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For this Question: Gruumm has set up N lasers with i<sup>th</sup> laser having attack power Attack[i], all pointing earth. The damage caused by lasers on earth is the XOR of the attack powers of all the lasers. Cruger wants to set up a shield of defensive strength K. As Cruger has budget constraints, K can only be a non-negative integer with value strictly less than the strength of the weakest laser (minimum value of Attack array) set by Gruumm. Due to the defense shield, the attack power of each laser reduces by K. Help Kruger find the number of values of K such that the damage caused by lasers on earth is 0.First line of input contains a single integer, N.
Second line of input contains N integers, denoting the array Attack.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= Attack[i] <= 10<sup>18</sup>Print a single integer denoting the number of values of K such that the damage on earth is 0.Sample Input 1
3
3 4 5
Sample Output 1
1
Explanation: Only valid value of K is 2. With K = 2, attack powers become [1, 2, 3] and (1 xor 2 xor 3) = 0
Sample Input 2
2
4 4
Sample Output 2
4
Explanation: valid values of K are [0, 1, 2, 3], I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
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>
/////////////
const int MAXN = 100000;
const int MAXBIT = 62;
int n;
long long a[MAXN];
int cnt_with_bit[MAXBIT], ones_pref[MAXBIT][MAXN+1];
long long d[MAXBIT][MAXN+1];
long long get_d (int bit, int greater) {
if (bit == MAXBIT)
return greater ? 0 : 1;
long long & my = d[bit][greater];
if (my != -1) return my;
my = 0;
for (int x_bit=0; x_bit<2; ++x_bit) {
int ngreater = 0, cur_xor = 0;
for (int tp=0; tp<2; ++tp)
for (int a_bit=0; a_bit<2; ++a_bit) {
int cnt = ones_pref[bit][greater],
total = tp==0 ? greater : n-greater;
if (tp)
cnt = cnt_with_bit[bit] - cnt;
if (!a_bit)
cnt = total - cnt;
if (x_bit > a_bit || x_bit == a_bit && tp == 0)
ngreater += cnt;
int res_bit = (a_bit - x_bit - (1-tp)) & 1;
if (res_bit)
cur_xor ^= cnt & 1;
}
if (!cur_xor)
my += get_d (bit+1, ngreater);
}
return my;
}
void calc_tables() {
for (int bit=0; bit<MAXBIT; ++bit) {
cnt_with_bit[bit] = 0;
for (int i=0; i<n; ++i)
if (a[i] & (1ll << bit))
++cnt_with_bit[bit];
}
vector<int> order (n);
for (int i=0; i<n; ++i)
order[i] = i;
for (int bit=0; bit<MAXBIT; ++bit) {
vector<int> v[2];
for (int i=0; i<n; ++i) {
ones_pref[bit][i] = (int) v[1].size();
v[(a[order[i]] >> bit) & 1].push_back (order[i]);
}
ones_pref[bit][n] = (int) v[1].size();
order = v[0];
order.insert (order.end(), v[1].begin(), v[1].end());
}
}
signed main() {
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
cin >> n;
for (int i=0; i<n; ++i){
cin >> a[i];
}
calc_tables();
memset (d, -1, sizeof d);
long long ans = get_d (0, 0);
long long mi = * min_element (a, a+n);
long long x = 0;
for (int i=0; i<n; ++i)
x ^= a[i] - mi;
if (!x)
--ans;
cout << ans << endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length, breadth, and height of a cuboid. Your task is to calculate its Perimeter.
Note:- Formula for the perimeter of the cuboid is 4(Length + Breadth + height)<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>Perimeter()</b> that takes integers L, B, and H as parameters.
Constraints:-
1 <= L, B, H <= 100Return the length of the Cuboid.Sample Input:-
L = 3, B = 5, H = 1
Sample Output:-
36
Sample Input:-
L = 1, B = 1, H = 1
Sample Output:-
12, I have written this Solution Code:
L,B,H=input().split()
a=4*(int(L)+int(B)+int(H))
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length, breadth, and height of a cuboid. Your task is to calculate its Perimeter.
