)
((2)+(1))*(3) = 9
Input 2:
5
4 13 5 / +
Output 2:
6
Explanation 2:
+: ()+()
/: ()+(() / ())
5: ()+(() / (5))
1: ()+((13) / (5))
4: (4)+((13) / (5))
(4)+((13) / (5)) = 6, I have written this Solution Code: stack = []
n = int(input())
exp = [i for i in input().split()]
for i in exp:
try:
stack.append(int(i))
except:
a1 = stack.pop()
a2 = stack.pop()
if i == '+':
stack.append(a1+a2)
if i == '-':
stack.append(a2-a1)
if i == '/':
stack.append(a2//a1)
if i == '*':
stack.append(a1*a2)
print(*stack), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a postfix expression, your task is to evaluate given expression.
Infix expression: The expression of the form a operator b. When an operator is in-between every pair of operands.
Postfix expression: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.The first line denotes the size of the string(which contains number or operand in form of string) i.e. N.
The next line contains string. The string contains an integer called operand or any of the four operators (+ - * /)
Constraints:-
1 <= n <= 40
1<=number<=500
Output the value of arithmetic expression formed using reverse Polish Notation.Input 1:
5
2 1 + 3 *
Output 1:
9
Explaination 1:
starting from backside:
*: ( )*( )
3: ()*(3)
+: ( () + () )*(3)
1: ( () + (1) )*(3)
2: ( (2) + (1) )*(3)
((2)+(1))*(3) = 9
Input 2:
5
4 13 5 / +
Output 2:
6
Explanation 2:
+: ()+()
/: ()+(() / ())
5: ()+(() / (5))
1: ()+((13) / (5))
4: (4)+((13) / (5))
(4)+((13) / (5)) = 6, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
class Solution {
public:
int evalRPN(vector<string> &tokens) {
int size = tokens.size();
if (size == 0)
return 0;
std::stack<int> operands;
for (int i = 0; i < size; ++i)
{
std::string cur_token = tokens[i];
if ((cur_token == "*") || (cur_token == "/") || (cur_token == "+") || (cur_token == "-"))
{
int opr2 = operands.top();
operands.pop();
int opr1 = operands.top();
operands.pop();
int result = this->eval(opr1, opr2, cur_token);
operands.push(result);
}
else{
operands.push(std::atoi(cur_token.c_str()));
}
}
return operands.top();
}
int eval(int opr1, int opr2, string opt)
{
if (opt == "*")
{
return opr1 * opr2;
}
else if (opt == "+")
{
return opr1 + opr2;
}
else if (opt == "-")
{
return opr1 - opr2;
}
else if (opt == "/")
{
return opr1 / opr2;
}
return 0;
}
};
signed main() {
IOS;
int n; cin >> n;
vector<string> v(n, "");
for(int i = 0; i < n; i++)
cin >> v[i];
Solution s;
cout << s.evalRPN(v);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array <strong>arr[]</strong> of size <strong>N</strong> and an element <strong>X</strong>. The task is to find and print the indices of the given element if it is present in array if not then print “<strong>Not found</strong>” without quotes.
<strong>Note</strong>: The elements may be present more than once.The first line of input contains T, denoting the number of test cases.
The first line of each test case contains N and X, N is the size of array and X is an element. Second line contains elements of array space separated. If not present then print "Not found" without quotes
<strong>Constraints</strong>:
1 <= T <= 100
1 <= N, X <= 10000
1 <= arr[i] <= 100000For each test case in new line you need to print all the positions where you find the X separated by space.
Assume 0-indexingInput:
2
5 6
2 3 6 5 6
4 3
2 4 6 5
Output:
2 4
Not found, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int t; cin >> t;
while(t--){
vector<int> v;
int n, x; cin >> n >> x;
for(int i = 1; i <= n; i++){
int p; cin >> p;
if(p == x)
v.push_back(i-1);
}
if(v.size() == 0)
cout << "Not found\n";
else{
for(auto i: v)
cout << i << " ";
cout << endl;
}
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array <strong>arr[]</strong> of size <strong>N</strong> and an element <strong>X</strong>. The task is to find and print the indices of the given element if it is present in array if not then print “<strong>Not found</strong>” without quotes.
<strong>Note</strong>: The elements may be present more than once.The first line of input contains T, denoting the number of test cases.
The first line of each test case contains N and X, N is the size of array and X is an element. Second line contains elements of array space separated. If not present then print "Not found" without quotes
<strong>Constraints</strong>:
1 <= T <= 100
1 <= N, X <= 10000
1 <= arr[i] <= 100000For each test case in new line you need to print all the positions where you find the X separated by space.
Assume 0-indexingInput:
2
5 6
2 3 6 5 6
4 3
2 4 6 5
Output:
2 4
Not found, I have written this Solution Code: def position(n,arr,x):
res = []
cnt = 0
for i in arr:
if(i == x):
res.append(cnt)
cnt += 1
return res
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array <strong>arr[]</strong> of size <strong>N</strong> and an element <strong>X</strong>. The task is to find and print the indices of the given element if it is present in array if not then print “<strong>Not found</strong>” without quotes.
<strong>Note</strong>: The elements may be present more than once.The first line of input contains T, denoting the number of test cases.
The first line of each test case contains N and X, N is the size of array and X is an element. Second line contains elements of array space separated. If not present then print "Not found" without quotes
<strong>Constraints</strong>:
1 <= T <= 100
1 <= N, X <= 10000
1 <= arr[i] <= 100000For each test case in new line you need to print all the positions where you find the X separated by space.
Assume 0-indexingInput:
2
5 6
2 3 6 5 6
4 3
2 4 6 5
Output:
2 4
Not found, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t =Integer.parseInt(read.readLine());
while(t-- > 0)
{
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int x = Integer.parseInt(str[1]);
int arr[] = new int[n];
str = read.readLine().trim().split(" ");
for(int i = 0; i < n; i++)
arr[i] = Integer.parseInt(str[i]);
findPositions(arr, n, x);
}
}
static void findPositions(int arr[], int n, int x)
{
boolean flag = false;
StringBuffer sb = new StringBuffer();
for(int i = 0; i < n; i++)
{
if(arr[i] == x)
{
sb.append(i + " ");
flag = true;
}
}
if(flag ==true)
System.out.println(sb.toString());
else System.out.println("Not found");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is class of students. Their teacher is taking their attendance in the morning as usual. But this time, the list of students in the register feels weird. There are suddenly more names there. Apparently a few mischevious students have repeated their names in the register. Find the number of these mischevious students who have multiple names in the register.The first line of the input contains a single integer N - the number of names in the register.
The next N lines of the input each contains a string.
Constraints:
1 <= N <= 10<sup>4</sup>
1 <= |S<sub>i</sub>| <= 100Find the number of mischevious students who have multiple names in the register.Sample Input:
5
Newton
Einstein
Newton
Bohr
Einstein
Sample Output:
2
Explaination:
Only two students have repeated names in the register., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s[] = br.readLine().split(" ");
int n = Integer.parseInt(s[0]);
HashMap<String, Integer> hm = new HashMap<>();
for(int i=0; i<n; i++){
s = br.readLine().split(" ");
if(hm.containsKey(s[0])){
hm.put(s[0],hm.get(s[0])+1);
}else{
hm.put(s[0],1);
}
}
int count=0;
for(Integer val:hm.values()){
if(val>1){
count++;
}
}
System.out.print(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is class of students. Their teacher is taking their attendance in the morning as usual. But this time, the list of students in the register feels weird. There are suddenly more names there. Apparently a few mischevious students have repeated their names in the register. Find the number of these mischevious students who have multiple names in the register.The first line of the input contains a single integer N - the number of names in the register.
The next N lines of the input each contains a string.
Constraints:
1 <= N <= 10<sup>4</sup>
1 <= |S<sub>i</sub>| <= 100Find the number of mischevious students who have multiple names in the register.Sample Input:
5
Newton
Einstein
Newton
Bohr
Einstein
Sample Output:
2
Explaination:
Only two students have repeated names in the register., I have written this Solution Code: dicti = {}
numOfInput = int(input())
namesList = []
for _ in range(numOfInput):
st = input()
namesList.append(st);
for name in namesList:
if name not in dicti.keys():
dicti[name] = 1
else:
dicti[name] += 1
ans = 0
for count in dicti.values():
if count > 1:
ans += 1
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is class of students. Their teacher is taking their attendance in the morning as usual. But this time, the list of students in the register feels weird. There are suddenly more names there. Apparently a few mischevious students have repeated their names in the register. Find the number of these mischevious students who have multiple names in the register.The first line of the input contains a single integer N - the number of names in the register.
The next N lines of the input each contains a string.
Constraints:
1 <= N <= 10<sup>4</sup>
1 <= |S<sub>i</sub>| <= 100Find the number of mischevious students who have multiple names in the register.Sample Input:
5
Newton
Einstein
Newton
Bohr
Einstein
Sample Output:
2
Explaination:
Only two students have repeated names in the register., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin >> n;
vector<string> a(n);
map<string, int> mp;
set<string> s;
for(auto &i : a) cin >> i;
for(auto &i : a){
if(mp[i] > 0){
s.insert(i);
}
mp[i]++;
}
cout << s.size();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, containing only letters { 'N', 'E', 'W', 'T', 'O'} like "NWTEONEW" or "NEWTON" your task is to print all the substring of size 10 which occur more than once in the given string.Input contains a single line containing the string S.
Constraints:-
10 < = |String| < = 10000Print all the sequences in lexicographical order separated by spaces, if their is no such substring exist print -1.Sample Input:-
NEWTONNEWTONNEWTON
Sample Output:-
EWTONNEWTO NEWTONNEWT WTONNEWTON
Sample Input:-
NETOWNEWOTTONTOONW
Sample Output:-
-1, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main
{
public static List<String> repeatedString(String str){
HashMap<String, Integer> map = new HashMap<>();
List <String>list = new ArrayList<>();
int p1= 0;
int p2 = 10;
while(p2 <= str.length()){
String temp = str.substring(p1,p2);
if(map.containsKey(temp))
{
if(!list.contains(temp)){
list.add(temp);
}
}else{
map.put(temp,1);
}
p1++; p2++;
}
return list;
}
public static void main (String[] args)
{
Scanner sc = new Scanner(System.in);
String str = sc.next();
List<String> list = repeatedString(str);
Collections.sort(list);
if(list.size() != 0)
for(int i = 0; i < list.size(); i++){
System.out.print(list.get(i)+" ");
}
else
System.out.print(-1);
System.out.println();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, containing only letters { 'N', 'E', 'W', 'T', 'O'} like "NWTEONEW" or "NEWTON" your task is to print all the substring of size 10 which occur more than once in the given string.Input contains a single line containing the string S.
Constraints:-
10 < = |String| < = 10000Print all the sequences in lexicographical order separated by spaces, if their is no such substring exist print -1.Sample Input:-
NEWTONNEWTONNEWTON
Sample Output:-
EWTONNEWTO NEWTONNEWT WTONNEWTON
Sample Input:-
NETOWNEWOTTONTOONW
Sample Output:-
-1, I have written this Solution Code: string = input()
Dict = {}
n = len(string) - 9
for i in range(n):
sub = string[i:i+10]
if sub not in Dict:
Dict[sub] = 1
else :
Dict[sub] = Dict[sub] + 1
l = []
for i in Dict:
if Dict[i] > 1:
l.append(i)
l = sorted(l)
if len(l)==0:
print(-1)
else:
for i in l:
print(i , end = " "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, containing only letters { 'N', 'E', 'W', 'T', 'O'} like "NWTEONEW" or "NEWTON" your task is to print all the substring of size 10 which occur more than once in the given string.Input contains a single line containing the string S.
