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For this Question: You must be familiar with various types of operators. One of the most commonly used operators in any language are increment and decrement operators. Given two numbers X and Y. Your task is to print the value of X decremented by 1 and value of Y after incremented by 1.The first line of the input contains two integers X and Y.
<b>Constraints:</b>
1 ≤ X, Y ≤ 10000You need to perform the task as mentioned in the question, and finally, print the result separated by a space.Sample Input 1:
4 5
Sample Output 1:
3 6
Sample Input 2:
5 6
Sample Output 2:
4 7, I have written this Solution Code: def incremental_decremental(x, y):
x -= 1
y += 1
print(x, y, end =' ')
def main():
input1 = input().split()
x = int(input1[0])
y = int(input1[1])
#z = int(input1[2])
incremental_decremental(x, y)
if __name__ == '__main__':
main(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After saving the planet of Xander in 2014 from Ronan, the guardians of the galaxy decided to play a game on it. Between the vast furious oceans, they discover a row of buildings. The buildings are of different heights arranged at each index along a number line. Gamora starts at building 0 and a height of 0. Groot gets the task to <b>determine the minimum thrust that Gamora needs at the start of the game so that she can jump to the top of each building without her thrust going below zero</b>. Write a code to help Groot with his task.
The units of height relate directly to units of thrust. Gamora’s thrust level is calculated as follows:
• If gamora's thrust is less than the height of the building, her new thrust = gamora's thrust – (height – gamora's thrust)
• If gamora's thrust is greater than equal to the height of the building, her new thrust = gamora's thrust + (gamora's thrust - height)The first line contains an integer n, the number of buildings.
The second line n space-separated integers, arr[1], arr[2]…arr[n], the heights of the building.
<b>Constraints:-</b>
1 ≤ n ≤ 10<sup>5</sup>
1 ≤ arr[i] ≤ 10<sup>5</sup>The output contains a single integer denoting the minimum thrust Gamora needs at the start of the game.Sample Input:-
5
3 4 3 2 4
Sample Output:-
4
Explanation:-
If we take 3 then:-
at index 1:- 3 + 3-3 = 3
at index 2:- 3 - (4-3) = 2
at index 3:- 2 - (3-2) = 1
at index 4:- 1 - (2-1) = 0
Sample Input:-
3
4 4 4
Sample Output:-
4, 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();
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i]= sc.nextLong();
}
long l = 0;
long r= 1000000000;
long x=0;
while (l!=r){
x= (l+r)/2;
if(checkThrust(arr,x)) {
r=x;
}
else {
l=x+1;
}
}
System.out.print(l);
}
static boolean checkThrust(long[] arr,long r){
long thrust = r;
for (int i = 0; i < arr.length; i++) {
thrust = 2*thrust-arr[i];
if(thrust>=1000000000000L) return true;
if(thrust<0) return false;
}
return true;
}
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: After saving the planet of Xander in 2014 from Ronan, the guardians of the galaxy decided to play a game on it. Between the vast furious oceans, they discover a row of buildings. The buildings are of different heights arranged at each index along a number line. Gamora starts at building 0 and a height of 0. Groot gets the task to <b>determine the minimum thrust that Gamora needs at the start of the game so that she can jump to the top of each building without her thrust going below zero</b>. Write a code to help Groot with his task.
The units of height relate directly to units of thrust. Gamora’s thrust level is calculated as follows:
• If gamora's thrust is less than the height of the building, her new thrust = gamora's thrust – (height – gamora's thrust)
• If gamora's thrust is greater than equal to the height of the building, her new thrust = gamora's thrust + (gamora's thrust - height)The first line contains an integer n, the number of buildings.
The second line n space-separated integers, arr[1], arr[2]…arr[n], the heights of the building.
<b>Constraints:-</b>
1 ≤ n ≤ 10<sup>5</sup>
1 ≤ arr[i] ≤ 10<sup>5</sup>The output contains a single integer denoting the minimum thrust Gamora needs at the start of the game.Sample Input:-
5
3 4 3 2 4
Sample Output:-
4
Explanation:-
If we take 3 then:-
at index 1:- 3 + 3-3 = 3
at index 2:- 3 - (4-3) = 2
at index 3:- 2 - (3-2) = 1
at index 4:- 1 - (2-1) = 0
Sample Input:-
3
4 4 4
Sample Output:-
4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
const long long linf = 0x3f3f3f3f3f3f3f3fLL;
const int N = 111111;
int n;
int h[N];
int check(long long x) {
long long energy = x;
for(int i = 1; i <= n; i++) {
energy += energy - h[i];
if(energy >= linf) return 1;
if(energy < 0) return 0;
}
return 1;
}
int main() {
cin >> n;
for(int i = 1; i <= n; i++) {
cin >> h[i];
}
long long L = 0, R = linf;
long long ans=0;
while(L < R) {
long long M = (L + R) / 2;
if(check(M)) {
R = M;
ans=M;
} else {
L = M + 1;
}
}
cout << ans << endl;
return 0;
}
, 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, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, I have written this Solution Code: // mat is the matrix/ 2d array
// n,m are dimensions
function goodCell(mat, n, m) {
// write code here
// do not console.log
// return the answer as a number
let cnt = 0;
for (let i = 1; i < n - 1; i++) {
for (let j = 1; j < m - 1; j++) {
if (mat[i - 1][j] == 1 && mat[i + 1][j] == 1 && mat[i][j - 1] == 1 && mat[i][j + 1] == 1) {
cnt++;
}
}
}
return cnt
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean Matrix of size N*M, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, I have written this Solution Code: N, M= list(map(int,input().split()))
mat =[]
for i in range(N):
List =list(map(int,input().split()))[:M]
mat.append(List)
count =0
for i in range(1,N-1):
for j in range(1,M-1):
if (mat[i][j-1] == 1 and mat[i][j+1] == 1 and mat[i-1][j] == 1 and mat[i+1][j] == 1):
count +=1
print(count), 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, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, 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 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);
}
signed main(){
int n,m;
cin>>n>>m;
int a[n][m];
FOR(i,n){
FOR(j,m){
cin>>a[i][j];}}
int sum=0,sum1=0;;
FOR1(i,1,n-1){
FOR1(j,1,m-1){
if(a[i-1][j]==1 && a[i+1][j]==1 && a[i][j-1]==1 && a[i][j+1]==1){
sum++;
}
}
}
out1(sum);
}
, 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, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, 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 m= sc.nextInt();
int a[][]= new int[n][m];
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
a[i][j]=sc.nextInt();}}
int cnt=0;
for(int i=1;i<n-1;i++){
for(int j=1;j<m-1;j++){
if(a[i-1][j]==1 && a[i+1][j]==1 && a[i][j-1]==1 && a[i][j+1]==1){
cnt++;
}
}
}
System.out.print(cnt);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three distinct positive integers, A, B and C, in the input. Your task is to find their median.
The median of a set of integers is the number at the middle position if they are arranged in ascending order.The input consists of a single line containing three integers A, B and C.
<b> Constraints: </b>
1 ≤ A, B, C ≤ 10<sup>9</sup>
A, B, C are pairwise distinct.
Print a single integer, the median of the given numbers.Sample Input 1:
5 1 3
Sample Output 1:
3
Sample Input 2:
1 2 3
Sample Output 2:
2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
try{
InputStreamReader r=new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(r);
String a = in.readLine();
String[] nums = a.split(" ");
long[] l = new long[3];
for(int i=0; i<3; i++){
l[i] = Long.parseLong(nums[i]);
}
Arrays.sort(l);
System.out.print(l[1]);
}
catch(Exception e){
System.out.println(e);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three distinct positive integers, A, B and C, in the input. Your task is to find their median.
The median of a set of integers is the number at the middle position if they are arranged in ascending order.The input consists of a single line containing three integers A, B and C.
<b> Constraints: </b>
1 ≤ A, B, C ≤ 10<sup>9</sup>
A, B, C are pairwise distinct.