Note:- Formula for the perimeter of the cuboid is 4(Length + Breadth + height)<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>Perimeter()</b> that takes integers L, B, and H as parameters.
Constraints:-
1 <= L, B, H <= 100Return the length of the Cuboid.Sample Input:-
L = 3, B = 5, H = 1
Sample Output:-
36
Sample Input:-
L = 1, B = 1, H = 1
Sample Output:-
12, I have written this Solution Code: static int Perimeter(int L, int B, int H){
return 4*(L+B+H);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new type of game in which she can jump in either the left direction or in the right direction. Also after each jump the range of her jump increases by 1 unit. i.e if starts from 1 in the next jump she has to jump 2 units then 3 units and so on.
Given the number of jumps as N, the range of the first jump to be 1. What will be the minimum distance Sara can be at from the starting point.<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>minDistanceCoveredBySara()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 1000Return the minimum distance Sara can be at from the starting point.Sample Input:-
3
Sample Output:-
0
Explanation:-
First jump:- right
Second jump:- right
Third jump:- left
Total distance covered = 1+2-3 = 0
Sample Input:-
5
Sample Output:-
1, I have written this Solution Code: int minDistanceCoveredBySara(int N){
if(N%4==1 || N%4==2){return 1;}
return 0;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new type of game in which she can jump in either the left direction or in the right direction. Also after each jump the range of her jump increases by 1 unit. i.e if starts from 1 in the next jump she has to jump 2 units then 3 units and so on.
Given the number of jumps as N, the range of the first jump to be 1. What will be the minimum distance Sara can be at from the starting point.<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>minDistanceCoveredBySara()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 1000Return the minimum distance Sara can be at from the starting point.Sample Input:-
3
Sample Output:-
0
Explanation:-
First jump:- right
Second jump:- right
Third jump:- left
Total distance covered = 1+2-3 = 0
Sample Input:-
5
Sample Output:-
1, I have written this Solution Code: int minDistanceCoveredBySara(int N){
if(N%4==1 || N%4==2){return 1;}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new type of game in which she can jump in either the left direction or in the right direction. Also after each jump the range of her jump increases by 1 unit. i.e if starts from 1 in the next jump she has to jump 2 units then 3 units and so on.
Given the number of jumps as N, the range of the first jump to be 1. What will be the minimum distance Sara can be at from the starting point.<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>minDistanceCoveredBySara()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 1000Return the minimum distance Sara can be at from the starting point.Sample Input:-
3
Sample Output:-
0
Explanation:-
First jump:- right
Second jump:- right
Third jump:- left
Total distance covered = 1+2-3 = 0
Sample Input:-
5
Sample Output:-
1, I have written this Solution Code: def minDistanceCoveredBySara(N):
if N%4==1 or N%4==2:
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new type of game in which she can jump in either the left direction or in the right direction. Also after each jump the range of her jump increases by 1 unit. i.e if starts from 1 in the next jump she has to jump 2 units then 3 units and so on.
Given the number of jumps as N, the range of the first jump to be 1. What will be the minimum distance Sara can be at from the starting point.<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>minDistanceCoveredBySara()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 1000Return the minimum distance Sara can be at from the starting point.Sample Input:-
3
Sample Output:-
0
Explanation:-
First jump:- right
Second jump:- right
Third jump:- left
Total distance covered = 1+2-3 = 0
Sample Input:-
5
Sample Output:-
1, I have written this Solution Code: static int minDistanceCoveredBySara(int N){
if(N%4==1 || N%4==2){return 1;}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called a bad string, if it follows the two conditions:
1. All odd positions of the string contain lowercase letters.
2. All even positions of the string contain uppercase letters.
Determine whether a string S is bad.The first line contains the string S.
<b>Constraints</b>
S consists of uppercase and lowercase English letters.