Constraints:-
10 < = |String| < = 10000Print all the sequences in lexicographical order separated by spaces, if their is no such substring exist print -1.Sample Input:-
NEWTONNEWTONNEWTON
Sample Output:-
EWTONNEWTO NEWTONNEWT WTONNEWTON
Sample Input:-
NETOWNEWOTTONTOONW
Sample Output:-
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 200001
#define MOD 1000000007
#define read(type) readInt<type>()
#define int long long
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
unordered_map<string,int> m;
signed main(){
string s;
cin>>s;
string p=s.substr(0,10);
m[p]++;
for(int i=10;i<s.length();i++){
p=s.substr(i-10+1,10);
m[p]++;
}
vector<string> v;
for(auto it=m.begin();it!=m.end();it++){
if(it->second>1){v.EB(it->first);}
}
if(v.size()==0){out(-1);return 0;}
sort(v.begin(),v.end());
FOR(i,v.size()){
out1(v[i]);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the sum of all the digits of the number
Note: Use a recursive method to solve this problem.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Sum()</b> that takes the integer N as a parameter.
Constraints:
1 ≤ T ≤ 100
0 ≤ N ≤ 1000000000Return sum of digits.Sample Input
2
25
28
Sample Output
7
10, I have written this Solution Code: // n is the input number
function recSum(n) {
// write code here
// do not console.log
// return the answer as a number
if (n < 10) return n;
return n % 10 + recSum(Math.floor(n / 10))
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the sum of all the digits of the number
Note: Use a recursive method to solve this problem.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Sum()</b> that takes the integer N as a parameter.
Constraints:
1 ≤ T ≤ 100
0 ≤ N ≤ 1000000000Return sum of digits.Sample Input
2
25
28
Sample Output
7
10, I have written this Solution Code: static long Sum(long n)
{ if(n==0){return 0;}
return n%10+Sum(n/10);
} , In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Anya owns N triplets of integers (A<sub>i</sub>, B<sub>i</sub>, C<sub>i</sub>), for each i from 1 to N. She asks you to find a sequence of integers which satisfies the following conditions:
1. The sequence contains at least C<sub>i</sub> distinct integers from the closed interval [A<sub>i</sub>, B<sub>i</sub>], for each i from 1 to N.
2. Out of all sequences satisfying the first condition, choose a sequence with the minimum possible number of elements.
For simplicity, she asks you to just print the length of such a sequence.The first line of the input contains a single integer N denoting the number of triplets.
Then N lines follow, where the i<sup>th</sup> line contains three integers A<sub>i</sub>, B<sub>i</sub>, C<sub>i</sub> for each i from 1 to N.
<b> Constraints: </b>
1 ≤ N ≤ 2000
1 ≤ A<sub>i</sub> ≤ B<sub>i</sub> ≤ 2000
0 ≤ C<sub>i</sub> ≤ B<sub>i</sub> - A<sub>i</sub> + 1Print a single integer — the minimum possible sequence length.Sample Input 1:
1
1 3 3
Sample Output 1:
3
Sample Explanation 1:
Since there are only 3 elements in the closed interval [1,3], and we need to take 3 of them, clearly the smallest possible length is 3.
Sample Input 2:
2
1 3 1
3 5 1
Sample Output 2:
1
Sample Explanation 2:
We can take the sequence consisting of a single element {3}., I have written this Solution Code: l = [0]*2001
k = []
n = int(input())
for i in range(n):
a,b,c = map(int,input().split())
k.append([b,(a,c)])
k.sort()
for b,aa in k:
a = aa[0]
c = aa[1]
cnt = 0
for i in range(b,a-1,-1):
if l[i]:
cnt+=1
if cnt>=c:
continue
else:
for i in range(b,a-1,-1):
if not l[i]:
l[i]=1
cnt+=1
if cnt==c:
break
print(sum(l)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Anya owns N triplets of integers (A<sub>i</sub>, B<sub>i</sub>, C<sub>i</sub>), for each i from 1 to N. She asks you to find a sequence of integers which satisfies the following conditions:
1. The sequence contains at least C<sub>i</sub> distinct integers from the closed interval [A<sub>i</sub>, B<sub>i</sub>], for each i from 1 to N.
2. Out of all sequences satisfying the first condition, choose a sequence with the minimum possible number of elements.
For simplicity, she asks you to just print the length of such a sequence.The first line of the input contains a single integer N denoting the number of triplets.
Then N lines follow, where the i<sup>th</sup> line contains three integers A<sub>i</sub>, B<sub>i</sub>, C<sub>i</sub> for each i from 1 to N.
<b> Constraints: </b>
1 ≤ N ≤ 2000
1 ≤ A<sub>i</sub> ≤ B<sub>i</sub> ≤ 2000
0 ≤ C<sub>i</sub> ≤ B<sub>i</sub> - A<sub>i</sub> + 1Print a single integer — the minimum possible sequence length.Sample Input 1:
1
1 3 3
Sample Output 1:
3
Sample Explanation 1:
Since there are only 3 elements in the closed interval [1,3], and we need to take 3 of them, clearly the smallest possible length is 3.
Sample Input 2:
2
1 3 1
3 5 1
Sample Output 2:
1
Sample Explanation 2:
We can take the sequence consisting of a single element {3}., I have written this Solution Code: //Author: Xzirium
//Time and Date: 00:28:35 28 December 2021
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t clk;
clk = clock();
//-----------------------------------------------------------------------------------------------------------//
READV(N);
vector<pair<pll,ll>> Z;
FORI(i,0,N)
{
READV(a);
READV(b);
READV(c);
Z.pb({{b,a},c});
}
sort(Z.begin(),Z.end());
ordered_set1<ll> unused;
FORI(i,1,2001)
{
unused.insert(i);
}
ll ans=0;
FORI(i,0,N)
{
ll a=Z[i].fi.se;
ll b=Z[i].fi.fi;
ll c=Z[i].se;
ll curr=b-a+1-(unused.order_of_key(a-1)-unused.order_of_key(b));
while(curr<c)
{
ans++;
curr++;
auto it=unused.lower_bound(b);
unused.erase(it);
}
}
cout<<ans<<endl;
//-----------------------------------------------------------------------------------------------------------//
clk = clock() - clk;
cerr << fixed << setprecision(6) << "Time: " << ((double)clk)/CLOCKS_PER_SEC << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary string S. The task is to count the number of substrings that start and end with 1. For example, if the input string is “00100101”, then there are three substrings “1001”, “100101” and “101”.User Task:
Since this will be a functional problem, you don't have to take input. You just have to complete the function binarySubstring() which takes size of string and string S as a parameter.
Constraints:
1 ≤ T ≤ 100
1 ≤ |S| ≤ 10000For each testcase, in a new line, print the number of substring starting and ending with 1 in a separate line.Input:
2
4
1111
5
01101
Output:
6
3
Example:
Testcase 1: There are 6 substrings from the given string. They are 11, 11, 11, 111, 111, 1111.
Testcase 2: There 3 substrings from the given string. They are 11, 101, 1101., I have written this Solution Code: t=int(input())
for i in range(t):
ans=0
n=int(input())
s=input()
c=s.count('1')
if(c>=2):
for i in range(1,c):
ans=ans+(c-i)
print(ans)
else:
print(0), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary string S. The task is to count the number of substrings that start and end with 1. For example, if the input string is “00100101”, then there are three substrings “1001”, “100101” and “101”.User Task:
Since this will be a functional problem, you don't have to take input. You just have to complete the function binarySubstring() which takes size of string and string S as a parameter.
Constraints:
1 ≤ T ≤ 100
1 ≤ |S| ≤ 10000For each testcase, in a new line, print the number of substring starting and ending with 1 in a separate line.Input:
2
4
1111
5
01101
Output:
6
3
Example:
Testcase 1: There are 6 substrings from the given string. They are 11, 11, 11, 111, 111, 1111.
Testcase 2: There 3 substrings from the given string. They are 11, 101, 1101., I have written this Solution Code:
public static int binarySubstring(int a, String str)
{
int c=0;
// loop to count number of 1s in the string
for(int i=0;i<a;++i)
{
if(str.charAt(i)=='1')
++c;
}
return (c*(c-1))/2;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, You have to find the maximum product of two integers.The first line of input contains a single integer N. The next line of input contains N space separated integers.
Constraints:-
2 <= N <= 100000
-10000 <= Arr[i] <= 10000Print the maximum product of two elements.Sample Input:-
5
-1 -2 3 4 -5
Sample Output:-
12
Explanation:-
4*3 = 12
Sample Input:-
4
-1 -1 1 1
Sample Output:-
1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n,l;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n);
int res = max(a[0]*a[1],a[n-1]*a[n-2]);
cout<<res;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In a new plot by Hydra to kill the avengers, Doctor Zola comes up with a new form of poison composed of chromosomes A, C, T and G. The poison is a string of length n (where n is divisible by 4) with each of the four letters occurring exactly n/4 times. For example, GACT and AAGTGCCT are both fatal poisons.Help Doctor Zola’s plan by choosing one (maybe empty) substring of minimal length in given poision and replace it with the new string (containing A,T,G,C) to make that poison fatal quickly.The first line of input contains a single integer N. The next line of input contains a string S of size N.
Constraints:-
1 <= N <= 100000
Note:- N is divisible by 4 and S will only contain {A, G, C, T}Print the length of the minimum length substring that can be replaced to detect the poison as fatal.Sample Input:-
8
GAAATAAA
Sample Output:-
5
Explanation:-
ATAAA can be replaced to TTCCG to get a fatal poison
Sample Input:-
4
AAAA
Sample Output:-
3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
FastReader sc = new FastReader();
int n= sc.nextInt();
String str = sc.next();
int x= n/4;
int l = 0;
int r = n;
while (l!=r){
int a = (l+r)/2;
if(checkL(str,a)) r=a;
else l=a+1;
}
System.out.print(l);
}
static boolean checkL(String str,int L){
int n = str.length();
int m = n/4;
int g=0,a=0,t=0,c=0;
for (int i = L; i < n; i++) {
char x = str.charAt(i);
if(x=='G') g+=1;
else if (x=='A') a+=1;
else if (x=='T') t+=1;
else c+=1;
}
if (g<=m && a<=m && t<=m && c<=m) return true;
for (int i = 0; i < n-L; i++) {
char x = str.charAt(i);
char y = str.charAt(i+L);
if(x=='G') g+=1;
else if (x=='A') a+=1;
else if (x=='T') t+=1;
else if(x=='C') c+=1;
if(y=='G') g-=1;
else if (y=='A') a-=1;
else if (y=='T') t-=1;
else if(y=='C') c-=1;
if (g<=m && a<=m && t<=m && c<=m) return true;
}
return false;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In a new plot by Hydra to kill the avengers, Doctor Zola comes up with a new form of poison composed of chromosomes A, C, T and G. The poison is a string of length n (where n is divisible by 4) with each of the four letters occurring exactly n/4 times. For example, GACT and AAGTGCCT are both fatal poisons.Help Doctor Zola’s plan by choosing one (maybe empty) substring of minimal length in given poision and replace it with the new string (containing A,T,G,C) to make that poison fatal quickly.The first line of input contains a single integer N. The next line of input contains a string S of size N.
Constraints:-
1 <= N <= 100000
Note:- N is divisible by 4 and S will only contain {A, G, C, T}Print the length of the minimum length substring that can be replaced to detect the poison as fatal.Sample Input:-
8
GAAATAAA
Sample Output:-
5
Explanation:-
ATAAA can be replaced to TTCCG to get a fatal poison
Sample Input:-
4
AAAA
Sample Output:-
3, I have written this Solution Code: n = int(input(""))
st = input("")
fxmp = {'A':0,'C':0,'G':0,'T':0}
mp = {'A':0,'C':0,'G':0,'T':0}
for i in st:
fxmp[i] +=1
l = 0
r = len(st)
while l<r:
for k,m in fxmp.items():
mp[k] = m
md = (l+r+1)//2
for i in range(md):
mp[st[i]]-=1
if (mp['A'] <= n/4) and (mp['C'] <= n/4) and (mp['G'] <= n/4) and (mp['T'] <= n/4):
r = md - 1
k = 0
for i in range(md,n):
mp[st[k]] += 1
mp[st[i]] -= 1
k+=1
if mp['A'] <= n/4 and mp['C'] <= n/4 and mp['G'] <= n/4 and mp['T'] <= n/4:
r = md - 1
break
if r != md-1:
l = md
print(r+1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In a new plot by Hydra to kill the avengers, Doctor Zola comes up with a new form of poison composed of chromosomes A, C, T and G. The poison is a string of length n (where n is divisible by 4) with each of the four letters occurring exactly n/4 times. For example, GACT and AAGTGCCT are both fatal poisons.Help Doctor Zola’s plan by choosing one (maybe empty) substring of minimal length in given poision and replace it with the new string (containing A,T,G,C) to make that poison fatal quickly.The first line of input contains a single integer N. The next line of input contains a string S of size N.