Print a single integer, the median of the given numbers.Sample Input 1:
5 1 3
Sample Output 1:
3
Sample Input 2:
1 2 3
Sample Output 2:
2, I have written this Solution Code:
//#define ASC
//#define DBG_LOCAL
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
#define int long long
// #define int __int128
#define all(X) (X).begin(), (X).end()
#define pb push_back
#define endl '\n'
#define fi first
#define se second
// const int mod = 1e9 + 7;
const int mod=998'244'353;
const long long INF = 2e18 + 10;
// const int INF=1e9+10;
#define readv(x, n) \
vector<int> x(n); \
for (auto &i : x) \
cin >> i;
template <typename T>
using v = vector<T>;
template <typename T>
using vv = vector<vector<T>>;
template <typename T>
using vvv = vector<vector<vector<T>>>;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>> vvvi;
typedef vector<vector<vector<vector<int>>>> vvvvi;
typedef vector<vector<double>> vvd;
typedef pair<int, int> pii;
int multiply(int a, int b, int in_mod) { return (int)(1LL * a * b % in_mod); }
int mult_identity(int a) { return 1; }
const double PI = acosl(-1);
auto power(auto a, auto b, const int in_mod)
{
auto prod = mult_identity(a);
auto mult = a % 2;
while (b != 0)
{
if (b % 2)
{
prod = multiply(prod, mult, in_mod);
}
if(b/2)
mult = multiply(mult, mult, in_mod);
b /= 2;
}
return prod;
}
auto mod_inv(auto q, const int in_mod)
{
return power(q, in_mod - 2, in_mod);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define stp cout << fixed << setprecision(20);
void solv()
{
int A ,B, C;
cin>>A>>B>>C;
vector<int> values;
values.push_back(A);
values.push_back(B);
values.push_back(C);
sort(all(values));
cout<<values[1]<<endl;
}
void solve()
{
int t = 1;
// cin>>t;
for(int i = 1;i<=t;i++)
{
// cout<<"Case #"<<i<<": ";
solv();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#else
#ifdef ASC
namespace fs = std::filesystem;
std::string path = "./";
string filename;
for (const auto & entry : fs::directory_iterator(path)){
if( entry.path().extension().string() == ".in"){
filename = entry.path().filename().stem().string();
}
}
if(filename != ""){
string input_file = filename +".in";
string output_file = filename +".out";
if (fopen(input_file.c_str(), "r"))
{
freopen(input_file.c_str(), "r", stdin);
freopen(output_file.c_str(), "w", stdout);
}
}
#endif
#endif
// auto clk = clock();
// -------------------------------------Code starts here---------------------------------------------------------------------
signed t = 1;
// cin >> t;
for (signed test = 1; test <= t; test++)
{
// cout<<"Case #"<<test<<": ";
// cout<<endl;
solve();
}
// -------------------------------------Code ends here------------------------------------------------------------------
// clk = clock() - clk;
#ifndef ONLINE_JUDGE
// cerr << fixed << setprecision(6) << "\nTime: " << ((float)clk) / CLOCKS_PER_SEC << "\n";
#endif
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three distinct positive integers, A, B and C, in the input. Your task is to find their median.
The median of a set of integers is the number at the middle position if they are arranged in ascending order.The input consists of a single line containing three integers A, B and C.
<b> Constraints: </b>
1 ≤ A, B, C ≤ 10<sup>9</sup>
A, B, C are pairwise distinct.
Print a single integer, the median of the given numbers.Sample Input 1:
5 1 3
Sample Output 1:
3
Sample Input 2:
1 2 3
Sample Output 2:
2, I have written this Solution Code: lst = list(map(int, input().split()))
lst.sort()
print(lst[1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four integers a, b, c, and d. Find the value of a<sup>(b<sup>(c<sup>d</sup>)</sup>)</sup> modulo 1000000007.
(Fact: 0<sup>0</sup> = 1)The first and the only line of input contains 4 integers a, b, c, and d.
Constraints
1 <= a, b, c, d <= 12Output a single integer, the answer modulo 1000000007.Sample Input
2 2 2 2
Sample Output
65536
Explanation
2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536.
Sample Input
0 7 11 1
Sample Output
0, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.math.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader sc= new BufferedReader(new InputStreamReader(System.in));
int a=0, b=0, c=0, d=0;
long z=0;
String[] str;
str = sc.readLine().split(" ");
a= Integer.parseInt(str[0]);
b= Integer.parseInt(str[1]);
c= Integer.parseInt(str[2]);
d= Integer.parseInt(str[3]);
BigInteger m = new BigInteger("1000000007");
BigInteger n = new BigInteger("1000000006");
BigInteger zero = new BigInteger("0");
BigInteger ans, y;
if(d==0){
z =1;
}else{
z = (long)Math.pow(c, d);
}
if(b==0){
y= zero;
}else{
y = (BigInteger.valueOf(b)).modPow((BigInteger.valueOf(z)), n);
}
if(y == zero){
System.out.println("1");
}else if(a==0){
System.out.println("0");
}else{
ans = (BigInteger.valueOf(a)).modPow(y, m);
System.out.println(ans);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four integers a, b, c, and d. Find the value of a<sup>(b<sup>(c<sup>d</sup>)</sup>)</sup> modulo 1000000007.
(Fact: 0<sup>0</sup> = 1)The first and the only line of input contains 4 integers a, b, c, and d.
Constraints
1 <= a, b, c, d <= 12Output a single integer, the answer modulo 1000000007.Sample Input
2 2 2 2
Sample Output
65536
Explanation
2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536.
Sample Input
0 7 11 1
Sample Output
0, I have written this Solution Code: nan, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four integers a, b, c, and d. Find the value of a<sup>(b<sup>(c<sup>d</sup>)</sup>)</sup> modulo 1000000007.
(Fact: 0<sup>0</sup> = 1)The first and the only line of input contains 4 integers a, b, c, and d.
Constraints
1 <= a, b, c, d <= 12Output a single integer, the answer modulo 1000000007.Sample Input
2 2 2 2
Sample Output
65536
Explanation
2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536.
Sample Input
0 7 11 1
Sample Output
0, 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
int powmod(int a, int b, int c = MOD){
int ans = 1;
while(b){
if(b&1){
ans = (ans*a)%c;
}
a = (a*a)%c;
b >>= 1;
}
return ans;
}
void solve(){
int a, b, c, d; cin>>a>>b>>c>>d;
int x = pow(c, d);
int y = powmod(b, x, MOD-1);
int ans = powmod(a, y, MOD);
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: You are given two arrays - value and frequency both containing N elements.
There is also a third array C which is currently empty. Then you perform N insertion operation in the array. For ith operation you insert value[i] to the end of the array frequency[i] number of times.
Finally you have to tell the kth smallest element in the array C.First line of input contains N.
Second line contains N integers denoting array - value
Third line contains N integers denoting array - frequency
Fourth line contains single integer K.
Constraints
1 <= N, value[i], frequency[i] <= 100000
1 <= k <= frequency[1] + frequency[2] +frequency[3] +........ + frequency[N]
Output a single integer which is the kth smallest element of the array C.Sample input 1
5
1 2 3 4 5
1 1 1 2 2
3
Sample output 1
3
Explanation 1:
Array C constructed is 1 2 3 4 4 5 5
Third smallest element is 3
Sample input 2
3
2 1 3
3 3 2
2
sample output 2
1
Explanation 2:
Array C constructed is 2 2 2 1 1 1 3 3
Second smallest element is 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
StringTokenizer st = new StringTokenizer(br.readLine());
int[] val = new int[n];
for(int i=0; i<n; i++){
val[i] = Integer.parseInt(st.nextToken());
}
st = new StringTokenizer(br.readLine());
int[] freq = new int[n];
for(int i=0; i<n; i++){
freq[i] = Integer.parseInt(st.nextToken());
}
int k = Integer.parseInt(br.readLine());
for (int i=0; i<n; i++) {
for (int j=i+1; j<n; j++) {
if (val[j] < val[i]) {
int temp = val[i];
val[i] = val[j];
val[j] = temp;
int temp1 = freq[i];
freq[i] = freq[j];
freq[j] = temp1;
}
}
}
int element=0;
for(int i=0; i<n; i++){
for(int j=0; j<freq[i]; j++){
element++;
int value = val[i];
if(element==k){
System.out.print(value);
break;
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two arrays - value and frequency both containing N elements.
There is also a third array C which is currently empty. Then you perform N insertion operation in the array. For ith operation you insert value[i] to the end of the array frequency[i] number of times.
Finally you have to tell the kth smallest element in the array C.First line of input contains N.
Second line contains N integers denoting array - value
Third line contains N integers denoting array - frequency
Fourth line contains single integer K.
Constraints
1 <= N, value[i], frequency[i] <= 100000
1 <= k <= frequency[1] + frequency[2] +frequency[3] +........ + frequency[N]
Output a single integer which is the kth smallest element of the array C.Sample input 1
5
1 2 3 4 5
1 1 1 2 2
3
Sample output 1
3
Explanation 1:
Array C constructed is 1 2 3 4 4 5 5
Third smallest element is 3
Sample input 2
3
2 1 3
3 3 2
2
sample output 2
1
Explanation 2:
Array C constructed is 2 2 2 1 1 1 3 3
Second smallest element is 1, I have written this Solution Code: def myFun():
n = int(input())
arr1 = list(map(int,input().strip().split()))
arr2 = list(map(int,input().strip().split()))
k = int(input())
arr = []
for i in range(n):
arr.append((arr1[i], arr2[i]))
arr.sort()
c = 0
for i in arr:
k -= i[1]
if k <= 0:
print(i[0])
return
myFun()
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two arrays - value and frequency both containing N elements.
There is also a third array C which is currently empty. Then you perform N insertion operation in the array. For ith operation you insert value[i] to the end of the array frequency[i] number of times.
Finally you have to tell the kth smallest element in the array C.First line of input contains N.
Second line contains N integers denoting array - value
Third line contains N integers denoting array - frequency
Fourth line contains single integer K.
Constraints
1 <= N, value[i], frequency[i] <= 100000
1 <= k <= frequency[1] + frequency[2] +frequency[3] +........ + frequency[N]
Output a single integer which is the kth smallest element of the array C.Sample input 1
5
1 2 3 4 5
1 1 1 2 2
3
Sample output 1
3
Explanation 1:
Array C constructed is 1 2 3 4 4 5 5
Third smallest element is 3
Sample input 2
3
2 1 3
3 3 2
2
sample output 2
1
Explanation 2:
Array C constructed is 2 2 2 1 1 1 3 3
Second smallest element is 1, I have written this Solution Code: #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 inf 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;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
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()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int N;ll K;
cin>>N;
int c=0;
pair<int, ll> A[N];
for(int i=0;i<N;++i){
cin >> A[i].first ;
}
for(int i=0;i<N;++i){
cin >> A[i].second ;
}
cin>>K;
sort(A, A+N);
for(int i=0;i<N;++i){
K -= A[i].second;
if(K <= 0){
cout << A[i].first << endl;;
break;
}
}
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, super primes are those integers from 1 to N whose multiples (other than themselves) do no exist in the range [1, N].