The length of S is between 1 and 1000 (inclusive).If S is bad, print Yes; otherwise, print No.Sample Input 1
dIfFiCuLt
Sample Output 1
Yes
The odd- positioned characters are all lowercase and the even- positioned characters are all uppercase, so it is bad.
Sample Input 2
eASY
Sample Output 2
No
The 3- rd character is not lowercase, so it is not bad.
Sample Input 3
a
Sample Output 3
Yes, I have written this Solution Code: import java.util.*;
import java.math.*;
import java.io.*;
public class Main{
static FastReader fr;
static void solve(){
String s = fr.next();
int ans = 0;
for(int i=s.length()-1; i>=0; i--){
if(Character.toLowerCase(s.charAt(i))==s.charAt(i) && i%2!=0){
ans = 1;
break;
}
if(Character.toUpperCase(s.charAt(i))==s.charAt(i) && i%2!=1){
ans = 1;
break;
}
}
System.out.println((ans==0?"Yes":"No"));
}
public static void main(String[] args){
fr = new FastReader();
int t = 1;
for(int i=1; i<=t; i++){
solve();
}
}
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());
}
float nextFloat(){
return Float.parseFloat(next());
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called a bad string, if it follows the two conditions:
1. All odd positions of the string contain lowercase letters.
2. All even positions of the string contain uppercase letters.
Determine whether a string S is bad.The first line contains the string S.
<b>Constraints</b>
S consists of uppercase and lowercase English letters.
The length of S is between 1 and 1000 (inclusive).If S is bad, print Yes; otherwise, print No.Sample Input 1
dIfFiCuLt
Sample Output 1
Yes
The odd- positioned characters are all lowercase and the even- positioned characters are all uppercase, so it is bad.
Sample Input 2
eASY
Sample Output 2
No
The 3- rd character is not lowercase, so it is not bad.
Sample Input 3
a
Sample Output 3
Yes, I have written this Solution Code: s=input()
l=len(s)
s1=""
s2=""
for i in range(1,l):
if i%2==0:
s2+=s[i]
else:
s1+=s[i]
if(s2.isupper() or s2.islower()):
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called a bad string, if it follows the two conditions:
1. All odd positions of the string contain lowercase letters.
2. All even positions of the string contain uppercase letters.
Determine whether a string S is bad.The first line contains the string S.
<b>Constraints</b>
S consists of uppercase and lowercase English letters.
The length of S is between 1 and 1000 (inclusive).If S is bad, print Yes; otherwise, print No.Sample Input 1
dIfFiCuLt
Sample Output 1
Yes
The odd- positioned characters are all lowercase and the even- positioned characters are all uppercase, so it is bad.
Sample Input 2
eASY
Sample Output 2
No
The 3- rd character is not lowercase, so it is not bad.
Sample Input 3
a
Sample Output 3
Yes, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
string s;
cin >> s;
bool flag = 0;
for(int i=0; i<s.size(); i++){
if(i % 2 == 0){
if(s.at(i) < 97){
flag = 1;
break;
}
}else{
if(s.at(i) >= 97){
flag = 1;
break;
}
}
}
if(flag) cout << "No" << endl;
else cout << "Yes" << endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a string. Your task is to compress the consecutive letters of the string
For example, given string is "AAACCCBBD", thus here
A's occurrence 3 times
C's occurrence 3 times
B's occurrence 2 times
D's occurrence 1 time
So after compressing string becomes "A3C3B2D1".The first line of input contains an integer T denoting the number of test cases. Each test case will have a string provided in the new line.
Constraints:
1 <= T <= 10
1 <= sizeof(String) <= 10^6
All characters of String are upper case letters. (A-Z)
Sum of size of Strings over all testcases is <= 10^6For each testcase, in a new line, print the compressed string for each test case in a new line. Input:
2
AAACCCBBD
ABCD
Output:
A3C3B2D1
A1B1C1D1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void compress(String str, int l){
for (int i = 0; i < l; i++) {
int count = 1;
while (i < l - 1 && str.charAt(i) == str.charAt(i + 1)) {
count++;
i++;
}
System.out.print(str.charAt(i));
System.out.print(count);
}
System.out.println();
}
public static void main (String[] args) throws IOException{
BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
int test = Integer.parseInt(rd.readLine());
while(test-->0){
String s = rd.readLine();
int len = s.length();
compress(s,len);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a string. Your task is to compress the consecutive letters of the string
For example, given string is "AAACCCBBD", thus here
A's occurrence 3 times
C's occurrence 3 times
B's occurrence 2 times
D's occurrence 1 time
So after compressing string becomes "A3C3B2D1".The first line of input contains an integer T denoting the number of test cases. Each test case will have a string provided in the new line.