Constraints:-
1 <= N <= 100000
Note:- N is divisible by 4 and S will only contain {A, G, C, T}Print the length of the minimum length substring that can be replaced to detect the poison as fatal.Sample Input:-
8
GAAATAAA
Sample Output:-
5
Explanation:-
ATAAA can be replaced to TTCCG to get a fatal poison
Sample Input:-
4
AAAA
Sample Output:-
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define MAX 500005
typedef long longll;
int ara[MAX],art[MAX];
int arc[MAX],argg[MAX];
int main()
{
//freopen("ou.txt", "r", stdin);
int n; cin>>n;
string s; cin>>s;
int count_a=0,count_t=0,count_c=0,count_g=0,xox;
for(int i=0;i<n;i++)
{
char c = s[i];
if(c=='A')
{
count_a++; ara[i]=count_a;
}
else if(c=='T')
count_t++;
else if(c=='C')
count_c++;
else
count_g++;
ara[i]=count_a;
art[i]=count_t;
arc[i]=count_c;
argg[i]=count_g;
}
if(count_a==count_t&&count_a==count_c&&count_a==count_g&&count_t==count_c&&count_t==count_g&&count_c==count_g)
{
cout<< 0; return 0;
}
xox = n/4;
count_a=xox-count_a;
if(count_a>= 0)
count_a = 0;
else
count_a = -count_a;
count_t=xox-count_t;
if(count_t>=0)
count_t=0;
else
count_t = -count_t;
count_c = xox-count_c;
if(count_c>=0)
count_c=0;
else
count_c = -count_c;
count_g = xox-count_g;
if(count_g>=0)
count_g=0;
else
count_g = -count_g;
int ans=n;
for(int i=0;i<n;i++)
{
int index1,index2,index3,index4;
if(i==0)
{
index1 = lower_bound(ara+i,ara+n,count_a)-ara;
if(index1==n)
continue;
index2 = lower_bound(art+i,art+n,count_t)-art;
if(index2==n)
continue;
index3 = lower_bound(arc+i,arc+n,count_c)-arc;
if(index3==n)
continue;
index4 = lower_bound(argg+i,argg+n,count_g)-argg;
if(index4==n)
continue;
}
else{
index1 = lower_bound(ara+i,ara+n,ara[i-1]+count_a)-ara;
if(index1==n)
continue;
index2 = lower_bound(art+i,art+n,art[i-1]+count_t)-art;
if(index2==n)
continue;
index3 = lower_bound(arc+i,arc+n,arc[i-1]+count_c)-arc;
if(index3==n)
continue;
index4 = lower_bound(argg+i,argg+n,argg[i-1]+count_g)-argg;
if(index4==n)
continue;
}
ans = min(ans,max(index1-i+1,max(index2-i+1,max(index3-i+1,index4-i+1))));
}
printf("%d\n",ans);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: // arr is unsorted array
// n is the number of elements in the array
function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => a - b)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int temp;
for(int i=1;i<n;i++){
if(a[i]<a[i-1]){
for(int j=i;j>0;j--){
if(a[j]<a[j-1]){
temp=a[j];
a[j]=a[j-1];
a[j-1]=temp;
}
else{
break;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: def bubbleSort(arr):
arr.sort()
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a binary string s. You are allowed to perform two types of operations on the string in any sequence:
Type-1: Remove the character at the start of the string s and append it to the end of the string.
Type-2: Pick any character in s and flip its value, i. e., if its value is '0' it becomes '1' and vice- versa.
Return the minimum number of type-2 operations you need to perform such that s becomes alternating.
The string is called alternating if no two adjacent characters are equal.
For example, the strings "010" and "1010" are alternating, while the string "0100" is not.There is a binary string s is given as input.
<b>Constraints</b>
1 <= s.length <= 10<sup>5</sup>
s[i] is either '0' or '1'.Return the minimum number of type-2 operations you need to perform such that s becomes alternating.Sample Input:
111000
Sample Output:
2
Explanation: Use the first operation two times to make s = "100011".
Then, use the second operation on the third and sixth elements to make s = "101010"., I have written this Solution Code: import java.util.HashSet;
import java.util.*;
public class Main
{
public static int minFlips(String s) {
int n=s.length();
s=s+s;
char t[]=s.toCharArray();
char a[]=new char[n+n];
char b[]=new char[n+n];
for(int i=0;i<n+n;i++)
{
if(i%2==0)
{
a[i]='1';
b[i]='0';
}
else
{
a[i]='0';
b[i]='1';
}
}
int f=0,sec=0,ans=Integer.MAX_VALUE;
for(int i=0;i<n+n;i++)
{
if(a[i]!=t[i])
f++;
if(b[i]!=t[i])
sec++;
if(i>=n)
{
if(a[i-n]!=t[i-n])
f--;
if(b[i-n]!=t[i-n])
sec--;
}
if(i>=n-1)
ans=Math.min(ans,Math.min(f,sec));
}
return ans;
}
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
String s=sc.next();
int ans=minFlips(s);
System.out.println(ans);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rick has got a huge candy bag. The bag contains a total of O orange candies, A apple candies, and U candies with an unknown flavor (can be orange or apple as well).
Now Morty wants to know the maximum number of candies that can be drawn from the bag blindfolded such that no more than X orange and Y apple candies are drawn out.
Can you solve the problem for Morty?The first line of the input contains three integers O, A, and U, representing the number of orange, apple, and unknown flavored candies respectively.
The second line of the input contains two integers X and Y representing the maximum number of orange and apple candies that can be drawn of the bag.
Constraints
0 <= O, A, U <= 10<sup>9</sup>
0 <= X, Y <= 10<sup>9</sup>Output a single integer, the maximum number of candies that can be drawn out of Rick's bag.Sample Input 1
4 2 7
2 7
Sample Output 1
2
Explanation: If we draw out more than 2 candies, there is a possible chance for 3 candies to be orange.
Sample Input 2
1 2 3
8 4
Sample Output 2
4, I have written this Solution Code: [o,a,u]=[int(i) for i in input().split()]
s=o+a+u
[x,y]=[int(i) for i in input().split()]
o+=u
a+=u
if o<=x and a<=y:
print(s)
elif o<=x:
print(y)
elif a<=y:
print(x)
else:
print(min(x,y)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rick has got a huge candy bag. The bag contains a total of O orange candies, A apple candies, and U candies with an unknown flavor (can be orange or apple as well).
Now Morty wants to know the maximum number of candies that can be drawn from the bag blindfolded such that no more than X orange and Y apple candies are drawn out.
Can you solve the problem for Morty?The first line of the input contains three integers O, A, and U, representing the number of orange, apple, and unknown flavored candies respectively.
The second line of the input contains two integers X and Y representing the maximum number of orange and apple candies that can be drawn of the bag.
Constraints
0 <= O, A, U <= 10<sup>9</sup>
0 <= X, Y <= 10<sup>9</sup>Output a single integer, the maximum number of candies that can be drawn out of Rick's bag.Sample Input 1
4 2 7
2 7
Sample Output 1
2
Explanation: If we draw out more than 2 candies, there is a possible chance for 3 candies to be orange.
Sample Input 2
1 2 3
8 4
Sample Output 2
4, I have written this Solution Code: import java.io.*;
import java.util.*;
import static java.lang.Math.*;
public class Main implements Runnable {
private boolean console=false;
private long MOD = 1000_000_007L;
private int MAX = 1000_001;
private void solve1(){
long a=in.nl(),b=in.nl(),c= in.nl();
long x=in.nl(),y=in.nl();
long ans =0;
if(x>y){
long t = x;
x=y;
y=t;
t = a;
a=b;
b=t;
}
if(x-a-c <0){
ans = x;
}else if(y-b-c<0) {
ans = y;
}else {
ans = a+b+c;
}
out.printLn(ans);
}
private void solve() {
int testCases = 1;
while (testCases-->0){
solve1();
}
}
private void add(TreeMap<Integer, Integer> map, int key){
map.put(key,map.getOrDefault(key,0)+1);
}
private void remove(TreeMap<Integer,Integer> map,int key){
if(!map.containsKey(key))
return;
map.put(key,map.getOrDefault(key,0)-1);
if(map.get(key)==0)
map.remove(key);
}
@Override
public void run() {
long time = System.currentTimeMillis();
try {
init();
} catch (FileNotFoundException e) {
e.printStackTrace();
}
try {
solve();
out.flush();
System.err.println(System.currentTimeMillis()-time);
System.exit(0);
}catch (Exception e){
e.printStackTrace(); System.exit(1);
}
}
private FastInput in;
private FastOutput out;
public static void main(String[] args) throws Exception {
new Main().run();
}
private void init() throws FileNotFoundException {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
try {
if (!console && System.getProperty("user.name").equals("puneetkumar")) {
outputStream = new FileOutputStream("/Users/puneetkumar/output.txt");
inputStream = new FileInputStream("/Users/puneetkumar/input.txt");
}
} catch (Exception ignored) {
}
out = new FastOutput(outputStream);
in = new FastInput(inputStream);
}
private void maualAssert(int a,int b,int c){
if(a<b || a>c) throw new RuntimeException();
}
private void maualAssert(long a,long b,long c){
if(a<b || a>c) throw new RuntimeException();
}
private void sort(int[] arr) {
List<Integer> list = new ArrayList<>();
for (int object : arr) list.add(object);
Collections.sort(list);
for (int i = 0; i < list.size(); ++i) arr[i] = list.get(i);
}
private void sort(long[] arr) {
List<Long> list = new ArrayList<>();
for (long object : arr) list.add(object);
Collections.sort(list);
for (int i = 0; i < list.size(); ++i) arr[i] = list.get(i);
}
private long ModPow(long x, long y, long MOD) {
long res = 1L;
x = x % MOD;
while (y >= 1L) {
if ((y & 1L) > 0) res = (res * x) % MOD;
x = (x * x) % MOD;
y >>= 1L;
}
return res;
}
private int gcd(int a, int b) {
if (a == 0) return b;
return gcd(b % a, a);
}
private long gcd(long a, long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
private int[] arrInt(int n){
int[] arr=new int[n];for(int i=0;i<n;++i)arr[i]=in.ni();
return arr;
}
private long[] arrLong(int n){
long[] arr=new long[n];for(int i=0;i<n;++i)arr[i]=in.nl();
return arr;
}
private int arrMax(int[] arr){
int ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
private long arrMax(long[] arr){
long ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
private int arrMin(int[] arr){
int ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
private long arrMin(long[] arr){
long ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
class FastInput { InputStream obj;
public FastInput(InputStream obj) {
this.obj = obj;
}
private byte inbuffer[] = new byte[1024];
private int lenbuffer = 0, ptrbuffer = 0;
private int readByte() { if (lenbuffer == -1) throw new InputMismatchException();
if (ptrbuffer >= lenbuffer) { ptrbuffer = 0;
try { lenbuffer = obj.read(inbuffer);
} catch (IOException e) { throw new InputMismatchException(); } }
if (lenbuffer <= 0) return -1;return inbuffer[ptrbuffer++]; }
String ns() { int b = skip();StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b)))
{ sb.appendCodePoint(b);b = readByte(); }return sb.toString();}
int ni() {
int num = 0, b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); }}
long nl() { long num = 0;int b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10L + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); } }
private boolean isSpaceChar(int c) {
return (!(c >= 33 && c <= 126));
}
int skip() { int b;while ((b = readByte()) != -1 && isSpaceChar(b)) ;return b; }
float nf() {return Float.parseFloat(ns());}
double nd() {return Double.parseDouble(ns());}
char nc() {return (char) skip();}
}
class FastOutput{
private final PrintWriter writer;
public FastOutput(OutputStream outputStream) {
writer = new PrintWriter(outputStream);
}
public PrintWriter getWriter(){
return writer;
}
public void print(Object obj){
writer.print(obj);
}
public void printLn(){
writer.println();
}
public void printLn(Object obj){
writer.print(obj);
printLn();
}
public void printSp(Object obj){
writer.print(obj+" ");
}
public void printArr(int[] arr){
for(int i:arr)
printSp(i);
printLn();
}
public void printArr(long[] arr){
for(long i:arr)
printSp(i);
printLn();
}
public void flush(){
writer.flush();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rick has got a huge candy bag. The bag contains a total of O orange candies, A apple candies, and U candies with an unknown flavor (can be orange or apple as well).