Your task is to generate all super primes <= N in sorted order.
</b>Note: Super primes are not related to primes in any way.</b><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>SuperPrime()</b> that takes the integer N as a parameter.
Constraints:-
2 < = N <= 1000Printing will be done by the driver code you just need to complete the generator function.Sample Input:-
5
Sample Output:-
3 4 5
Sample Input:-
4
Sample Output:-
3 4, I have written this Solution Code: public static void SuperPrimes(int n){
int x = n/2+1;
for(int i=x ; i<=n ; i++){
out.printf("%d ",i);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, super primes are those integers from 1 to N whose multiples (other than themselves) do no exist in the range [1, N].
Your task is to generate all super primes <= N in sorted order.
</b>Note: Super primes are not related to primes in any way.</b><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>SuperPrime()</b> that takes the integer N as a parameter.
Constraints:-
2 < = N <= 1000Printing will be done by the driver code you just need to complete the generator function.Sample Input:-
5
Sample Output:-
3 4 5
Sample Input:-
4
Sample Output:-
3 4, I have written this Solution Code: def SuperPrimes(N):
for i in range (int(N/2)+1,N+1):
yield i
return
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given number N, Consider a table of N X N in which elements at the intersection of ith row and jth column are the product of i and j, i. e i x j.
Also given a positive integer X. Your task is to count the number of elements in the table that contain the integer X.<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>OccurenceOfX()</b> that takes the integer N and the integer X as parameter.
Constraints:-
1 <= N <= 10^5
1 <= X <= 10^9Return the count of X.Sample Input:-
5 5
Sample Output:-
2
Explanation:-
table :-
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
Count of occurrence of X is :- 2
Sample Input:-
10 13
Sample Output:-
0, I have written this Solution Code: def OccurenceOfX(N,X):
cnt=0
for i in range(1, N+1):
if(X%i==0 and X/i<=N):
cnt=cnt+1
return cnt, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given number N, Consider a table of N X N in which elements at the intersection of ith row and jth column are the product of i and j, i. e i x j.
Also given a positive integer X. Your task is to count the number of elements in the table that contain the integer X.<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>OccurenceOfX()</b> that takes the integer N and the integer X as parameter.
Constraints:-
1 <= N <= 10^5
1 <= X <= 10^9Return the count of X.Sample Input:-
5 5
Sample Output:-
2
Explanation:-
table :-
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
Count of occurrence of X is :- 2
Sample Input:-
10 13
Sample Output:-
0, I have written this Solution Code:
int OccurenceOfX(int N,long X){
int cnt=0,i;
for( i=1;i<=N;i++){
if(X%i==0 && X/i<=N){cnt++;}}
return cnt;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given number N, Consider a table of N X N in which elements at the intersection of ith row and jth column are the product of i and j, i. e i x j.
Also given a positive integer X. Your task is to count the number of elements in the table that contain the integer X.<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>OccurenceOfX()</b> that takes the integer N and the integer X as parameter.
Constraints:-
1 <= N <= 10^5
1 <= X <= 10^9Return the count of X.Sample Input:-
5 5
Sample Output:-
2
Explanation:-
table :-
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
Count of occurrence of X is :- 2
Sample Input:-
10 13
Sample Output:-
0, I have written this Solution Code:
int OccurenceOfX(int N,long X){
int cnt=0,i;
for( i=1;i<=N;i++){
if(X%i==0 && X/i<=N){cnt++;}}
return cnt;
}
int main()
{
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given number N, Consider a table of N X N in which elements at the intersection of ith row and jth column are the product of i and j, i. e i x j.
Also given a positive integer X. Your task is to count the number of elements in the table that contain the integer X.<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>OccurenceOfX()</b> that takes the integer N and the integer X as parameter.
Constraints:-
1 <= N <= 10^5
1 <= X <= 10^9Return the count of X.Sample Input:-
5 5
Sample Output:-
2
Explanation:-
table :-
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
Count of occurrence of X is :- 2
Sample Input:-
10 13
Sample Output:-
0, I have written this Solution Code: public static int OccurenceOfX(int N,int X){
int cnt=0,i;
for( i=1;i<=N;i++){
if(X%i==0 && X/i<=N){cnt++;}}
return cnt;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*1).<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: def factorial(n):
if(n == 1):
return 1
return n * factorial(n-1)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*1).<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: static int Factorial(int N)
{
if(N==0){
return 1;}
return N*Factorial(N-1);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*1).<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: // n is the input number
function factorial(n) {
// write code here
// do not console.log
// return the answer as a number
if (n == 1 ) return 1;
return n * factorial(n-1)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<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>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code: static int numberOfDiagonal(int N){
if(N<=3){return 0;}
return (N*(N-3))/2;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<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>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code:
int numberOfDiagonals(int n){
if(n<=3){return 0;}
return (n*(n-3))/2;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<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>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code: def numberOfDiagonals(n):
if n <=3:
return 0
return (n*(n-3))//2
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<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>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code:
int numberOfDiagonals(int n){
if(n<=3){return 0;}
return (n*(n-3))/2;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K and a queue of integers, your task is to reverse the order of the first K elements of the queue, leaving the other elements in the same relative order.User task:
Since this will be a functional problem, you don't have to take input. You just have to complete the functions:
<b>ReverseK()</b>:- that takes the Queue and the integer K as parameters.
Constraints:
1 ≤ K ≤ N ≤ 10000
1 ≤ elements ≤ 10000
You need to return the modified Queue.Input 1:
5 3
1 2 3 4 5
Output 1:
3 2 1 4 5
Input 2:
5 5
1 2 3 4 5
Output 2:
5 4 3 2 1, I have written this Solution Code: static Queue<Integer> ReverseK(Queue<Integer> queue, int k) {
Stack<Integer> stack = new Stack<Integer>();
// Push the first K elements into a Stack
for (int i = 0; i < k; i++) {
stack.push(queue.peek());
queue.remove();
}
// Enqueue the contents of stack at the back
// of the queue
while (!stack.empty()) {
queue.add(stack.peek());
stack.pop();
}
// Remove the remaining elements and enqueue
// them at the end of the Queue
for (int i = 0; i < queue.size() - k; i++) {
queue.add(queue.peek());
queue.remove();
}
return queue;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Create a new class for a bank account (BankAccount)
Create fields for the balance(balance), customer name(name).
Create constructor with two parameter as balance and name
Create two additional methods
1. To allow the customer to deposit funds named depositFund() with one parameter as fund to be added (this should increment the balance field).
2. To allow the customer to withdraw funds named withdrawFund() with one parameter as fund to be withdrawn from account and should return boolean as if withdraw went successful else false. This should deduct from the balance field, but not allow
the withdrawal to complete if their are insufficient funds.
Note: Each methods and variable should be publicYou don't have to take any input, You only have to write class <b>BankAccount</b>.Output will be printed by tester, "Correct" if your code is perfectly fine otherwise "Wrong".Sample Input:
class BankAccount{
// if your code works fine for the testert
}
Sample Output:
Correct, I have written this Solution Code: class BankAccount{
public int balance;
public String name;
BankAccount(int _balance,String _name){
balance=_balance;
name =_name;
}
public void depositFund(int fund){
balance += fund;
}
public boolean withdrawFund(int fund){
if(fund <= balance){
balance -= fund;
return true;
}
return false;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Create a new class for a bank account (BankAccount)
Create fields for the balance(balance), customer name(name).
Create constructor with two parameter as balance and name
Create two additional methods
1. To allow the customer to deposit funds named depositFund() with one parameter as fund to be added (this should increment the balance field).
2. To allow the customer to withdraw funds named withdrawFund() with one parameter as fund to be withdrawn from account and should return boolean as if withdraw went successful else false. This should deduct from the balance field, but not allow
the withdrawal to complete if their are insufficient funds.
Note: Each methods and variable should be publicYou don't have to take any input, You only have to write class <b>BankAccount</b>.Output will be printed by tester, "Correct" if your code is perfectly fine otherwise "Wrong".Sample Input:
class BankAccount{
// if your code works fine for the testert
}
Sample Output:
Correct, I have written this Solution Code: class Bank_Account:
def __init__(self,b,n):
self.balance=b
self.name=n
def depositFund(self,b):
self.balance += b
def withdrawFund(self,f):
if self.balance>=f:
self.balance-=f
return True
else:
return False
, 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: 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: Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie.
Each child i has a greed factor greed[i], which is the minimum size of a cookie that the child will be content with; and each cookie j has a size sz[j]. If sz[j] >= greed[i], we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.There are two number n(number of childrens) and m(number of cookies) are given in first line of input.
in second line, n space seperated integers, greed of children are given in a row..
in third line, m space seperated integers, sizes of cookies are given in a row..
1 <= greed. length <= 3 * 104
0 <= sz. length <= 3 * 104
1 <= greed[i], sz[j] <= 231 - 1Your goal is to maximize the number of your content children and output the maximum number and print what to report in case of exceptions.Sample Input:
3 2
1 2 3
1 1
Sample Output:
1
Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3.