Constraints:
1 <= T <= 10
1 <= sizeof(String) <= 10^6
All characters of String are upper case letters. (A-Z)
Sum of size of Strings over all testcases is <= 10^6For each testcase, in a new line, print the compressed string for each test case in a new line. Input:
2
AAACCCBBD
ABCD
Output:
A3C3B2D1
A1B1C1D1, I have written this Solution Code:
def compress(st):
n = len(st)
i = 0
while i < n:
count = 1
while (i < n-1 and st[i] == st[i + 1]):
count += 1
i += 1
i += 1
print(st[i-1] +str(count),end="")
t=int(input())
for i in range(t):
s=input()
compress(s)
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a string. Your task is to compress the consecutive letters of the string
For example, given string is "AAACCCBBD", thus here
A's occurrence 3 times
C's occurrence 3 times
B's occurrence 2 times
D's occurrence 1 time
So after compressing string becomes "A3C3B2D1".The first line of input contains an integer T denoting the number of test cases. Each test case will have a string provided in the new line.
Constraints:
1 <= T <= 10
1 <= sizeof(String) <= 10^6
All characters of String are upper case letters. (A-Z)
Sum of size of Strings over all testcases is <= 10^6For each testcase, in a new line, print the compressed string for each test case in a new line. Input:
2
AAACCCBBD
ABCD
Output:
A3C3B2D1
A1B1C1D1, 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 = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
void solve(){
string s; cin >> s;
int c = 1;
char p = 0;
int n = s.length();
for(int i = 1; i < n; i++){
if(s[i] != s[i-1]){
cout << s[i-1] << c;
c = 1;
}
else
c++;
}
cout << s[n-1] << c << endl;
}
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, 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 an array of size n consisting of unique integers and an integer k. A function f is defined as follows:
f(i) = Minimum j such that subarray a<sub>i</sub>, a<sub>i+1</sub>, ., a<sub>j</sub> have k elements strictly greater than a<sub>i</sub>
and -1 (if no such j exists)
Your task is to calculate f(i) for all 1 ≤ i ≤ n.The first line contains an integer t (1 <= T <= 10<sup>5</sup>) - the number of test cases.
The first line of each test case contains two integers n and k (1 <= n <= 10<sup>5</sup>, 1 <= k <= n).
The second line of each test case contains n integers a<sub>1</sub>, a<sub>2</sub>, …a<sub>n</sub> (−10<sup>9</sup> <= a<sub>i</sub> <= 10<sup>9</sup>).
It is guaranteed that elements in a given array are unique.
It is guaranteed that the sum of n over all test cases does not exceed 2 * 10<sup>5</sup>.For each test case, print a separate line containing space-separated n integers f(1), f(2),. , f(n).Sample Input 1:
1
3 1
2 1 3
Sample Output 1:
2 2 -1
Explanation:
For index 0, the element at index 2 is the first element greater than 2.
For index 1, the element at index 2 is the first element greater than 1.
For index 2, no such index exists.
, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(),k = sc.nextInt();
FenwickTree tree = new FenwickTree(n+1);
int arr[] = sc.readArray(n);
compress(arr);
ArrayList<Pair> q = new ArrayList<Main.Pair>();
for(int i = 1; i<=n; i++)q.add(new Pair(arr[i-1], i));
Collections.sort(q);
int[] res = new int[n];
for(int i = 1; i<=n; i++) {
Pair curr = q.get(i-1);
int l = curr.y,r = n;
int ans = -1;
while(l<=r) {
int mid = (l+r)/2;
int len = (mid-curr.y+1);
len-=tree.find(curr.y,mid);
if(len > k) {
ans = mid-1;
r = mid-1;
}
else {
l = mid+1;
}
}
res[curr.y-1] = ans;
tree.add(curr.y, 1);
}
for(int e : res)out.print(e+" ");
out.println();
}
static class FenwickTree
{
public int[] tree;
public int size;
public FenwickTree(int size)
{
this.size = size;
tree = new int[size+5];
}
public void add(int i, int v)
{
while(i <= size)
{
tree[i] += v;
i += i&-i;
}
}
public int find(int i)
{
int res = 0;
while(i >= 1)
{
res += tree[i];
i -= i&-i;
}
return res;
}
public int find(int l, int r)
{
return find(r)-find(l-1);
}
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 2_000_00;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of size n consisting of unique integers and an integer k. A function f is defined as follows:
f(i) = Minimum j such that subarray a<sub>i</sub>, a<sub>i+1</sub>, ., a<sub>j</sub> have k elements strictly greater than a<sub>i</sub>
and -1 (if no such j exists)
Your task is to calculate f(i) for all 1 ≤ i ≤ n.The first line contains an integer t (1 <= T <= 10<sup>5</sup>) - the number of test cases.
The first line of each test case contains two integers n and k (1 <= n <= 10<sup>5</sup>, 1 <= k <= n).
The second line of each test case contains n integers a<sub>1</sub>, a<sub>2</sub>, …a<sub>n</sub> (−10<sup>9</sup> <= a<sub>i</sub> <= 10<sup>9</sup>).
It is guaranteed that elements in a given array are unique.
It is guaranteed that the sum of n over all test cases does not exceed 2 * 10<sup>5</sup>.For each test case, print a separate line containing space-separated n integers f(1), f(2),. , f(n).Sample Input 1:
1
3 1
2 1 3
Sample Output 1:
2 2 -1
Explanation:
For index 0, the element at index 2 is the first element greater than 2.
For index 1, the element at index 2 is the first element greater than 1.
For index 2, no such index exists.
, I have written this Solution Code: #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.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>;
#define endl '\n'
#define pb push_back
#define ub upper_bound
#define lb lower_bound
#define fi first
#define se second
#define int long long
typedef long long ll;
typedef long double ld;
#define pii pair<int,int>
#define sz(x) ((ll)x.size())
#define fr(a,b,c) for(int a=b; a<=c; a++)
#define frev(a,b,c) for(int a=c; a>=b; a--)
#define rep(a,b,c) for(int a=b; a<c; a++)
#define trav(a,x) for(auto &a:x)
#define all(con) con.begin(),con.end()
#define done(x) {cout << x << endl;return;}
#define mini(x,y) x = min(x,y)
#define maxi(x,y) x = max(x,y)
const ll infl = 0x3f3f3f3f3f3f3f3fLL;
const int infi = 0x3f3f3f3f;
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
//const int mod = 998244353;
const int mod = 1e9 + 7;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<vector<int>> vvi;
typedef vector<pair<int, int>> vpii;
typedef map<int, int> mii;
typedef set<int> si;
typedef set<pair<int,int>> spii;
typedef queue<int> qi;
uniform_int_distribution<int> rng(0, 1e9);
// DEBUG FUNCTIONS START
void __print(int x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V> void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T> void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void deb() {cerr << "\n";}
template <typename T, typename... V> void deb(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; deb(v...);}
// DEBUG FUNCTIONS END
const int N = 2e5 + 5;
void solve(){
int n, k;
cin >> n >> k;
vi a(n);
rep(i,0,n){
cin >> a[i];
}
vi b = a;
sort(all(b));
vi pos(n);
ordered_set<int> s;
rep(i,0,n){
a[i] = lb(all(b), a[i]) - b.begin();
pos[a[i]] = i;
s.insert(i);
}
vi res(n, -1);
rep(i,0,n){
int position = pos[i];
int idx = s.order_of_key(position);