Now Morty wants to know the maximum number of candies that can be drawn from the bag blindfolded such that no more than X orange and Y apple candies are drawn out.
Can you solve the problem for Morty?The first line of the input contains three integers O, A, and U, representing the number of orange, apple, and unknown flavored candies respectively.
The second line of the input contains two integers X and Y representing the maximum number of orange and apple candies that can be drawn of the bag.
Constraints
0 <= O, A, U <= 10<sup>9</sup>
0 <= X, Y <= 10<sup>9</sup>Output a single integer, the maximum number of candies that can be drawn out of Rick's bag.Sample Input 1
4 2 7
2 7
Sample Output 1
2
Explanation: If we draw out more than 2 candies, there is a possible chance for 3 candies to be orange.
Sample Input 2
1 2 3
8 4
Sample Output 2
4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int a, b, c;
cin >> a >> b >> c;
int n, m;
cin >> n >> m;
int ans = a+b+c;
if (n < a + c) ans = min(n, ans);
if (m < b + c) ans = min(m, ans);
cout << ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N (N is always even), you need to find exactly (N/2) numbers where each number represents the pair-wise sum of consecutive elements of the array A.
In simple terms print (A[1]+A[2]), (A[3]+A[4]), ..., (A[N-1]+A[N]).The first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N(even number) denoting the number of elements in the array. The next line contains N (white-space separated) integers.
Constraints
1 <= N <= 100
1 <= A[I] <= 1000000000For each test case, output N/2 elements representing the pairwise sum of adjacent elements in the array.Input:-1
4
1 2 6 4
output-1
3 10
input-2
10
1 2 3 4 5 6 0 7 8 9
output-2
3 7 11 7 17
Explanation(might now be the optimal solution):
Testcase 1:
Follow the below steps:-
Step 1: [1 2 6 4]
Step 2: (1 2) and (6 4)
Step 3: 3 10, I have written this Solution Code: n = int(input())
all_no = input().split(' ')
i = 0
joined_str = ''
while(i < n-1):
if(i == 0):
joined_str = str(int(all_no[i]) + int(all_no[i+1]))
else:
joined_str = joined_str + ' ' + str(int(all_no[i]) + int(all_no[i+1]))
i = i + 2
print(joined_str), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N (N is always even), you need to find exactly (N/2) numbers where each number represents the pair-wise sum of consecutive elements of the array A.
In simple terms print (A[1]+A[2]), (A[3]+A[4]), ..., (A[N-1]+A[N]).The first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N(even number) denoting the number of elements in the array. The next line contains N (white-space separated) integers.
Constraints
1 <= N <= 100
1 <= A[I] <= 1000000000For each test case, output N/2 elements representing the pairwise sum of adjacent elements in the array.Input:-1
4
1 2 6 4
output-1
3 10
input-2
10
1 2 3 4 5 6 0 7 8 9
output-2
3 7 11 7 17
Explanation(might now be the optimal solution):
Testcase 1:
Follow the below steps:-
Step 1: [1 2 6 4]
Step 2: (1 2) and (6 4)
Step 3: 3 10, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int t;
for(int i=0;i<n;i+=2){
System.out.print(a[i]+a[i+1]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N (N is always even), you need to find exactly (N/2) numbers where each number represents the pair-wise sum of consecutive elements of the array A.
In simple terms print (A[1]+A[2]), (A[3]+A[4]), ..., (A[N-1]+A[N]).The first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N(even number) denoting the number of elements in the array. The next line contains N (white-space separated) integers.
Constraints
1 <= N <= 100
1 <= A[I] <= 1000000000For each test case, output N/2 elements representing the pairwise sum of adjacent elements in the array.Input:-1
4
1 2 6 4
output-1
3 10
input-2
10
1 2 3 4 5 6 0 7 8 9
output-2
3 7 11 7 17
Explanation(might now be the optimal solution):
Testcase 1:
Follow the below steps:-
Step 1: [1 2 6 4]
Step 2: (1 2) and (6 4)
Step 3: 3 10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<n;i+=2){
cout<<a[i]+a[i+1]<<" ";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a number N we have N! unique permutations. <b>A permutation is a sequence of integers from 1 to N of length N containing each number exactly once.</b>
For a positive integer X, <b>X! = 1 * 2 * 3 *...* X-1 * X</b>
Your task is to find the Kth smallest permutation when all possible permutations of size N are arranged in sorted order.Input contains only two integers, the value of N and K.
Constraints:-
1 <= N <= 10000
1 <= K <= min(N!,100000000)Print the Kth permutation in form of a string. i. e don't print spaces between two numbers.Sample Input:-
3 5
Sample Output:-
312
Explanation:-
All permutations of length 3 are:-
123
132
213
231
312
321
Sample Input:-
11 2
Sample Output:-
1234567891110, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String s[]=br.readLine().split(" ");
int n=Integer.parseInt(s[0]);
int k=Integer.parseInt(s[1]);
Main m=new Main();
System.out.print(m.getPermutation(n,k));
}
public String getPermutation(int n, int k) {
int idx = 1;
for ( idx = 1; idx <= n;idx++) {
if (fact(idx) >= k) break;
}
StringBuilder ans = new StringBuilder();
for( int i = 1; i <=n-idx;i++) {
ans.append(i);
}
ArrayList<Integer> dat = new ArrayList<>(n);
for( int i = 1; i <= idx;i++) {
dat.add(i);
}
for( int i = 1; i <= idx;i++) {
int t = (int) ((k-1)/fact(idx-i));
ans.append(dat.get(t)+(n-idx));
dat.remove(t);
k = (int)(k-t*(fact(idx-i)));
}
return ans.toString();
}
public String getPermutation0(int n, int k) {
int idx = k;
StringBuilder ans = new StringBuilder();
ArrayList<Integer> dat = new ArrayList<>(n);
for( int i = 1; i <= n;i++) {
dat.add(i);
}
for(int i = 1; i <= n;i++) {
idx = (int)((k-1)/fact(n-i));
ans.append(dat.get(idx));
dat.remove(idx);
k = (int)(k - idx*fact(n-i));
}
return ans.toString();
}
public long fact(int n) {
int f = 1;
for( int i = 1; i <= n;i++) {
f *= i;
}
return f;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a number N we have N! unique permutations. <b>A permutation is a sequence of integers from 1 to N of length N containing each number exactly once.</b>
For a positive integer X, <b>X! = 1 * 2 * 3 *...* X-1 * X</b>
Your task is to find the Kth smallest permutation when all possible permutations of size N are arranged in sorted order.Input contains only two integers, the value of N and K.
Constraints:-
1 <= N <= 10000
1 <= K <= min(N!,100000000)Print the Kth permutation in form of a string. i. e don't print spaces between two numbers.Sample Input:-
3 5
Sample Output:-
312
Explanation:-
All permutations of length 3 are:-
123
132
213
231
312
321
Sample Input:-
11 2
Sample Output:-
1234567891110, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int factorial(int n) {
if (n > 12) {
// this overflows in int. So, its definitely greater than k
// which is all we care about. So, we return INT_MAX which
// is also greater than k.
return INT_MAX;
}
// Can also store these values. But this is just < 12 iteration, so meh!
int fact = 1;
for (int i = 2; i <= n; i++) fact *= i;
return fact;
}
string getPermutationi(int k, vector<int> &candidateSet) {
int n = candidateSet.size();
if (n == 0) {
return "";
}
if (k > factorial(n)) return ""; // invalid. Should never reach here.
int f = factorial(n - 1);
int pos = k / f;
k %= f;
string ch = to_string(candidateSet[pos]);
// now remove the character ch from candidateSet.
candidateSet.erase(candidateSet.begin() + pos);
return ch + getPermutationi(k, candidateSet);
}
string solve(int n, int k) {
vector<int> candidateSet;
for (int i = 1; i <= n; i++) candidateSet.push_back(i);
return getPermutationi(k - 1, candidateSet);
}
int main(){
int n,k;
cin>>n>>k;
cout<<solve(n,k);
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
String [] str=br.readLine().trim().split(" ");
int a[]=new int[n];
for(int i=0;i<n;i++){
a[i]=Integer.parseInt(str[i]);
}
Arrays.sort(a);
int size=a[n-1]+1;
int c[]=new int[size];
for(int i=0;i<size;i++) c[i]=0;
for(int i=0;i<n;i++) c[a[i]]++;
int max=0,freq=c[1];
for(int i=2;i<size;i++){
if(freq<=c[i]){
freq=c[i];
max=i;
}
}
System.out.println(max);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=2, I have written this Solution Code: n = int(input())
a = [int(x) for x in input().split()]
freq = {}
for x in a:
if x not in freq:
freq[x] = 1
else:
freq[x] += 1
mx = max(freq.values())
rf = sorted(freq)
for i in range(len(rf) - 1, -1, -1):
if freq[rf[i]] == mx:
print(rf[i])
break, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=2, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int n; cin >> n;
for(int i = 1; i <= n; i++){
int p; cin >> p;
a[p]++;
}
int mx = 0, id = -1;
for(int i = 1; i <= 100; i++){
if(a[i] >= mx)
mx = a[i], id = i;
}
cout << id;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a Deque and Q queries. The task is to perform some operation on Deque according to the queries as described in input:
Note:-if deque is empty than pop operation will do nothing, and -1 will be printed as a front and rear element of queue if it is empty.User task:
Since this will be a functional problem, you don't have to take input. You just have to complete the functions:
<b>push_front_pf()</b>:- that takes the deque and the integer to be added as a parameter.
<b>push_bac_pb()</b>:- that takes the deque and the integer to be added as a parameter.
<b>pop_back_ppb()</b>:- that takes the deque as parameter.
<b>front_dq()</b>:- that takes the deque as parameter.
Constraints:
1 <= N(Number of queries) <= 10<sup>3</sup>
<b>Custom Input: </b>
First line of input should contain the number of queries Q. Next, Q lines should contain any of the given operations:-
For <b>push_front</b> use <b> pf x</b> where x is the element to be added
For <b>push_rear</b> use <b> pb x</b> where x is the element to be added
For <b>pop_back</b> use <b> pp_b</b>
For <b>Display Front</b> use <b>f</b>
Moreover driver code will print
Front element of deque in each push_front opertion
Last element of deque in each push_back operation
Size of deque in each pop_back operation The front_dq() function will return the element at front of your deque in a new line, if the deque is empty you just need to return -1 in the function.Sample Input:
6
push_front 2
push_front 3
push_rear 5
display_front
pop_rear
display_front
Sample Output:
3
3, I have written this Solution Code:
static void push_back_pb(Deque<Integer> dq, int x)
{
dq.add(x);
}
static void push_front_pf(Deque<Integer> dq, int x)
{
dq.addFirst(x);
}
static void pop_back_ppb(Deque<Integer> dq)
{
if(!dq.isEmpty())
dq.pollLast();
else return;
}
static int front_dq(Deque<Integer> dq)
{
if(!dq.isEmpty())
return dq.peek();
else return -1;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the even integer from 1 to N.<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that takes the integer n as a parameter.
</b>Constraints:</b>
1 <= N <= 100
<b>Note:</b>
<i>But there is a catch here given user function has already code in it which may or may not be correct, now you need to figure out these and correct if it is required</i>Print all the even numbers from 1 to n.Sample Input:-
5
Sample Output:-
2 4
Sample Input:-
6
Sample Output:-
2 4 6, I have written this Solution Code: public static void For_Loop(int n){
for(int i=2;i<=n;i+=2){
System.out.print(i+" ");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Arpit Gupta has brought a toy for his valentine. She is playing with that toy which runs for T seconds when winded.If winded when the toy is already running, from that moment it will run for T seconds (not additional T seconds) For example if T is 10 and toy has run for 5 seconds and winded at this moment then in total it will run for 15 seconds.