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1., I have written this Solution Code: [n,m] = [x for x in input().split()]
ch = [int(x) for x in input().split()]
co = [int(x) for x in input().split()]
count = 0
ch = sorted(ch)
for c in ch:
try:
avco = min([x for x in co if x>=c])
co.remove(avco)
count = count+1
except:
continue
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie.
Each child i has a greed factor greed[i], which is the minimum size of a cookie that the child will be content with; and each cookie j has a size sz[j]. If sz[j] >= greed[i], we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.There are two number n(number of childrens) and m(number of cookies) are given in first line of input.
in second line, n space seperated integers, greed of children are given in a row..
in third line, m space seperated integers, sizes of cookies are given in a row..
1 <= greed. length <= 3 * 104
0 <= sz. length <= 3 * 104
1 <= greed[i], sz[j] <= 231 - 1Your goal is to maximize the number of your content children and output the maximum number and print what to report in case of exceptions.Sample Input:
3 2
1 2 3
1 1
Sample Output:
1
Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3.
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1., I have written this Solution Code: import java.lang.reflect.Array;
import java.util.*;
class Main {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int n,m;
int[] v;
try{
n=sc.nextInt();
m=sc.nextInt();
v=new int[n];
for(int i=0;i<n;i++){
int a=sc.nextInt();
v[i]=a;
}
PriorityQueue<Integer> pq=new PriorityQueue<>();
for(int i=0;i<m;i++){
int a=sc.nextInt();
pq.add(a);
}
int ans=0;
for(int i=0;i<n;i++){
int val=v[i];
while(pq.size()>0&&pq.peek()<val){
pq.poll();
}
Iterator it = pq.iterator();
if(it.hasNext()){
ans++;
pq.poll();
}
}
System.out.println(ans);
}
catch(Exception e){
System.out.println(e);
}
}
}
, 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 return the sum of all of its divisors.<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>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code: public static long SumOfDivisors(long N){
long sum=0;
long c=(long)Math.sqrt(N);
for(long i=1;i<=c;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){sum+=N/i;}
}
}
return sum;
}
, 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 return the sum of all of its divisors.<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>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
long long SumOfDivisors(long long N){
long long sum=0;
long sq=sqrt(N);
for(long i=1;i<=sq;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){
sum+=N/i;
}
}
}
return sum;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<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>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
long long SumOfDivisors(long long N){
long long sum=0;
long sq=sqrt(N);
for(long i=1;i<=sq;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){
sum+=N/i;
}
}
}
return sum;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<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>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
def SumOfDivisors(num) :
# Final result of summation of divisors
result = 0
# find all divisors which divides 'num'
i = 1
while i<= (math.sqrt(num)) :
# if 'i' is divisor of 'num'
if (num % i == 0) :
# if both divisors are same then
# add it only once else add both
if (i == (num / i)) :
result = result + i;
else :
result = result + (i + num/i);
i = i + 1
# Add 1 to the result as 1 is also
# a divisor
return (result); , 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: 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: Numbers are awesome, larger numbers are more awesome!
Given an array A of size N, you need to find the maximum sum that can be obtained from the elements of the array (the selected elements need not be contiguous). You may even decide to take no element to get a sum of 0.The first line of the input contains the size of the array.
The next line contains N (white-space separated) integers denoting the elements of the array.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>4</sup>
-10<sup>7</sup> ≤ A[i] ≤10<sup>7</sup>For each test case, output one integer representing the maximum value of the sum that can be obtained using the various elements of the array.Input 1:
5
1 2 1 -1 1
output 1:
5
input 2:
5
0 0 -1 0 0
output 2:
0
<b>Explanation 1:</b>
In order to maximize the sum among [ 1 2 1 -1 1], we need to only consider [ 1 2 1 1] and neglect the [-1]., 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 n; cin >> n;
int sum = 0;
for(int i = 1; i <= n; i++){
int p; cin >> p;
if(p > 0)
sum += p;
}
cout << sum;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Numbers are awesome, larger numbers are more awesome!
Given an array A of size N, you need to find the maximum sum that can be obtained from the elements of the array (the selected elements need not be contiguous). You may even decide to take no element to get a sum of 0.The first line of the input contains the size of the array.
The next line contains N (white-space separated) integers denoting the elements of the array.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>4</sup>
-10<sup>7</sup> ≤ A[i] ≤10<sup>7</sup>For each test case, output one integer representing the maximum value of the sum that can be obtained using the various elements of the array.Input 1:
5
1 2 1 -1 1
output 1:
5
input 2:
5
0 0 -1 0 0
output 2:
0
<b>Explanation 1:</b>
In order to maximize the sum among [ 1 2 1 -1 1], we need to only consider [ 1 2 1 1] and neglect the [-1]., I have written this Solution Code: n=int(input())
li = list(map(int,input().strip().split()))
sum=0
for i in li:
if i>0:
sum+=i
print(sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Numbers are awesome, larger numbers are more awesome!
Given an array A of size N, you need to find the maximum sum that can be obtained from the elements of the array (the selected elements need not be contiguous). You may even decide to take no element to get a sum of 0.The first line of the input contains the size of the array.
The next line contains N (white-space separated) integers denoting the elements of the array.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>4</sup>
-10<sup>7</sup> ≤ A[i] ≤10<sup>7</sup>For each test case, output one integer representing the maximum value of the sum that can be obtained using the various elements of the array.Input 1:
5
1 2 1 -1 1
output 1:
5
input 2:
5
0 0 -1 0 0
output 2:
0
<b>Explanation 1:</b>
In order to maximize the sum among [ 1 2 1 -1 1], we need to only consider [ 1 2 1 1] and neglect the [-1]., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
int n = Integer.parseInt(br.readLine());
String str[] = br.readLine().split(" ");
long arr[] = new long[n];
long sum=0;
for(int i=0;i<n;i++){
arr[i] = Integer.parseInt(str[i]);
if(arr[i]>0){
sum+=arr[i];
}
}
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a double linked list consisting of N nodes, your task is to reverse the linked list and return the head of the modified list.<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>Reverse()</b> that takes head node of the linked list as a parameter.
Constraints:
1 <= N <= 10^3
1<=value<=100Return the head of the modified linked list.Input:
6
1 2 3 4 5 6
Output:
6 5 4 3 2 1
Explanation:
After reversing the list, elements are as 6 <-> 5 <-> 4 <-> 3 <-> 2 <-> 1., I have written this Solution Code: public static Node Reverse(Node head) {
Node temp = null;
Node current = head;
while (current != null) {
temp = current.prev;
current.prev = current.next;
current.next = temp;
current = current.prev;
}
if (temp != null) {
head = temp.prev;
}
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer Q representing the number of queries and an integer K. There are two types of queries:
(i) 1 x : Add the number x to the stream
(ii) 2 : Print the sum of last K numbers of the stream. If there are less than K numbers in the stream, print the sum of current stream.
Process all the queries.First line contains two integers Q and K.
Next Q lines contains the queries.
Constraints
1 <= Q <= 10^5
1 <= x <= 10^5
1 <= K <= Q
There is atleast one query of 2nd type. For each query of type 2, print the sum of last K numbers of the stream on a new line.Sample Input 1:
5 2
1 4
2
1 1
1 3
2
Output
4
4
Explanation:
Initial Stream = {}
Add 4. Stream = {4}
Sum of last two elements = 4
Add 1. Stream = {4, 1}
Add 3. Stream = {4, 1, 3}
Sum of last two elements = 4
Sample Input 2:
3 1
1 1
2
2
Output
1
1
Explanation
Initial Stream = {}
Add 1. Stream = {1}
Sum of last element = 1
Sum of last element = 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));
int n,k;
String line = br.readLine();
String[] strs = line.trim().split("\\s+");
n=Integer.parseInt(strs[0]);
k=Integer.parseInt(strs[1]);
Long sum=0L;
Queue<Integer>x=new LinkedList<Integer>();
while(n>0){
n--;
line = br.readLine();
strs = line.trim().split("\\s+");
int y=Integer.parseInt(strs[0]);
if(y==1){
y=Integer.parseInt(strs[1]);
x.add(y);
sum += y;
if (x.size() > k) {
sum -= x.peek();
x.remove();
}
}
else{
System.out.print(sum+"\n");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer Q representing the number of queries and an integer K. There are two types of queries:
(i) 1 x : Add the number x to the stream
(ii) 2 : Print the sum of last K numbers of the stream. If there are less than K numbers in the stream, print the sum of current stream.
Process all the queries.First line contains two integers Q and K.
Next Q lines contains the queries.
Constraints
1 <= Q <= 10^5
1 <= x <= 10^5
1 <= K <= Q
There is atleast one query of 2nd type. For each query of type 2, print the sum of last K numbers of the stream on a new line.Sample Input 1:
5 2
1 4
2
1 1
1 3
2
Output
4
4
Explanation:
Initial Stream = {}
Add 4. Stream = {4}
Sum of last two elements = 4
Add 1. Stream = {4, 1}
Add 3. Stream = {4, 1, 3}
Sum of last two elements = 4
Sample Input 2:
3 1
1 1
2
2
Output
1
1
Explanation
Initial Stream = {}
Add 1. Stream = {1}
Sum of last element = 1
Sum of last element = 1, I have written this Solution Code: /**
* author: tourist1256
* created: 2022-06-14 14:26:47
**/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t random_long(long long l = LLONG_MIN, long long r = LLONG_MAX) {
uniform_int_distribution<int64_t> generator(l, r);
return generator(rng);
}
int32_t main() {
int Q, K;
cin >> Q >> K;
deque<int> st;
int sum = 0;
while (Q--) {
int x;
cin >> x;
if (x == 1) {
int y;
cin >> y;
if (st.size() == K) {
sum -= st.back();
st.pop_back();
st.push_front(y);
sum += y;
} else {
st.push_front(y);
sum += y;
}
} else {
cout << sum << "\n";
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N places, numbered 1,2,…,N. For each i (1≤i≤N), the height of place i is hi.