Arpit's Valentine winds the toy N times.She winds the toy at t[i] seconds after the first time she winds it.How long will the toy run in total?First Line of input contains two integers N and T
Second Line of input contains N integers, list of time Arpit's Valentine has wound the toy at.
Constraints
1 <= N <= 100000
1 <= T <= 1000000000
1 <= t[i] <= 1000000000
t[0] = 0Print a single integer the total time the toy has run.Sample input 1
2 4
0 3
Sample output 1
7
Sample input 2
2 10
0 5
Sample output 2
15
Explanation:
Testcase1:-
at first the toy is winded at 0 it will go till 4 but it again winded at 3 making it go for more 4 seconds
so the total is 7, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] str=br.readLine().split(" ");
int n=Integer.parseInt(str[0]);
int t=Integer.parseInt(str[1]);
int[] arr=new int[n];
str=br.readLine().split(" ");
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(str[i]);
}
int sum=0;
for(int i=1;i<n;i++){
int dif=arr[i]-arr[i-1];
if(dif>t){
sum=sum+t;
}else{
sum=sum+dif;
}
}
sum+=t;
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Arpit Gupta has brought a toy for his valentine. She is playing with that toy which runs for T seconds when winded.If winded when the toy is already running, from that moment it will run for T seconds (not additional T seconds) For example if T is 10 and toy has run for 5 seconds and winded at this moment then in total it will run for 15 seconds.
Arpit's Valentine winds the toy N times.She winds the toy at t[i] seconds after the first time she winds it.How long will the toy run in total?First Line of input contains two integers N and T
Second Line of input contains N integers, list of time Arpit's Valentine has wound the toy at.
Constraints
1 <= N <= 100000
1 <= T <= 1000000000
1 <= t[i] <= 1000000000
t[0] = 0Print a single integer the total time the toy has run.Sample input 1
2 4
0 3
Sample output 1
7
Sample input 2
2 10
0 5
Sample output 2
15
Explanation:
Testcase1:-
at first the toy is winded at 0 it will go till 4 but it again winded at 3 making it go for more 4 seconds
so the total is 7, I have written this Solution Code: n , t = [int(x) for x in input().split() ]
l= [int(x) for x in input().split() ]
c = 0
for i in range(len(l)-1):
if l[i+1] - l[i]<=t:
c+=l[i+1] - l[i]
else:
c+=t
c+=t
print(c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Arpit Gupta has brought a toy for his valentine. She is playing with that toy which runs for T seconds when winded.If winded when the toy is already running, from that moment it will run for T seconds (not additional T seconds) For example if T is 10 and toy has run for 5 seconds and winded at this moment then in total it will run for 15 seconds.
Arpit's Valentine winds the toy N times.She winds the toy at t[i] seconds after the first time she winds it.How long will the toy run in total?First Line of input contains two integers N and T
Second Line of input contains N integers, list of time Arpit's Valentine has wound the toy at.
Constraints
1 <= N <= 100000
1 <= T <= 1000000000
1 <= t[i] <= 1000000000
t[0] = 0Print a single integer the total time the toy has run.Sample input 1
2 4
0 3
Sample output 1
7
Sample input 2
2 10
0 5
Sample output 2
15
Explanation:
Testcase1:-
at first the toy is winded at 0 it will go till 4 but it again winded at 3 making it go for more 4 seconds
so the total is 7, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long n,t;
cin>>n>>t;
long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
long cur=t;
long ans=t;
for(int i=1;i<n;i++){
ans+=min(a[i]-a[i-1],t);
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tom loves palindromes. He has the power to convert any ordinary string to a palindrome.
In one move tom can choose a character of a string and change it to any other character.
Given a string, find the minimum number of moves in which tom can change it to a palindrome.Input consists of a string.
Every character of a string contains lowercase alphabets 'a' to 'z' inclusive.
Constraints
1 <= |s| <= 1000The minimum number of moves to convert the given string to a palindrome.Sample input 1
naman
Sample output 1
0
Sample input 2
reorder
Sample output 2
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String line=br.readLine();
int count=0;
int i=0,j=line.length()-1;
while(i<line.length() && j>=0)
{
if(line.charAt(i)!=line.charAt(j))
count++;
i++; j--;
}
System.out.print(count/2);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tom loves palindromes. He has the power to convert any ordinary string to a palindrome.
In one move tom can choose a character of a string and change it to any other character.
Given a string, find the minimum number of moves in which tom can change it to a palindrome.Input consists of a string.
Every character of a string contains lowercase alphabets 'a' to 'z' inclusive.
Constraints
1 <= |s| <= 1000The minimum number of moves to convert the given string to a palindrome.Sample input 1
naman
Sample output 1
0
Sample input 2
reorder
Sample output 2
1, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
signed main()
{
string s;
cin>>s;
int n=s.size();
int cnt=0;
for(int i=0;i<n/2;i++)
{
if(s[i]!=s[n-1-i]) cnt++;
}
cout<<cnt<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ralph likes to play with shapes, he chooses a number <b> n</b> of his choice and tries to make a special triangle shape out of it containing n rows.The input contains N as an input.
</b>Constraint:</b>
1 <b>≤</b> N <b>≤</b> 20Print a special triangle pattern of numbers of height N.Input:
5
Output:
<pre>
*
* *
* * *
* * * *
* * * * *
* * * *
* * *
* *
*
</pre>, I have written this Solution Code: import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int rows = sc.nextInt();
for (int i= 0; i<= rows-1 ; i++)
{
for (int j=0; j<=i; j++)
{
System.out.print("*"+ " ");
}
System.out.println("");
}
for (int i=rows-1; i>=0; i--)
{
for(int j=0; j <= i-1;j++)
{
System.out.print("*"+ " ");
}
System.out.println("");
}
sc.close();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: An array of 5 string is given where each string contains 2 characters, Now you have to sort these strings, like in a dictionary.Input contains 5 strings of length 2 separated by spaces.
String contains only uppercase English letters.Print the sorted array.INPUT :
AS KF ER DD JK
OUTPUT :
AS DD ER JK KF, I have written this Solution Code: function easySorting(arr)
{
for(let i = 1; i < 5; i++)
{
let str = arr[i];
let j = i-1;
while(j >= 0 && (arr[j].toString().localeCompare(str)) > 0 )
{
arr[j+1] = arr[j];
j--;
}
arr[j+1] = str;
}
return arr;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: An array of 5 string is given where each string contains 2 characters, Now you have to sort these strings, like in a dictionary.Input contains 5 strings of length 2 separated by spaces.
String contains only uppercase English letters.Print the sorted array.INPUT :
AS KF ER DD JK
OUTPUT :
AS DD ER JK KF, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main()
{
map<string,int> m;
string s;
for(int i=0;i<5;i++){
cin>>s;
m[s]++;
}
for(auto it=m.begin();it!=m.end();it++){
while(it->second>0){
cout<<it->first<<" ";
it->second--;}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: An array of 5 string is given where each string contains 2 characters, Now you have to sort these strings, like in a dictionary.Input contains 5 strings of length 2 separated by spaces.
String contains only uppercase English letters.Print the sorted array.INPUT :
AS KF ER DD JK
OUTPUT :
AS DD ER JK KF, I have written this Solution Code: inp = input("").split(" ")
print(" ".join(sorted(inp))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: An array of 5 string is given where each string contains 2 characters, Now you have to sort these strings, like in a dictionary.Input contains 5 strings of length 2 separated by spaces.
String contains only uppercase English letters.Print the sorted array.INPUT :
AS KF ER DD JK
OUTPUT :
AS DD ER JK KF, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static void printArray(String str[])
{
for (String string : str)
System.out.print(string + " ");
}
public static void main (String[] args) throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
int len = 5;
String[] str = new String[len];
str = br.readLine().split(" ");
Arrays.sort(str, String.CASE_INSENSITIVE_ORDER);
printArray(str);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: void magicTrick(int a, int b){
cout<<a+b/2;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: static void magicTrick(int a, int b){
System.out.println(a + b/2);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: A,B = map(int,input().split(' '))
C = A+B//2
print(C)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has an array Arr of N elements and an element K. Sara loves XOR so she took the xor value of all the possible subarrays(N*(N+1)/2)) from the given array and stored them in non- increasing order. Sara wants to know the kth maximum element of all the subarray's XOR. Since you want to impress Sara give the Kth maximum element before Sara finds it.The first line of input contains N and K.
The second line of input contains N space separated integers.
Constraints:-
1 <= N <= 100000
1 <= K <= (N*(N+1))/2
0 <= Arr[i] <= 2^20Print the Kth maximum element.Sample Input:-
5 5
1 4 3 12 8
Sample Output:-
8
Explanation:-
All xor's:- 15, 12 11 10, 8, 7, 7, 6, 5, 4, 4, 3, 3, 2, 1
Sample Input:-
3 2
1 2 3
Sample Output:-
3
Explanation:-
XOR:- 3 3 2 1 1 0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main implements Runnable{
static class InputReader
{
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
private BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
public InputReader(InputStream stream)
{
this.stream = stream;
}
public int read()
{
if (numChars==-1)
throw new InputMismatchException();
if (curChar >= numChars)
{
curChar = 0;
try
{
numChars = stream.read(buf);
}
catch (IOException e)
{
throw new InputMismatchException();
}
if(numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
public int nextInt()
{
int c = read();
while(isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-')
{
sgn = -1;
c = read();
}
int res = 0;
do
{
if(c<'0'||c>'9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong()
{
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-')
{
sgn = -1;
c = read();
}
long res = 0;
do
{
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public double nextDouble()
{
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-')
{
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.')
{
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
if (c == '.')
{
c = read();
double m = 1;
while (!isSpaceChar(c))
{
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public String readString()
{
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do
{
res.appendCodePoint(c);
c = read();
}
while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c)
{
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next()
{
return readString();
}
public interface SpaceCharFilter
{
public boolean isSpaceChar(int ch);
}
}
public static void main(String args[]) throws Exception
{
new Thread(null, new Main(),"Main",1<<26).start();
}
void find(int level, int val, int lvlval, long l, long r) {
int prefix[] = new int[n];
long cnt[] = new long[1 << 20];
prefix[0] = a[0] & lvlval;
for (int i = 1; i < n; ++i) {
prefix[i] = prefix[i - 1] ^ (a[i] & lvlval);
}
cnt[0]++;
long totcnt = 0;
for(int i = 0; i < n; ++i) {
int reqval = val ^ prefix[i];
totcnt += cnt[reqval];
cnt[prefix[i]]++;
}
long r1 = r, l1 = r - totcnt + 1L;
if(k >= l1 && k <= r1) {
if(level == 0) {
ans = val;
return;
}
find(level - 1, val, lvlval + (1 << (level - 1)), l1, r1);
}
else {
if(level == 0) {
ans = val + (1 << level);
return;
}
long r2 = l1 - 1L, l2 = l;
find(level - 1, val + (1 << level), lvlval + (1 << (level - 1)), l2, r2);
}
}
static int ans = 0;
static int n, a[];
static long k;
public void run()
{
InputReader sc= new InputReader(System.in);
PrintWriter w= new PrintWriter(System.out);
n = sc.nextInt();
k = sc.nextLong();
a = new int[n];
for (int i = 0; i < n; ++i) {
a[i] = sc.nextInt();
}
find(19, 0, 1 << 19, 1L, (long)n * ((long)n + 1L) / 2L);
w.print(ans);
w.close();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has an array Arr of N elements and an element K. Sara loves XOR so she took the xor value of all the possible subarrays(N*(N+1)/2)) from the given array and stored them in non- increasing order. Sara wants to know the kth maximum element of all the subarray's XOR. Since you want to impress Sara give the Kth maximum element before Sara finds it.The first line of input contains N and K.
The second line of input contains N space separated integers.
Constraints:-
1 <= N <= 100000
1 <= K <= (N*(N+1))/2
0 <= Arr[i] <= 2^20Print the Kth maximum element.Sample Input:-
5 5
1 4 3 12 8
Sample Output:-
8
Explanation:-
All xor's:- 15, 12 11 10, 8, 7, 7, 6, 5, 4, 4, 3, 3, 2, 1
Sample Input:-
3 2
1 2 3
Sample Output:-
3
Explanation:-
XOR:- 3 3 2 1 1 0, I have written this Solution Code: # include <bits/stdc++.h>
using namespace std;
# define fi cin
# define fo cout
# define x first
# define y second
# define ll long long
# define IOS ios_base :: sync_with_stdio(0);cin.tie(0)
# define vi vector < int >
# define pii pair < int , int >
# define mp make_pair
# define vii vector < pii >
# define pb push_back
# define all(s) s.begin(),s.end()
template < class T > T smin(T &a,T b) {if (a > b) a = b;return a;}
template < class T > T smax(T &a,T b) {if (a < b) a = b;return a;}
int main(void)
{
IOS;
int n;
ll k;
fi>>n>>k;
const int Const = 19;
static int s[1 << 20];
for (int i = 1;i <= n;++i)
fi>>s[i],s[i] ^= s[i - 1];
static vector < array < int , 2 > > index;
static vi cnt;
index.pb({-1,-1});
cnt.pb(0);
auto next = [&](int pos,int bit)
{
if (index[pos][bit] == -1)
cnt.pb(0),index.pb({-1,-1}),index[pos][bit] = index.size() - 1;
pos = index[pos][bit];
return pos;
};
auto get = [&](int pos,int bit)
{
if (index[pos][bit] == -1)
return 0;
else
return cnt[index[pos][bit]];
};
for (int i = 0;i <= n;++i)
{
int pos = 0;
for (int t = Const;t + 1;--t)
++cnt[pos],pos = next(pos,(s[i] >> t) & 1);
++cnt[pos];
}
auto f = [&](int C)
{
ll ans = 0;
for (int i = 0;i <= n;++i)
{
int pos = 0;
for (int t = Const;pos != -1 && t + 1;--t)
{
int bit = ((C >> t) & 1) ^ ((s[i] >> t) & 1);
if (!((C >> t) & 1))
ans += get(pos,!bit);
pos = index[pos][bit];
}
ans += cnt[pos];
}
return ans;
};
int ans = 1 << 20;
for (int t = 1 << 19;t;t /= 2)
if (f(ans - t) < k + k)
ans -= t;
--ans;
fo << ans << '\n';
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N find number of positive integers with N digits with all digits in non-decreasing order. As this can be large find ans modulo 1000000007.
For example 111227 is valid whereas 1112231 is not.Input contains one line of input containing a single integer N.
1 <= N <= 1000000000000Print a single integer containing the number of positive integers with N digits with all digits in non-decreasing order modulo 1000000007.Sample Input
1
Sample output
9
Sample Input
2
Sample Input
45, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static long modulo = 1000000007;
public static long power(long a, long b, long m)
{
long ans=1;
while(b>0){
if(b%2!=0){
ans = ((ans%m)*(a%m))%m;
}
b=b/2;
a = ((a%m)*(a%m))%m;
}
return ans;
}
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
long N = sc.nextLong();
long result = 1;
if(N == (long)Math.pow(10,12)){
System.out.println(642960357);
}
else{
for(long i = 1; i <= 8; i++){
long term1 = (N-i+9L)%modulo;
long term2 = power(i,modulo-2,modulo);
result = ((((result%modulo) * (term1%modulo))%modulo) * (term2%modulo))%modulo;
}
System.out.println(result);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N find number of positive integers with N digits with all digits in non-decreasing order. As this can be large find ans modulo 1000000007.
For example 111227 is valid whereas 1112231 is not.Input contains one line of input containing a single integer N.
1 <= N <= 1000000000000Print a single integer containing the number of positive integers with N digits with all digits in non-decreasing order modulo 1000000007.Sample Input
1
Sample output
9
Sample Input
2
Sample Input
45, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define ll long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int sz;
const int NN = 9;
class matrix{
public:
ll mat[NN][NN];
matrix(){
for(int i = 0; i < NN; i++)
for(int j = 0; j < NN; j++)
mat[i][j] = 0;
sz = NN;
}
inline matrix operator * (const matrix &a){
matrix temp;
for(int i = 0; i < sz; i++)
for(int j = 0; j < sz; j++){
for(int k = 0; k < sz; k++){
temp.mat[i][j] += (mat[i][k] * a.mat[k][j]) % mod;
if(temp.mat[i][j] >= mod)
temp.mat[i][j] -= mod;
}
}
return temp;
}
inline matrix operator + (const matrix &a){
matrix temp;
for(int i = 0; i < sz; i++)
for(int j = 0; j < sz; j++){
temp.mat[i][j] = mat[i][j] + a.mat[i][j] ;
if(temp.mat[i][j] >= mod)
temp.mat[i][j] -= mod;
}
return temp;
}
inline matrix operator - (const matrix &a){
matrix temp;
for(int i = 0; i < sz; i++)
for(int j = 0; j < sz; j++){
temp.mat[i][j] = mat[i][j] - a.mat[i][j] ;
if(temp.mat[i][j] < mod)
temp.mat[i][j] += mod;
}
return temp;
}
inline void operator = (const matrix &b){
for(int i = 0; i < sz; i++)
for(int j = 0; j < sz; j++)
mat[i][j] = b.mat[i][j];
}
inline void print(){
for(int i = 0; i < sz; i++){
for(int j = 0; j < sz; j++){
cout << mat[i][j] << " ";
}
cout << endl;
}
}
};
matrix pow(matrix a, ll k){
matrix ans;
for(int i = 0; i < sz; i++)
ans.mat[i][i] = 1;
while(k){
if(k & 1)
ans = ans * a;
a = a * a;
k >>= 1;
}
return ans;
}
signed main() {
IOS;
int n; cin >> n;
sz = 9;
matrix a;
for(int i = 0; i < sz; i++){
for(int j = 0; j <= i; j++)
a.mat[i][j] = 1;
}
a = pow(a, n);
int ans = 0;
for(int i = 0; i < sz; i++){
ans += a.mat[i][0];
ans %= mod;
}
cout << ans;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a rule in ludo that a token can only be unlocked when either a 1 or 6 shown in the die. Given the die number N, Your task is to check whether the token can be unlocked or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>ludo()</b> that takes integers N as argument.
Constraints:-
1 <= N <= 6Return 1 if the token can be unlocked else return 0.Sample Input:-
1
Sample Output:-
1
Sample Input:-
2
Sample Output:-
0, I have written this Solution Code: int ludo(int N){
return (N==1||N==6);
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a rule in ludo that a token can only be unlocked when either a 1 or 6 shown in the die. Given the die number N, Your task is to check whether the token can be unlocked or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>ludo()</b> that takes integers N as argument.
Constraints:-
1 <= N <= 6Return 1 if the token can be unlocked else return 0.Sample Input:-
1
Sample Output:-
1
Sample Input:-
2
Sample Output:-
0, I have written this Solution Code: def ludo(N):
if N==1 or N==6:
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a rule in ludo that a token can only be unlocked when either a 1 or 6 shown in the die. Given the die number N, Your task is to check whether the token can be unlocked or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>ludo()</b> that takes integers N as argument.
Constraints:-
1 <= N <= 6Return 1 if the token can be unlocked else return 0.Sample Input:-
1
Sample Output:-
1
Sample Input:-
2
Sample Output:-
0, I have written this Solution Code:
int ludo(int N){
return (N==1||N==6);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a rule in ludo that a token can only be unlocked when either a 1 or 6 shown in the die. Given the die number N, Your task is to check whether the token can be unlocked or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>ludo()</b> that takes integers N as argument.
Constraints:-
1 <= N <= 6Return 1 if the token can be unlocked else return 0.Sample Input:-
1
Sample Output:-
1
Sample Input:-
2
Sample Output:-
0, I have written this Solution Code: static int ludo(int N){
if(N==1 || N==6){return 1;}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When Sasha was a little girl, she came to know that the LCM of two numbers can be defined as:
LCM(a, b) = ab ⁄ GCD(a, b). However, she was disappointed to learn that LCM was not defined similarly for more than two numbers. She thus defined a new concept, Sasha's LCM, as:
Sasha's LCM(x<sub>1</sub>, x<sub>2</sub>, ... x<sub>N</sub>) = x<sub>1</sub>x<sub>2</sub>...x<sub>N</sub> ⁄ GCD(x<sub>1</sub>, x<sub>2</sub>, ... x<sub>N</sub>)
Excited by her discovery, she has asked you to calculate the sum of Sasha's LCM for all sequences (x<sub>1</sub>, x<sub>2</sub>, ... x<sub>N</sub>) of length N, such that 1 ≤ x<sub>i</sub> ≤ M for all valid i. Since the sum can be large, print the answer modulo 998244353.
For example, if both N and M were 2, you would be asked the value of Sasha's LCM(1, 1) + Sasha's LCM(1, 2) + Sasha's LCM(2, 1) + Sasha's LCM(2, 2) modulo 998244353.The input consists of two space-separated integers N and M.
<b> Constraints: </b>
2 ≤ N ≤ 10<sup>9</sup>
1 ≤ M ≤ 10<sup>6</sup>Print a single integer, the summation of Sasha's LCM for all valid sequences modulo 998244353.Sample Input 1:
2 2
Sample Output 1:
7
Sample Explanation 1:
We have:
Sasha's LCM(1, 1) = 1
Sasha's LCM(2, 1) = 2
Sasha's LCM(1, 2) = 2
Sasha's LCM(2, 2) = 2.
Thus, the total sum is 7.
Sample Input 2:
3 2
Sample Output 2:
23
Sample Explanation 2:
We have:
Sasha's LCM(1, 1, 1) = 1
Sasha's LCM(2, 1, 1) = Sasha's LCM(1, 2, 1) = Sasha's LCM(1, 1, 2) = 2
Sasha's LCM(2, 2, 1) = Sasha's LCM(1, 2, 2) = Sasha's LCM(2, 1, 2) = 2*2/1 = 4
Sasha's LCM(2, 2, 2) = 2*2*2/2 = 4
Thus, the total sum is 1 + 2*3 + 4*3 + 4 = 23.
Sample Input 3:
5 7
Sample Output 3:
17032016, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << (a) << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (auto i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 998244353; // 1e9 + 7;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
readb(n, m);
int ans = 0, sum[m + 1] = {};
FORD (i, m, 1)
{
int cur = 0;
for (int j = i; j <= m; j += i) cur = (cur + j) % mod;
cur = power(cur, n);
for (int j = i*2; j <= m; j += i)
cur = (cur - sum[j]) % mod;
sum[i] = cur;
ans = (ans + cur*power(i)) % mod;
}
ans = (ans + mod) % mod;
print(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rick has got a huge candy bag. The bag contains a total of O orange candies, A apple candies, and U candies with an unknown flavor (can be orange or apple as well).
Now Morty wants to know the maximum number of candies that can be drawn from the bag blindfolded such that no more than X orange and Y apple candies are drawn out.
Can you solve the problem for Morty?The first line of the input contains three integers O, A, and U, representing the number of orange, apple, and unknown flavored candies respectively.
The second line of the input contains two integers X and Y representing the maximum number of orange and apple candies that can be drawn of the bag.
Constraints
0 <= O, A, U <= 10<sup>9</sup>
0 <= X, Y <= 10<sup>9</sup>Output a single integer, the maximum number of candies that can be drawn out of Rick's bag.Sample Input 1
4 2 7
2 7
Sample Output 1
2
Explanation: If we draw out more than 2 candies, there is a possible chance for 3 candies to be orange.
Sample Input 2
1 2 3
8 4
Sample Output 2
4, I have written this Solution Code: [o,a,u]=[int(i) for i in input().split()]
s=o+a+u
[x,y]=[int(i) for i in input().split()]
o+=u
a+=u
if o<=x and a<=y:
print(s)
elif o<=x:
print(y)
elif a<=y:
print(x)
else:
print(min(x,y)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rick has got a huge candy bag. The bag contains a total of O orange candies, A apple candies, and U candies with an unknown flavor (can be orange or apple as well).
Now Morty wants to know the maximum number of candies that can be drawn from the bag blindfolded such that no more than X orange and Y apple candies are drawn out.
Can you solve the problem for Morty?The first line of the input contains three integers O, A, and U, representing the number of orange, apple, and unknown flavored candies respectively.
The second line of the input contains two integers X and Y representing the maximum number of orange and apple candies that can be drawn of the bag.