There is a man who is initially on place 1. He will repeat the following action some number of times to reach place N:
If the man is currently at place i, he will go to place i+1 or place i+2 i.e. maximum 2 jumps allowed. Here, a cost of |hi−hj| is incurred, where j is the place where the man goes. Find the minimum possible total cost incurred before the man reaches place N.First line contains N size of array, second line contains elements of array.
Constraints
2 ≤ N ≤ 10^5
1 ≤ hi ≤ 10^4Print the minimum possible total cost incurred.Sample Input:
4
10 30 40 20
Sample Output:
30
Explanation:
Testcase 1: If we follow the path 1→ 2→ 4, the total cost incurred would be |10−30|+|30−20|=30., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static long minCostIncurred(int arr[],int n){
long costTillNow[]=new long[n];
costTillNow[0]=0;
costTillNow[1]=Math.abs(arr[1]-arr[0]);
for(int i=2;i<n;i++){
costTillNow[i]=Math.min(costTillNow[i-1]+(long)Math.abs(arr[i]-arr[i-1]),costTillNow[i-2]+(long)Math.abs(arr[i]-arr[i-2]));
}
return costTillNow[n-1];
}
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 arr[]=new int[n];
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(str[i]);
}
System.out.println(minCostIncurred(arr,n));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N places, numbered 1,2,…,N. For each i (1≤i≤N), the height of place i is hi.
There is a man who is initially on place 1. He will repeat the following action some number of times to reach place N:
If the man is currently at place i, he will go to place i+1 or place i+2 i.e. maximum 2 jumps allowed. Here, a cost of |hi−hj| is incurred, where j is the place where the man goes. Find the minimum possible total cost incurred before the man reaches place N.First line contains N size of array, second line contains elements of array.
Constraints
2 ≤ N ≤ 10^5
1 ≤ hi ≤ 10^4Print the minimum possible total cost incurred.Sample Input:
4
10 30 40 20
Sample Output:
30
Explanation:
Testcase 1: If we follow the path 1→ 2→ 4, the total cost incurred would be |10−30|+|30−20|=30., I have written this Solution Code:
// author-Shivam gupta
#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 MOD 1000000007
const int N = 3e5+5;
#define read(type) readInt<type>()
#define max1 100001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
using vl = vector<ll>;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
int a[max1];
bool b[max1];
int dp[max1];
int n,k;
ll solve(int i){
if(i+k>=n-1){return abs(a[i]-a[n-1]);}
else{
ll ans=LONG_MAX;
if(dp[i]==-1){
for(int j=i+1;j<=i+k;j++){
ans=min(solve(j)+abs(a[j]-a[i]),ans);
}
dp[i]=ans;
}
return dp[i];
}
}
int main()
{
cin>>n;
k=2;
FOR(i,n){b[i]=false;dp[i]=-1;cin>>a[i];}
cout<<solve(0);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. The task is to generate and print all binary numbers with decimal values from 1 to N.
Try solving this using a queue.The only line of input contains a single integer N.
1 <= N <= 50000Print all binary numbers with decimal values from 1 to N in a single line.Sample Input:
5
Sample Output:
1 10 11 100 101, I have written this Solution Code: def DecimalToBinary(num):
return "{0:b}".format(int(num))
n = int(input())
for i in range(1,n+1):
print(DecimalToBinary(i),end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. The task is to generate and print all binary numbers with decimal values from 1 to N.
Try solving this using a queue.The only line of input contains a single integer N.
1 <= N <= 50000Print all binary numbers with decimal values from 1 to N in a single line.Sample Input:
5
Sample Output:
1 10 11 100 101, I have written this Solution Code: import java.util.*;
import java.math.*;
import java.io.*;
class Main{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
StringBuilder s=new StringBuilder("");
for(int i=1;i<=n;i++){
s.append(Integer.toBinaryString(i)+" ");
}
System.out.print(s);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. The task is to generate and print all binary numbers with decimal values from 1 to N.
Try solving this using a queue.The only line of input contains a single integer N.
1 <= N <= 50000Print all binary numbers with decimal values from 1 to N in a single line.Sample Input:
5
Sample Output:
1 10 11 100 101, I have written this Solution Code: #include "bits/stdc++.h"
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 = 500 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
void solve(){
int n; cin >> n;
queue<string> q;
q.push("1");
while(n--){
string x = q.front();
q.pop();
cout << x << " ";
q.push(x + "0");
q.push(x + "1");
}
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , 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 |
For this Question: Given an array of N strings, find the longest common prefix among all strings present in the array.The fist line of each test contains a single integer N. Next line has space-separated N strings.
Constraints:-
1 <= N <= 10^3
1 <= |S| <= 10^3
Print the longest common prefix as a string in the given array. If no such prefix exists print "-1"(without quotes).Sample Input:
4
geeksforgeeks geeks geek geezer
Sample Output:
gee
Sampel Input:-
3
a b c
Sample Output:-
-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));
int n = Integer.parseInt(br.readLine());
String[] str = br.readLine().split(" ");
String ref = str[0];
for(String s: str){
if(s.length()<ref.length()){
ref = s;
}
}
for(int i=0; i<n; i++){
if(str[i].contains(ref) == false){
ref = ref.substring(0, ref.length()-1);
}
}
if(ref.length()>0)
System.out.println(ref);
else
System.out.println(-1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N strings, find the longest common prefix among all strings present in the array.The fist line of each test contains a single integer N. Next line has space-separated N strings.
Constraints:-
1 <= N <= 10^3
1 <= |S| <= 10^3
Print the longest common prefix as a string in the given array. If no such prefix exists print "-1"(without quotes).Sample Input:
4
geeksforgeeks geeks geek geezer
Sample Output:
gee
Sampel Input:-
3
a b c
Sample Output:-
-1, I have written this Solution Code: def longestPrefix( strings ):
size = len(strings)
if (size == 0):
return -1
if (size == 1):
return strings[0]
strings.sort()
end = min(len(strings[0]), len(strings[size - 1]))
i = 0
while (i < end and
strings[0][i] == strings[size - 1][i]):
i += 1
pre = strings[0][0: i]
if len(pre) > 1:
return pre
else:
return -1
N=int(input())
strings=input().split()
print( longestPrefix(strings) ), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N strings, find the longest common prefix among all strings present in the array.The fist line of each test contains a single integer N. Next line has space-separated N strings.
Constraints:-
1 <= N <= 10^3
1 <= |S| <= 10^3
Print the longest common prefix as a string in the given array. If no such prefix exists print "-1"(without quotes).Sample Input:
4
geeksforgeeks geeks geek geezer
Sample Output:
gee
Sampel Input:-
3
a b c
Sample Output:-
-1, I have written this Solution Code: function commonPrefixUtil(str1,str2)
{
let result = "";
let n1 = str1.length, n2 = str2.length;
// Compare str1 and str2
for (let i = 0, j = 0; i <= n1 - 1 && j <= n2 - 1; i++, j++) {
if (str1[i] != str2[j]) {
break;
}
result += str1[i];
}
return (result);
}
// n is number of individual space seperated strings inside strings variable,
// strings is the string which contains space seperated words.
function longestCommonPrefix(strings,n){
// write code here
// do not console,log answer
// return the answer as string
if(n===1) return strings;
const arr= strings.split(" ")
let prefix = arr[0];
for (let i = 1; i <= n - 1; i++) {
prefix = commonPrefixUtil(prefix, arr[i]);
}
if(!prefix) return -1;
return (prefix);
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N strings, find the longest common prefix among all strings present in the array.The fist line of each test contains a single integer N. Next line has space-separated N strings.
Constraints:-
1 <= N <= 10^3
1 <= |S| <= 10^3
Print the longest common prefix as a string in the given array. If no such prefix exists print "-1"(without quotes).Sample Input:
4
geeksforgeeks geeks geek geezer
Sample Output:
gee
Sampel Input:-
3
a b c
Sample Output:-
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
string a[n];
int x=INT_MAX;
for(int i=0;i<n;i++){
cin>>a[i];
x=min(x,(int)a[i].length());
}
string ans="";
for(int i=0;i<x;i++){
for(int j=1;j<n;j++){
if(a[j][i]==a[0][i]){continue;}
goto f;
}
ans+=a[0][i];
}
f:;
if(ans==""){cout<<-1;return 0;}
cout<<(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N and a number K, the task is to find the smallest number M such that the sum of the array becomes lesser than or equal to the number K when every element of that array is divided by the number M.
Note: Each result of the division is rounded to the nearest integer greater than or equal to that element. For example: 10/3 = 4 and 6/2 = 3The first line contains two integers N and K.