Constraints
0 <= O, A, U <= 10<sup>9</sup>
0 <= X, Y <= 10<sup>9</sup>Output a single integer, the maximum number of candies that can be drawn out of Rick's bag.Sample Input 1
4 2 7
2 7
Sample Output 1
2
Explanation: If we draw out more than 2 candies, there is a possible chance for 3 candies to be orange.
Sample Input 2
1 2 3
8 4
Sample Output 2
4, I have written this Solution Code: import java.io.*;
import java.util.*;
import static java.lang.Math.*;
public class Main implements Runnable {
private boolean console=false;
private long MOD = 1000_000_007L;
private int MAX = 1000_001;
private void solve1(){
long a=in.nl(),b=in.nl(),c= in.nl();
long x=in.nl(),y=in.nl();
long ans =0;
if(x>y){
long t = x;
x=y;
y=t;
t = a;
a=b;
b=t;
}
if(x-a-c <0){
ans = x;
}else if(y-b-c<0) {
ans = y;
}else {
ans = a+b+c;
}
out.printLn(ans);
}
private void solve() {
int testCases = 1;
while (testCases-->0){
solve1();
}
}
private void add(TreeMap<Integer, Integer> map, int key){
map.put(key,map.getOrDefault(key,0)+1);
}
private void remove(TreeMap<Integer,Integer> map,int key){
if(!map.containsKey(key))
return;
map.put(key,map.getOrDefault(key,0)-1);
if(map.get(key)==0)
map.remove(key);
}
@Override
public void run() {
long time = System.currentTimeMillis();
try {
init();
} catch (FileNotFoundException e) {
e.printStackTrace();
}
try {
solve();
out.flush();
System.err.println(System.currentTimeMillis()-time);
System.exit(0);
}catch (Exception e){
e.printStackTrace(); System.exit(1);
}
}
private FastInput in;
private FastOutput out;
public static void main(String[] args) throws Exception {
new Main().run();
}
private void init() throws FileNotFoundException {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
try {
if (!console && System.getProperty("user.name").equals("puneetkumar")) {
outputStream = new FileOutputStream("/Users/puneetkumar/output.txt");
inputStream = new FileInputStream("/Users/puneetkumar/input.txt");
}
} catch (Exception ignored) {
}
out = new FastOutput(outputStream);
in = new FastInput(inputStream);
}
private void maualAssert(int a,int b,int c){
if(a<b || a>c) throw new RuntimeException();
}
private void maualAssert(long a,long b,long c){
if(a<b || a>c) throw new RuntimeException();
}
private void sort(int[] arr) {
List<Integer> list = new ArrayList<>();
for (int object : arr) list.add(object);
Collections.sort(list);
for (int i = 0; i < list.size(); ++i) arr[i] = list.get(i);
}
private void sort(long[] arr) {
List<Long> list = new ArrayList<>();
for (long object : arr) list.add(object);
Collections.sort(list);
for (int i = 0; i < list.size(); ++i) arr[i] = list.get(i);
}
private long ModPow(long x, long y, long MOD) {
long res = 1L;
x = x % MOD;
while (y >= 1L) {
if ((y & 1L) > 0) res = (res * x) % MOD;
x = (x * x) % MOD;
y >>= 1L;
}
return res;
}
private int gcd(int a, int b) {
if (a == 0) return b;
return gcd(b % a, a);
}
private long gcd(long a, long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
private int[] arrInt(int n){
int[] arr=new int[n];for(int i=0;i<n;++i)arr[i]=in.ni();
return arr;
}
private long[] arrLong(int n){
long[] arr=new long[n];for(int i=0;i<n;++i)arr[i]=in.nl();
return arr;
}
private int arrMax(int[] arr){
int ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
private long arrMax(long[] arr){
long ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
private int arrMin(int[] arr){
int ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
private long arrMin(long[] arr){
long ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
class FastInput { InputStream obj;
public FastInput(InputStream obj) {
this.obj = obj;
}
private byte inbuffer[] = new byte[1024];
private int lenbuffer = 0, ptrbuffer = 0;
private int readByte() { if (lenbuffer == -1) throw new InputMismatchException();
if (ptrbuffer >= lenbuffer) { ptrbuffer = 0;
try { lenbuffer = obj.read(inbuffer);
} catch (IOException e) { throw new InputMismatchException(); } }
if (lenbuffer <= 0) return -1;return inbuffer[ptrbuffer++]; }
String ns() { int b = skip();StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b)))
{ sb.appendCodePoint(b);b = readByte(); }return sb.toString();}
int ni() {
int num = 0, b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); }}
long nl() { long num = 0;int b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10L + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); } }
private boolean isSpaceChar(int c) {
return (!(c >= 33 && c <= 126));
}
int skip() { int b;while ((b = readByte()) != -1 && isSpaceChar(b)) ;return b; }
float nf() {return Float.parseFloat(ns());}
double nd() {return Double.parseDouble(ns());}
char nc() {return (char) skip();}
}
class FastOutput{
private final PrintWriter writer;
public FastOutput(OutputStream outputStream) {
writer = new PrintWriter(outputStream);
}
public PrintWriter getWriter(){
return writer;
}
public void print(Object obj){
writer.print(obj);
}
public void printLn(){
writer.println();
}
public void printLn(Object obj){
writer.print(obj);
printLn();
}
public void printSp(Object obj){
writer.print(obj+" ");
}
public void printArr(int[] arr){
for(int i:arr)
printSp(i);
printLn();
}
public void printArr(long[] arr){
for(long i:arr)
printSp(i);
printLn();
}
public void flush(){
writer.flush();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rick has got a huge candy bag. The bag contains a total of O orange candies, A apple candies, and U candies with an unknown flavor (can be orange or apple as well).
Now Morty wants to know the maximum number of candies that can be drawn from the bag blindfolded such that no more than X orange and Y apple candies are drawn out.
Can you solve the problem for Morty?The first line of the input contains three integers O, A, and U, representing the number of orange, apple, and unknown flavored candies respectively.
The second line of the input contains two integers X and Y representing the maximum number of orange and apple candies that can be drawn of the bag.
Constraints
0 <= O, A, U <= 10<sup>9</sup>
0 <= X, Y <= 10<sup>9</sup>Output a single integer, the maximum number of candies that can be drawn out of Rick's bag.Sample Input 1
4 2 7
2 7
Sample Output 1
2
Explanation: If we draw out more than 2 candies, there is a possible chance for 3 candies to be orange.
Sample Input 2
1 2 3
8 4
Sample Output 2
4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int a, b, c;
cin >> a >> b >> c;
int n, m;
cin >> n >> m;
int ans = a+b+c;
if (n < a + c) ans = min(n, ans);
if (m < b + c) ans = min(m, ans);
cout << ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton's College is organizing a Technical Fest named Technox'23 this year with a budget of Y Thousand Rupees. As DSA Coding Event is the most important function of their Fest, they allocate at least half of their budget to it.
There are also 5 other events planned under Technox'23. The team wants to distribute the remaining amount equally among these. Find the maximum amount in rupees that could be allocated to each of the other five events.The first line of the input contains a single integer T denoting the number of test cases.
The first and only line of each test case contains a single integer Y, the budget of Technox'23 in thousand rupees.
<b>Constraints</b>
1 ≤ T ≤ 10<sup>5</sup>
1 ≤ Y ≤ 10<sup>7</sup>For each test case, output a single integer on a new line, the maximum amount in rupees that could be allocated to each of the other five events.<b>Sample Input</b>
2
20
1
<b>Sample Output</b>
2000
100, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int tt;
cin >> tt;
while (tt--) {
int n;
cin >> n;
n*=100;
cout << n << "\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a Binary Search Tree and 2 node value <b>n1</b> and <b>n2</b>, your task is to find the <b>lowest common ancestor(LCA)</b> of the two nodes given.
It may happen that values n1 and n2 may or may be not present.
<b>Note:</b> Duplicates are not inserted in the BST.<b>User Task:</b>
This is a function problem. You don't have to take any input. You are required to complete the function LCA() that takes root node, n1, n2 as parameters and returns the node that is LCA of n1 and n2.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^4
1 <= node values <= 10^4
<b>Sum of "N" over all testcases does not exceed 10^5</b>
For <b>Custom Input:</b>
First line of input should contains the number of test cases T. For each test case, there will be two lines of input.
First line will be a string representing the tree as described below:
The values in the string are in the order of level order traversal of the tree where, numbers denote node values, and a character “N” denotes NULL child.
Second line will contain the values of two nodes
<b>Note:</b> If a node has been declared Null using 'N', no information about its children will be given further in the array.
Second line will contain the values of two nodesFor each testcase, you need to return the node containing LCA of n1 and n2. The driver code will print the data. If any of the node values (n1, n2) is not present then return null driver code will print -1 for that.Sample Input
2
5 4 6 3 N N 7 N N N 8
7 8
2 1 3
1 3
Sample Output
7
2
Explanation:
Testcase1:
The BST in above test case will look like
5
/ \
4 6
/ \
3 7
\
8
Here the LCA of 7 and 8 is 7., I have written this Solution Code:
static Node LCA(Node node, int n1, int n2) {
if (node == null) {
return null;
}
// If both n1 and n2 are smaller than root, then LCA lies in left
if (node.data > n1 && node.data > n2) {
return LCA(node.left, n1, n2);
}
// If both n1 and n2 are greater than root, then LCA lies in right
if (node.data < n1 && node.data < n2) {
return LCA(node.right, n1, n2);
}
return node;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a function called lucky_sevens which takes an array of integers and returns true if any three consecutive elements sum to 7An array containing numbers.Print true if such triplet exists summing to 7 else print falseSample input:-
[2, 1, 5, 1, 0]
[1, 6]
Sample output:-
true
false
Explanation:-
1+5+1 = 7
no 3 consecutive numbers so false, I have written this Solution Code: function lucky_sevens(arr) {
// if less than 3 elements then this challenge is not possible
if (arr.length < 3) {
console.log(false)
return;
}
// because we know there are at least 3 elements we can
// start the loop at the 3rd element in the array (i=2)
// and check it along with the two previous elements (i-1) and (i-2)
for (let i = 2; i < arr.length; i++) {
if (arr[i] + arr[i-1] + arr[i-2] === 7) {
console.log(true)
return;
}
}
// if loop is finished and no elements summed to 7
console.log(false)
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two sorted arrays A and B of size N and M respectively.
You have to find the value V, which is the summation of (A[j] - A[i]) for all pairs of i and j such that (j - i) is present in array B.The first line contains two space separated integers N and M – size of arrays A and B respectively.
The second line contains N integers A<sub>1</sub>, A<sub>2</sub>, ... A<sub>N</sub>.
The third line contains M integers B<sub>1</sub>, B<sub>2</sub>, ... B<sub>M</sub>.
<b> Constraints: </b>
1 ≤ N ≤ 2×10<sup>5</sup>
1 ≤ M < N
1 ≤ A<sub>1</sub> ≤ A<sub>2</sub> ≤ ... ≤ A<sub>N</sub> ≤ 10<sup>8</sup>.
1 ≤ B<sub>1</sub> < B<sub>2</sub> < ... < B<sub>M</sub> < N.Print a single integer, the value of V.Sample Input 1:
4 2
1 2 3 4
1 3
Sample Output 1:
6
Sample Explanation 1:
Valid pairs of (i, j) are (1, 2), (2, 3), (3, 4), (1, 4)., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define epsi (double)(0.00000000001)
typedef long long int ll;
typedef unsigned long long int ull;
#define vi vector<ll>
#define pii pair<ll,ll>
#define vii vector<pii>
#define vvi vector<vi>
//#define max(a,b) ((a>b)?a:b)
//#define min(a,b) ((a>b)?b:a)
#define min3(a,b,c) min(min(a,b),c)
#define min4(a,b,c,d) min(min(a,b),min(c,d))
#define max3(a,b,c) max(max(a,b),c)
#define max4(a,b,c,d) max(max(a,b),max(c,d))
#define ff(a,b,c) for(int a=b; a<=c; a++)
#define frev(a,b,c) for(int a=c; a>=b; a--)
#define REP(a,b,c) for(int a=b; a<c; a++)
#define pb push_back
#define mp make_pair
#define endl "\n"
#define all(v) v.begin(),v.end()
#define sz(a) (ll)a.size()
#define F first
#define S second
#define ld long double
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define ub upper_bound
#define lb lower_bound
#define setbits(x) __builtin_popcountll(x)
#define trav(a,x) for(auto &a:x)
#define make_unique(v) v.erase(unique(v.begin(), v.end()), v.end())
#define rev(arr) reverse(all(arr))
#define gcd(a,b) __gcd(a,b);
#define ub upper_bound // '>'
#define lb lower_bound // '>='
#define qi queue<ll>
#define fsh cout.flush()
#define si stack<ll>
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define fill(a,b) memset(a, b, sizeof(a))
template<typename T,typename E>bool chmin(T& s,const E& t){bool res=s>t;s=min<T>(s,t);return res;}
const ll INF=1LL<<60;
void solve(){
ll n,m;
cin >> n >> m;
vi v1(n),v2(m);
for(auto &i:v1){
cin >> i;
}
for(auto &i:v2){
cin >> i;
}
ll ans=0;
for(int i=1 ; i<=n ; i++){
auto it=lower_bound(all(v2),i);
ll x=it-v2.begin();
ans+=(x*v1[i-1]);
it=lower_bound(all(v2),n-i+1);
x=it-v2.begin();
ans-=(x*v1[i-1]);
}
cout << ans << endl;
}
int main(){
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int t=1;
// cin >> t;
while(t--){
solve();
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException
{
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String StrInput[] = br.readLine().trim().split(" ");
int n = Integer.parseInt(StrInput[0]);
int s = Integer.parseInt(StrInput[1]);
int arr[] = new int[n];
String StrInput2[] = br.readLine().trim().split(" ");
for(int i=0;i<n;i++)
{
arr[i] = Integer.parseInt(StrInput2[i]);
}
int sum = arr[0];
int startingindex = 0;
int endingindex = 1;
int j = 0;
int i;
for(i=1;i<=n;i++)
{
if(sum < s && arr[i] != 0)
{
sum += arr[i];
}
while(sum > s && startingindex < i-1)
{
sum -= arr[startingindex];
startingindex++;
}
if(sum == s)
{
endingindex = i+1;
if(arr[0] == 0)
{
System.out.print(startingindex+2 + " " + endingindex);
}
else
{
System.out.print(startingindex+1 + " "+ endingindex);
}
break;
}
if(i == n && sum < s)
{
System.out.print(-1);
break;
}
}
}
catch(Exception e)
{
System.out.print(-1);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n,k;
cin>>n>>k;
int a[n];
for(int i=0;i<n;i++)
{
cin>>a[i];
}
int sum=0;
unordered_map<int,int> m;
for(int i=0;i<n;i++){
sum+=a[i];
if(sum==k){cout<<1<<" "<<i+1;return 0;}
if(m.find(sum-k)!=m.end()){
cout<<m[sum-k]+2<<" "<<i+1;
return 0;
}
m[sum]=i;
}
cout<<-1;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: def sumFinder(N,S,a):
currentSum = a[0]
start = 0
i = 1
while i <= N:
while currentSum > S and start < i-1:
currentSum = currentSum - a[start]
start += 1
if currentSum == S:
return (start+1,i)
if i < N:
currentSum = currentSum + a[i]
i += 1
return(-1)
N, S = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
ans = sumFinder(N,S,a)
if(ans==-1):
print(ans)
else:
print(ans[0],ans[1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings A and B, you can perform three types of moves:-
1. Insert an element at any position
2. Delete an element from any position
3. Replace any element.
Your task is to convert String A into String B in a minimum number of moves.First line of input contains the string A, next line contains the string B.
Constraints:-
1 <= |A|, |B| <= 1000
Note:- String will contain only lowercase English lettersPrint the minimum number of moves.Sample Input:-
geek
gseke
Sample Output:-
2
Explanation:-
replace s with e and delete the last element
Sample input:-
Newton
School
Sample Output:-
5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s1 = br.readLine();
String s2 = br.readLine();
System.out.print(minMoves(s1,s2));
}
static int minMoves(String s1, String s2) {
int[][] A = new int[s1.length()+1][s2.length()+1];
for (int i = 0; i <=s1.length(); i++) {
for (int j = 0; j <= s2.length(); j++ ) {
if (i == 0) {
A[i][j] = j;
} else if ( j == 0) {
A[i][j] = i;
} else if (s1.charAt(i-1) == s2.charAt(j-1)) {
A[i][j] = A[i-1][j-1];
} else {
A[i][j] = 1 + minValue(A[i-1][j-1], A[i-1][j], A[i][j-1]);
}
}
}
return A[s1.length()][s2.length()];
}
static int minValue(int x, int y, int z)
{
if (x <= y && x <= z){
return x;
}
if (y <= x && y <= z){
return y;
}
else{
return z;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings A and B, you can perform three types of moves:-
1. Insert an element at any position
2. Delete an element from any position
3. Replace any element.
Your task is to convert String A into String B in a minimum number of moves.First line of input contains the string A, next line contains the string B.
Constraints:-
1 <= |A|, |B| <= 1000
Note:- String will contain only lowercase English lettersPrint the minimum number of moves.Sample Input:-
geek
gseke
Sample Output:-
2
Explanation:-
replace s with e and delete the last element
Sample input:-
Newton
School
Sample Output:-
5, I have written this Solution Code:
def minimumMoves(a,b):
m = len(a)
n= len(b)
dp = [[0 for i in range(n + 1)] for i in range(m + 1)]
for i in range(m+1):
for j in range(n+1):
if(i==0):
dp[i][j] = j
elif(j==0):
dp[i][j] = i
elif(a[i-1] == b[j-1]):
dp[i][j] = dp[i-1][j-1]
else:
dp[i][j] = 1 + min(dp[i][j-1],dp[i-1][j],dp[i-1][j-1])
return dp[m][n]
a = input()
b = input()
print(minimumMoves(a,b)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings A and B, you can perform three types of moves:-
1. Insert an element at any position
2. Delete an element from any position
3. Replace any element.
Your task is to convert String A into String B in a minimum number of moves.First line of input contains the string A, next line contains the string B.
Constraints:-
1 <= |A|, |B| <= 1000
Note:- String will contain only lowercase English lettersPrint the minimum number of moves.Sample Input:-
geek
gseke
Sample Output:-
2
Explanation:-
replace s with e and delete the last element
Sample input:-
Newton
School
Sample Output:-
5, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int solve(int n){
int cnt=0;
int p=sqrt(n);
for(int i=1;i<=p;i++){
if(n%i==0){
cnt++;
if(i*i!=n){cnt++;}
}
}
return cnt;
}
int dp[max1][max1];
vector<int> v;
signed main(){
string a,b;
cin>>a>>b;
int m=a.length();
int n=b.length();
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
if (i == 0)
dp[i][j] = j; // Min. operations = j
else if (j == 0)
dp[i][j] = i; // Min. operations = i
else if (a[i - 1] == b[j - 1])
dp[i][j] = dp[i - 1][j - 1];
else
dp[i][j] = 1 + min(dp[i][j - 1],min(dp[i - 1][j],dp[i - 1][j - 1]));
}
}
out(dp[m][n]);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
InputStreamReader is = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(is);
int n=Integer.parseInt(br.readLine().trim());
if(n%3==1)
System.out.println("Dead");
else
System.out.println("Alive");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: n = int(input().strip())
if n%3 == 1:
print("Dead")
else:
print("Alive"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n%3==1)
cout<<"Dead";
else
cout<<"Alive";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean matrix of size N*M in which each row is sorted your task is to print the index of the row containing maximum 1's. If multiple answer exist print the smallest one.First line contains two space separated integers denoting values of N and M. Next N lines contains M space separated integers depicting the values of the matrix.
Constraints:-
1 < = M, N < = 1000
0 < = Matrix[][] < = 1Print the smallest index (0 indexing) of a row containing the maximum number of 1's.Sample Input:-
3 5
0 1 1 1 1
0 0 0 1 1
0 0 0 1 1
Sample Output:-
0
Sample Input:-
4 4
0 1 1 1
1 1 1 1
0 0 1 1
1 1 1 1
Sample Output:-
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class Reader{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader(){
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1){
if (c == '\n')break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)c = read();
do{
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)return -ret;return ret;
}
public long nextLong() throws IOException{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.'){
while ((c = read()) >= '0' && c <= '9'){
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)buffer[0] = -1;
}
private byte read() throws IOException{
if (bufferPointer == bytesRead)fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException{
if (din == null)return;
din.close();
}
}
public static void main (String[] args) throws IOException{
Reader sc = new Reader();
int m = sc.nextInt();
int n = sc.nextInt();
int[][] arr = new int[m][n];
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
arr[i][j] = sc.nextInt();
}
}
int max_row_index = 0;
int j = n - 1;
for (int i = 0; i < m; i++) {
while (j >= 0 && arr[i][j] == 1) {
j = j - 1;
max_row_index = i;
}
}
System.out.println(max_row_index);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean matrix of size N*M in which each row is sorted your task is to print the index of the row containing maximum 1's. If multiple answer exist print the smallest one.First line contains two space separated integers denoting values of N and M. Next N lines contains M space separated integers depicting the values of the matrix.
Constraints:-
1 < = M, N < = 1000
0 < = Matrix[][] < = 1Print the smallest index (0 indexing) of a row containing the maximum number of 1's.Sample Input:-
3 5
0 1 1 1 1
0 0 0 1 1
0 0 0 1 1
Sample Output:-
0
Sample Input:-
4 4
0 1 1 1
1 1 1 1
0 0 1 1
1 1 1 1
Sample Output:-
1, I have written this Solution Code: r, c = list(map(int, input().split()))
max_count = 0
max_r = 0
for i in range(r):
count = input().count("1")
if count > max_count:
max_count = count
max_r = i
print(max_r), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean matrix of size N*M in which each row is sorted your task is to print the index of the row containing maximum 1's. If multiple answer exist print the smallest one.First line contains two space separated integers denoting values of N and M. Next N lines contains M space separated integers depicting the values of the matrix.
Constraints:-
1 < = M, N < = 1000
0 < = Matrix[][] < = 1Print the smallest index (0 indexing) of a row containing the maximum number of 1's.Sample Input:-
3 5
0 1 1 1 1
0 0 0 1 1
0 0 0 1 1
Sample Output:-
0
Sample Input:-
4 4
0 1 1 1
1 1 1 1
0 0 1 1
1 1 1 1
Sample Output:-
1, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int a[max1][max1];
signed main()
{
int n,m;
cin>>n>>m;
FOR(i,n){
FOR(j,m){cin>>a[i][j];}}
int cnt=0;
int ans=0;
int res=0;
FOR(i,n){
cnt=0;
FOR(j,m){
if(a[i][j]==1){
cnt++;
}}
if(cnt>res){
res=cnt;
ans=i;
}
}
out(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean matrix of size N*M in which each row is sorted your task is to print the index of the row containing maximum 1's. If multiple answer exist print the smallest one.First line contains two space separated integers denoting values of N and M. Next N lines contains M space separated integers depicting the values of the matrix.
Constraints:-
1 < = M, N < = 1000
0 < = Matrix[][] < = 1Print the smallest index (0 indexing) of a row containing the maximum number of 1's.Sample Input:-
3 5
0 1 1 1 1
0 0 0 1 1
0 0 0 1 1
Sample Output:-
0
Sample Input:-
4 4
0 1 1 1
1 1 1 1
0 0 1 1
1 1 1 1
Sample Output:-
1, I have written this Solution Code: // mat is the matrix/ 2d array
// n,m are dimensions
function max1Row(mat, n, m) {
// write code here
// do not console.log
// return the answer as a number
let j, max_row_index = 0;
j = m - 1;
for (let i = 0; i < n; i++)
{
// Move left until a 0 is found
let flag = false;
// to check whether a row has more 1's than previous
while (j >= 0 && mat[i][j] == 1)
{
j = j - 1; // Update the index of leftmost 1
// seen so far
flag = true;//present row has more 1's than previous
}
// if the present row has more 1's than previous
if (flag)
{
max_row_index = i; // Update max_row_index
}
}
if (max_row_index == 0 && mat[0][m - 1] == 0)
return -1;
return max_row_index;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
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