Next line contains N integers denoting the elements of arr[]
<b>Constraints:</b>
1 <= N <= 100000
1 <= arr[i] <= 100000
1 <= K <= 100000000Print a single integer the value of smallest number MSample Input:
4 6
2 3 4 9
Sample Output:
4
Explanation:
When every element is divided by 4 - 2/4 + 3/4 + 4/4 + 9/4 = 1 + 1 + 1 + 3 = 6
When every element is divided by 3 - 2/3 + 3/3 + 4/3 + 9/3 = 1 + 1 + 2 + 3 = 7 which is greater than K.
Hence the smallest integer which makes the sum less than or equal to K = 6 is 4., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
String s[] = br.readLine().split(" ");
int n = Integer.parseInt(s[0]);
int k = Integer.parseInt(s[1]);
String str[] = br.readLine().split(" ");
int[] arr = new int[n];
for(int i=0;i<n;i++){
arr[i] = Integer.parseInt(str[i]);
}
System.out.print(minDivisor(arr,n,k));
}
static int minDivisor(int arr[],int N, int limit) {
int low = 0, high = 1000000000;
while (low < high)
{
int mid = (low + high) / 2;
int sum = 0;
for(int i = 0; i < N; i++)
{
sum += Math.ceil((double) arr[i] / (double) mid);
}
if(sum <= limit){
high = mid;
}
else{
low = mid + 1;
}
}
return low;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N and a number K, the task is to find the smallest number M such that the sum of the array becomes lesser than or equal to the number K when every element of that array is divided by the number M.
Note: Each result of the division is rounded to the nearest integer greater than or equal to that element. For example: 10/3 = 4 and 6/2 = 3The first line contains two integers N and K.
Next line contains N integers denoting the elements of arr[]
<b>Constraints:</b>
1 <= N <= 100000
1 <= arr[i] <= 100000
1 <= K <= 100000000Print a single integer the value of smallest number MSample Input:
4 6
2 3 4 9
Sample Output:
4
Explanation:
When every element is divided by 4 - 2/4 + 3/4 + 4/4 + 9/4 = 1 + 1 + 1 + 3 = 6
When every element is divided by 3 - 2/3 + 3/3 + 4/3 + 9/3 = 1 + 1 + 2 + 3 = 7 which is greater than K.
Hence the smallest integer which makes the sum less than or equal to K = 6 is 4., I have written this Solution Code: from math import ceil
def kSum(arr, n, k):
low = 0
high = 10 ** 9
while (low < high):
mid = (low + high) // 2
sum = 0
for i in range(int(n)):
sum += ceil( int(arr[i]) / mid)
if (sum <= int(k)):
high = mid
else:
low = mid + 1
return low
n, k =input().split()
arr = input().split()
print( kSum(arr, n, k) ), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N and a number K, the task is to find the smallest number M such that the sum of the array becomes lesser than or equal to the number K when every element of that array is divided by the number M.
Note: Each result of the division is rounded to the nearest integer greater than or equal to that element. For example: 10/3 = 4 and 6/2 = 3The first line contains two integers N and K.
Next line contains N integers denoting the elements of arr[]
<b>Constraints:</b>
1 <= N <= 100000
1 <= arr[i] <= 100000
1 <= K <= 100000000Print a single integer the value of smallest number MSample Input:
4 6
2 3 4 9
Sample Output:
4
Explanation:
When every element is divided by 4 - 2/4 + 3/4 + 4/4 + 9/4 = 1 + 1 + 1 + 3 = 6
When every element is divided by 3 - 2/3 + 3/3 + 4/3 + 9/3 = 1 + 1 + 2 + 3 = 7 which is greater than K.
Hence the smallest integer which makes the sum less than or equal to K = 6 is 4., I have written this Solution Code: #include "bits/stdc++.h"
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 = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N], n, k;
bool f(int x){
int sum = 0;
for(int i = 1; i <= n; i++)
sum += (a[i]-1)/x + 1;
return (sum <= k);
}
void solve(){
cin >> n >> k;
for(int i = 1; i <= n; i++)
cin >> a[i];
int l = 0, h = N;
while(l+1 < h){
int m = (l + h) >> 1;
if(f(m))
h = m;
else
l = m;
}
cout << h;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*1).<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: def factorial(n):
if(n == 1):
return 1
return n * factorial(n-1)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*1).<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: static int Factorial(int N)
{
if(N==0){
return 1;}
return N*Factorial(N-1);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*1).<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: // n is the input number
function factorial(n) {
// write code here
// do not console.log
// return the answer as a number
if (n == 1 ) return 1;
return n * factorial(n-1)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alongside acting as his personal assistant F. R. I. D. A. Y also manages the Avengers Tower. With all the heroes of the universe arriving at the place it becomes quite messy to cater to the needs of all the guests. Assuming every super hero arrives by a train, how many platforms are needed on the Avengers tower to ensure no train gets delayed. Given arrival and departure times of all trains that reach Avengers Tower. Consider that all the trains arrive on the same day and leave on the same day. Arrival and departure time can never be the same for a train but we can have the arrival time of one train equal to the departure time of the other. At any given instance of time, the same platform cannot be used for both departure of a train and arrival of another train. In such cases, we need different platforms.The first line of input has the number of visitors.
It is followed by the entry of two books in the form of two arrays.
It is not mentioned which Array contains entry time and which has exit time but it is sure all entry time is in one array and all exit time is in other array.
eg: 9:00 is written as 900.
if ab:cd is the time then it will be written as ab*100 + cd.
Constraints:-
1 ≤ n ≤ 10000The output contains a single Line the minimum number of platforms needed.Sample Input:-
6
900 940 950 1100 1500 1800
910 1200 1120 1130 1900 2000
Sample Output:-
3
Sample Input:-
3
0900 1235 1100
1000 1200 1240
Sample Output:-
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int differentPlatforms(int arrivalTime[], int departureTime[], int visitors) {
Arrays.sort(arrivalTime);
Arrays.sort(departureTime);
int platformNeeded = 1, result = 1;
int i = 1, j = 0;
while (i < visitors && j < visitors) {
if (arrivalTime[i] <= departureTime[j]) {
platformNeeded++;
i++;
}
else if (arrivalTime[i] > departureTime[j]) {
platformNeeded--;
j++;
}
if (platformNeeded > result)
result = platformNeeded;
}
return result;
}
public static void main (String[] args) {
Scanner input = new Scanner(System.in);
int visitors = input.nextInt();
int[] arrivalTime = new int[visitors];
int[] departureTime = new int[visitors];
for(int i = 0; i < visitors; i++) {
arrivalTime[i] = input.nextInt();
}
for(int i = 0; i < visitors; i++) {
departureTime[i] = input.nextInt();
}
if(arrivalTime[0]<departureTime[0]){
System.out.print(differentPlatforms(arrivalTime, departureTime, visitors));
}else{
System.out.print(differentPlatforms(departureTime, arrivalTime, visitors));
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alongside acting as his personal assistant F. R. I. D. A. Y also manages the Avengers Tower. With all the heroes of the universe arriving at the place it becomes quite messy to cater to the needs of all the guests. Assuming every super hero arrives by a train, how many platforms are needed on the Avengers tower to ensure no train gets delayed. Given arrival and departure times of all trains that reach Avengers Tower. Consider that all the trains arrive on the same day and leave on the same day. Arrival and departure time can never be the same for a train but we can have the arrival time of one train equal to the departure time of the other. At any given instance of time, the same platform cannot be used for both departure of a train and arrival of another train. In such cases, we need different platforms.The first line of input has the number of visitors.
It is followed by the entry of two books in the form of two arrays.
It is not mentioned which Array contains entry time and which has exit time but it is sure all entry time is in one array and all exit time is in other array.
eg: 9:00 is written as 900.
if ab:cd is the time then it will be written as ab*100 + cd.
Constraints:-
1 ≤ n ≤ 10000The output contains a single Line the minimum number of platforms needed.Sample Input:-
6
900 940 950 1100 1500 1800
910 1200 1120 1130 1900 2000
Sample Output:-
3
Sample Input:-
3
0900 1235 1100
1000 1200 1240
Sample Output:-
1, I have written this Solution Code: #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#define mp make_pair
#define pb push_back
#define pf push_front
#define ss second
#define ff first
#define sz(x) (int)x.size()
#define newl "\n"
#define vi vector<int>
#define pii pair<int, int>
#define vii vector<pii>
#define vl vector<ll>
#define pll pair<ll, ll>
#define vll vector<pll>
#define coutp cout << fixed << setprecision(12)
#define mem(x, val) memset(x, val, sizeof(x))
#define fastio \
ios_base::sync_with_stdio(false); \
cin.tie(NULL); \
cout.tie(NULL);
#define all(v) (v).begin(), (v).end()
const ld pi = 3.14159265359;
ll INF = 1e18 + 10;
ll MOD = 998244353;
ll mod = 1e9 + 9;
inline ll add(ll a, ll b, ll m)
{
if ((a + b) >= m)
return (a + b) % m;
return a + b;
}
inline ll mul(ll a, ll b, ll m)
{
if ((a * b) < m)
return a * b;
return (a * b) % m;
}
ll power(ll x, ll y, ll m)
{
ll res = 1;
x = x % m;
if (x == 0)
return 0;
while (y > 0)
{
if (y & 1)
res = (res * x) % m;
y = y >> 1;
x = (x * x) % m;
}
return res;
}
void solve()
{
int n;
cin>>n;
vi a(n),b(n);
for(int i=0;i<n;i++) {
cin>>a[i];
}
for(int i=0;i<n;i++) {
cin>>b[i];
}
if(a[0]>b[0]) swap(a,b);
vi c(2500,0);
for(int i=0;i<n;i++) {
c[a[i]]++;
c[b[i]+1]--;
}
int ans = c[0];
for(int i=1;i<2500;i++) {
c[i]+=c[i-1];
ans = max(ans,c[i]);
}
cout<<ans;
}
int main()
{
fastio;
int t;
t = 1;
//cin >> t;
//int test=t;
while (t-- > 0)
{
//cout<<"Case #"<<(test-t)<<": ";
solve();
cout << newl;
}
return 0;
}, 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: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 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));
int N = Integer.parseInt(br.readLine());
double arr[] = new double[N];
String str[] = br.readLine().trim().split(" ");
for(int i=0;i<N;i++)
arr[i]=Integer.parseInt(str[i]);
double resistance=0;
int equResistance=0;
for(int i=0;i<N;i++)
arr[i]=Integer.parseInt(str[i]);
for(int i=0;i<N;i++)
{
resistance=resistance+(1/arr[i]);
}
equResistance = (int)Math.floor((1/resistance));
System.out.println(equResistance);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, I have written this Solution Code: r = input("")
r = int(r)
n = input("").split()
resistance=0.0
for i in range(0,r):
resistor = float(n[i])
resistance = resistance + (1/resistor)
print(int(1/resistance)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, 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 n; cin >> n;
double s = 0;
for(int i = 1; i <= n; i++){
double p; cin >> p;
s = s + (1/p);
}
s = 1/s;
cout << floor(s);
return 0;
}, 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 value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: static int equationSum(int n)
{
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: def equationSum(N) :
re=(N*(N+1))//2
re=re*re
re=re+9*((N*(N+1))//2)
re=re+4*N
return re
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: int equationSum(int n){
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}
, 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 value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: int equationSum(int n){
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers N and M, your task is to print the number of primes present between N and M (both included).
<b>Note</b>:- You have already provided a function that will check if the given number is prime or not. To use the given function you need to call <b>check_prime(x)</b> where x is the number you want to check. If the given number is prime the function will return 1 else it returns 0.
<b>Note</b>:- Do not close your main class.The input contains two space- separated integers depicting the values of N and M.
Constraints:-
1 <= N <= M <= 10000Print the count of prime numbers in the given range.Sample Input:-
1 10
Sample Output:-
4
Sample Input:-
8 10
Sample Output:-
0, I have written this Solution Code: from math import sqrt
def isPrime(n):
if (n <= 1):
return False
for i in range(2, int(sqrt(n))+1):
if (n % i == 0):
return False
return True
x=input().split()
n=int(x[0])
m=int(x[1])
count = 0
for i in range(n,m):
if isPrime(i):
count = count +1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers N and M, your task is to print the number of primes present between N and M (both included).
<b>Note</b>:- You have already provided a function that will check if the given number is prime or not. To use the given function you need to call <b>check_prime(x)</b> where x is the number you want to check. If the given number is prime the function will return 1 else it returns 0.
<b>Note</b>:- Do not close your main class.The input contains two space- separated integers depicting the values of N and M.
Constraints:-
1 <= N <= M <= 10000Print the count of prime numbers in the given range.Sample Input:-
1 10
Sample Output:-
4
Sample Input:-
8 10
Sample Output:-
0, 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 m = sc.nextInt();
int cnt=0;
for(int i=n;i<=m;i++){
if(check_prime(i)==1){cnt++;}
}
System.out.println(cnt);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of N positive integers. The task is to find a subsequence with maximum sum such that there should be no adjacent elements from the array in the subsequence.First line of input contains T, the number of test case.
First line of test case contains size of array N. Next line contains N space separated integers representing the elements of the array.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase, print the maximum sum of the subsequenceSample Input:-
2
3
1 2 3
3
1 20 3
Sample Output:-
4
20
Explanation:
Testcase 1: Elements 1 and 3 form a subsequence with maximum sum and no elements in the subsequence are adjacent in the array.
Testcase 2: Element 20 from the array forms a subsequence with maximum sum., I have written this Solution Code: import java.io.*;
import java.util.*;
public 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') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
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 long maxCoins(int coins[]) {
int n = coins.length;
long prev2 = 0;
long prev1 = coins[0];
for (int i = 1; i < n; i++) {
long inc = prev2 + coins[i];
long exc = prev1;
long ans = Math.max(inc, exc);
prev2=prev1;
prev1 = ans;
}
return prev1;
}
public static void main(String[] args) throws IOException {
Reader sc = new Reader();
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int arr[] = new int[n];
for (int i = 0; i < arr.length; i++) {
arr[i] = sc.nextInt();
}
System.out.println(maxCoins(arr));
}
sc.close();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of N positive integers. The task is to find a subsequence with maximum sum such that there should be no adjacent elements from the array in the subsequence.First line of input contains T, the number of test case.
First line of test case contains size of array N. Next line contains N space separated integers representing the elements of the array.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase, print the maximum sum of the subsequenceSample Input:-
2
3
1 2 3
3
1 20 3
Sample Output:-
4
20
Explanation:
Testcase 1: Elements 1 and 3 form a subsequence with maximum sum and no elements in the subsequence are adjacent in the array.
Testcase 2: Element 20 from the array forms a subsequence with maximum sum., I have written this Solution Code:
def rob(nums) -> int:
rob1, rob2 =0, 0
for n in nums:
temp = max(n + rob1, rob2)
rob1 = rob2
rob2 = temp
return rob2
t=int(input())
while t!=0:
t-=1
n=int(input())
arr = list(map(int, input().split()))
print(rob(arr)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of N positive integers. The task is to find a subsequence with maximum sum such that there should be no adjacent elements from the array in the subsequence.First line of input contains T, the number of test case.
First line of test case contains size of array N. Next line contains N space separated integers representing the elements of the array.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase, print the maximum sum of the subsequenceSample Input:-
2
3
1 2 3
3
1 20 3
Sample Output:-
4
20
Explanation:
Testcase 1: Elements 1 and 3 form a subsequence with maximum sum and no elements in the subsequence are adjacent in the array.
Testcase 2: Element 20 from the array forms a subsequence with maximum sum., 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 dp[N][2];
signed main() {
IOS;
int t; cin >> t;
while(t--){
memset(dp, 0, sizeof dp);
int n; cin >> n;
for(int i = 1; i <= n; i++){
int p; cin >> p;
dp[i][1] = p + dp[i-1][0];
dp[i][0] = max(dp[i-1][0], dp[i-1][1]);
}
cout << max(dp[n][0], dp[n][1]) << endl;;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given an array of N numbers. Your task is to first reverse the array (the first number becomes last, 2nd number becomes 2nd from the last, and so on) and then print the sum of the numbers at even indices and print the product of the numbers at odd indices.
<b>Note :</b> 1 based indexing is followed .The first line contains a single integer N: the number of elements
The second line contains N different integers separated by spaces
<b>constraints:-</b>
1 <= N <= 35
1 <= a[i] <= 10Two space separated integers representing sum of the numbers at even places and the product of the numbers at odd places.
Sample Input 1:
6
1 2 3 4 5 6
Sample Output 1:
9 48
Sample Input 2:
3
1 2 3
Sample Output 2:
2 3
<b>Explanation 1:-</b>
After reversing 1 2 3 4 5 6 becomes 6 5 4 3 2 1
Hence the sum of the numbers at even indices: 5+3+1=9
product of the numbers at odd indices: 6*4*2=48
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
long x=1,y=0;
if(n&1){
for(int i=0;i<n;i++){
if(i%2==0){
x*=a[i];
}
else{
y+=a[i];
}
}
}
else{
for(int i=0;i<n;i++){
if(i%2==0){
y+=a[i];
}
else{
x*=a[i];
}
}
}
cout<<y<<" "<<x;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given an array of N numbers. Your task is to first reverse the array (the first number becomes last, 2nd number becomes 2nd from the last, and so on) and then print the sum of the numbers at even indices and print the product of the numbers at odd indices.
<b>Note :</b> 1 based indexing is followed .The first line contains a single integer N: the number of elements
The second line contains N different integers separated by spaces
<b>constraints:-</b>
1 <= N <= 35
1 <= a[i] <= 10Two space separated integers representing sum of the numbers at even places and the product of the numbers at odd places.
Sample Input 1:
6
1 2 3 4 5 6
Sample Output 1:
9 48
Sample Input 2:
3
1 2 3
Sample Output 2:
2 3
<b>Explanation 1:-</b>
After reversing 1 2 3 4 5 6 becomes 6 5 4 3 2 1
Hence the sum of the numbers at even indices: 5+3+1=9
product of the numbers at odd indices: 6*4*2=48
, I have written this Solution Code: n = int(input())
lst = list(map(int, input().strip().split()))[:n]
lst.reverse()
lst_1 = 0
for i in range(1, n, 2):
lst_1 += lst[i]
lst_2 = 1
for i in range(0, n, 2):
lst_2 *= lst[i]
print(lst_1,lst_2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given an array of N numbers. Your task is to first reverse the array (the first number becomes last, 2nd number becomes 2nd from the last, and so on) and then print the sum of the numbers at even indices and print the product of the numbers at odd indices.
<b>Note :</b> 1 based indexing is followed .The first line contains a single integer N: the number of elements
The second line contains N different integers separated by spaces
<b>constraints:-</b>
1 <= N <= 35
1 <= a[i] <= 10Two space separated integers representing sum of the numbers at even places and the product of the numbers at odd places.
Sample Input 1:
6
1 2 3 4 5 6
Sample Output 1:
9 48
Sample Input 2:
3
1 2 3
Sample Output 2:
2 3
<b>Explanation 1:-</b>
After reversing 1 2 3 4 5 6 becomes 6 5 4 3 2 1
Hence the sum of the numbers at even indices: 5+3+1=9
product of the numbers at odd indices: 6*4*2=48
, 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 Arr[] = new int[n];
for(int i=0;i<n;i++){
Arr[i] = sc.nextInt();
}
if(n%2==1){
long sum=0;
long product=1;
for(int i=0;i<n;i++){
if(i%2==0){
product*=Arr[i];
}
else{
sum+=Arr[i];
}
}
System.out.print(sum+" "+product);
}
else{
long sum=0;
long product=1;
for(int i=0;i<n;i++){
if(i%2==1){
product*=Arr[i];
}
else{
sum+=Arr[i];
}
}
System.out.print(sum+" "+product);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of positive integers. The task is to find inversion count of array.
Inversion Count : For an array, inversion count indicates how far (or close) the array is from being sorted. If array is already sorted then inversion count is 0. If array is sorted in reverse order that inversion count is the maximum.
Formally, two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j.
Asked in Adobe, Amazon, Microsoft.The first line of each test case is N, the size of the array. The second line of each test case contains N elements.
Constraints:-
1 ≤ N ≤ 10^5
1 ≤ a[i] ≤ 10^5Print the inversion count of array.Sample Input:
5
2 4 1 3 5
Sample Output:
3
Explanation:
Testcase 1: The sequence 2, 4, 1, 3, 5 has three inversions (2, 1), (4, 1), (4, 3)., 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 input[]=br.readLine().split("\\s");
int a[]=new int[n];
for(int i=0;i<n;i++)
{
a[i]=Integer.parseInt(input[i]);
}
System.out.print(implementMergeSort(a,0,n-1));
}
public static long implementMergeSort(int arr[], int start, int end)
{
long count=0;
if(start<end)
{
int mid=start+(end-start)/2;
count +=implementMergeSort(arr,start,mid);
count +=implementMergeSort(arr,mid+1,end);
count +=merge(arr,start,end,mid);
}
return count;
}
public static long merge(int []a,int start,int end,int mid)
{
int i=start;
int j=mid+1;
int k=0;
int len=end-start+1;
int c[]=new int[len];
long inv_count=0;
while(i<=mid && j<=end)
{
if(a[i]<=a[j])
{
c[k++]=a[i];
i++;
}
else
{
c[k++]=a[j];
j++;
inv_count +=(mid-i)+1;
}
}
while(i<=mid)
{
c[k++]=a[i++];
}
while(j<=end)
{
c[k++]=a[j++];
}
for(int l=0;l<len;l++)
a[start+l]=c[l];
return inv_count;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of positive integers. The task is to find inversion count of array.
Inversion Count : For an array, inversion count indicates how far (or close) the array is from being sorted. If array is already sorted then inversion count is 0. If array is sorted in reverse order that inversion count is the maximum.
Formally, two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j.
Asked in Adobe, Amazon, Microsoft.The first line of each test case is N, the size of the array. The second line of each test case contains N elements.
Constraints:-
1 ≤ N ≤ 10^5
1 ≤ a[i] ≤ 10^5Print the inversion count of array.Sample Input:
5
2 4 1 3 5
Sample Output:
3
Explanation:
Testcase 1: The sequence 2, 4, 1, 3, 5 has three inversions (2, 1), (4, 1), (4, 3)., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
#define int long long
long long _mergeSort(int arr[], int temp[], int left, int right);
long long merge(int arr[], int temp[], int left, int mid, int right);
/* This function sorts the input array and returns the
number of inversions in the array */
long long mergeSort(int arr[], int array_size)
{
int temp[array_size];
return _mergeSort(arr, temp, 0, array_size - 1);
}
/* An auxiliary recursive function that sorts the input array and
returns the number of inversions in the array. */
long long _mergeSort(int arr[], int temp[], int left, int right)
{
int mid, inv_count = 0;
if (right > left) {
/* Divide the array into two parts and
call _mergeSortAndCountInv()
for each of the parts */
mid = (right + left) / 2;
/* Inversion count will be sum of
inversions in left-part, right-part
and number of inversions in merging */
inv_count += _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid + 1, right);
/*Merge the two parts*/
inv_count += merge(arr, temp, left, mid + 1, right);
}
return inv_count;
}
/* This funt merges two sorted arrays
and returns inversion count in the arrays.*/
long long merge(int arr[], int temp[], int left,
int mid, int right)
{
int i, j, k;
long long inv_count = 0;
i = left; /* i is index for left subarray*/
j = mid; /* j is index for right subarray*/
k = left; /* k is index for resultant merged subarray*/
while ((i <= mid - 1) && (j <= right)) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
}
else {
temp[k++] = arr[j++];
/* this is tricky -- see above
explanation/diagram for merge()*/
inv_count = inv_count + (mid - i);
}
}
/* Copy the remaining elements of left subarray
(if there are any) to temp*/
while (i <= mid - 1)
temp[k++] = arr[i++];
/* Copy the remaining elements of right subarray
(if there are any) to temp*/
while (j <= right)
temp[k++] = arr[j++];
/*Copy back the merged elements to original array*/
for (i = left; i <= right; i++)
arr[i] = temp[i];
return inv_count;
}
signed main()
{
int n;
cin>>n;
int a[n];
unordered_map<int,int> m;
for(int i=0;i<n;i++){
cin>>a[i];
if(m.find(a[i])==m.end()){
m[a[i]]=i;
}
}
cout<<mergeSort(a,n);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a chessboard of size N x N, where the top left square is black. Each square contains a value. Find the sum of the values of all black squares and all white squares.
Remember that in a chessboard black and white squares are alternate.The first line of input will be the N size of the matrix. Then next N lines will consist of elements of the matrix. Each row will contain N elements since it is a square matrix.
<b>Constraints:-</b>
1 ≤ N ≤ 800
1 ≤ Matrix[i][j] ≤ 100000
Print two lines, the first line containing the sum of black squares and the second line containing the sum of white squares.Input 1:
3
1 2 3
4 5 6
7 8 9
Output 1:
25
20
Sample Input 2:
4
1 2 3 4
6 8 9 10
11 12 13 14
15 16 17 18
Sample Output 2:
80
79
<b>Explanation 1</b>
The black square contains 1, 3, 5, 7, 9; sum = 25
The white square contains 2, 4, 6, 8; sum = 20, I have written this Solution Code: n=int(input())
bSum=0
wSum=0
for i in range(n):
c=(input().split())
for j in range(len(c)):
if(i%2==0):
if(j%2==0):
bSum+=int(c[j])
else:
wSum+=int(c[j])
else:
if(j%2==0):
wSum+=int(c[j])
else:
bSum+=int(c[j])
print(bSum)
print(wSum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a chessboard of size N x N, where the top left square is black. Each square contains a value. Find the sum of the values of all black squares and all white squares.
Remember that in a chessboard black and white squares are alternate.The first line of input will be the N size of the matrix. Then next N lines will consist of elements of the matrix. Each row will contain N elements since it is a square matrix.
<b>Constraints:-</b>
1 ≤ N ≤ 800
1 ≤ Matrix[i][j] ≤ 100000
Print two lines, the first line containing the sum of black squares and the second line containing the sum of white squares.Input 1:
3
1 2 3
4 5 6
7 8 9
Output 1:
25
20
Sample Input 2:
4
1 2 3 4
6 8 9 10
11 12 13 14
15 16 17 18
Sample Output 2:
80
79
<b>Explanation 1</b>
The black square contains 1, 3, 5, 7, 9; sum = 25
The white square contains 2, 4, 6, 8; sum = 20, 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 {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int mat[][] = new int[N][N];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
mat[i][j] = sc.nextInt();
}
alternate_Matrix_Sum(mat,N);
}
static void alternate_Matrix_Sum(int mat[][], int N)
{
long sum =0, sum1 = 0;
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
if((i+j)%2 == 0)
sum += mat[i][j];
else sum1 += mat[i][j];
}
}
System.out.println(sum);
System.out.print(sum1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: static int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: def mythOrFact(N):
prime=1
cnt=N+1
for i in range (2,N):
if N%i==0:
prime=0
cnt=cnt+i
x = 3*N
x=x//2
if(cnt <= x and prime==1) or (cnt>x and prime==0):
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Take input from standard input while you do not get 0 as an input. Print all the inputs separated by space. It is guaranteed that the number of integers are less than 100000.The input will contain a series of integers in one line each. Input should be taken while you have not get a 0 as an input.
0 <= input <= 10Print the input integers seperated by space.Sample Input
6
5
5
0
Sample Output
6 5 5 0
Sample Input
9
3
5
7
6
9
8
3
2
7
7
3
5
0
Sample Output
9 3 5 7 6 9 8 3 2 7 7 3 5 0, I have written this Solution Code: n=1
index=0
li=[]
while n!=0:
n=int(input())
li.append(n)
index=index+1
#li = list(map(int,input().strip().split()))
for i in range(0,len(li)-1):
print(li[i],end=" ")
print(li[len(li)-1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
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