if(sz(s) > idx + k){
res[position] = *s.find_by_order(idx + k);
}
s.erase(position);
}
rep(i,0,n){
cout << res[i] << " ";
}
cout << endl;
}
signed main(){
ios_base::sync_with_stdio(0), cin.tie(0);
cout << fixed << setprecision(15);
//freopen ("03.txt","r",stdin);
//freopen ("3.txt","w",stdout);
int t = 1;
cin >> t;
while (t--)
solve();
return 0;
}
int powm(int a, int b){
int res = 1;
while (b) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code:
import java.io.*;
import java.util.*;
class Main{
public static void main(String[] args)throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str[]=br.readLine().split(" ");
long l=Long.parseLong(str[0]);
long r=Long.parseLong(str[1]);
long k=Long.parseLong(str[2]);
long count=(r/k)-((l-1)/k);
System.out.print(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code: l,r,k=[int(x)for x in input().split()]
print(r//k - (l-1)//k), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
unsigned long long x,l,r,k;
cin>>l>>r>>k;
x=l/k;
if(l%k==0){x--;}
cout<<r/k-x;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Chinmay went to meet Bunty with 2x2 sweet because Bunty is a sweet child, and he saw Bunty playing with a right-angled isosceles triangle with base B. Chinmay came up with an intriguing notion and challenged Bunty to solve it; if he succeeds, he will be rewarded. Since Chinmay only purchased 'x' amount of sweets, he requests the greatest number of sweets that can fit in the triangle he's playing, with one side of the sweet parallel to the triangle's base.
Note: The base is the triangle's shortest side.The first line includes a space between "X" as the quantity of sweet Chinmay purchased and "B" as the base of the triangle.
<b>Constraints:</b>
1<b>≤</b> B <b>≤</b> 10<sup>4</sup>
1<b>≤</b> X <b>≤</b> 10<sup>9</sup>Output the maximum number of sweet he can able to fit in the triangle.Sample Input:
5 4
Sample Output:
1
Explanation:
In the triangle with base 4, Bunty can able to fit only one sweet in it., I have written this Solution Code: import java.util.*;
class Main{
public static void main(String []args){
Scanner scanner = new Scanner(System.in);
int x = scanner.nextInt();
int b = scanner.nextInt();
int n1 = (b / 2) - 1;
int a = (n1 * (n1 + 1)) / 2;
System.out.println((x >= a?a:x));
}
}, In this Programming Language: Java, 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 |
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:
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=100001;
int a;
for(int i=0;i<n;i++){
a=sc.nextInt();
System.out.print(a+" ");
if(a==0){break;}
}
}
}, In this Programming Language: Java, 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: #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;
while(cin >> n){
cout << n << " ";
if(n == 0) break;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to input three numbers from user and find maximum between three numbers using ternary operatorThe first line of the input contains three numbers a, b and c.
<b>Constraints<b>
1<= a, b, c <= 1e9Print the maximum NumberSample Input
12 14 15
Sample Output
15, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String s[]=br.readLine().split(" ");
int a=Integer.parseInt(s[0]);
int b=Integer.parseInt(s[1]);
int c=Integer.parseInt(s[2]);
int max = (a>b) ? a : (b>c) ? b : c;
System.out.print(max);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to input three numbers from user and find maximum between three numbers using ternary operatorThe first line of the input contains three numbers a, b and c.
<b>Constraints<b>
1<= a, b, c <= 1e9Print the maximum NumberSample Input
12 14 15
Sample Output
15, I have written this Solution Code: a,b,c=map(int,input().split())
print(max(a,b,c)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to input three numbers from user and find maximum between three numbers using ternary operatorThe first line of the input contains three numbers a, b and c.
<b>Constraints<b>
1<= a, b, c <= 1e9Print the maximum NumberSample Input
12 14 15
Sample Output
15, I have written this Solution Code: #include <stdio.h>
int main()
{
int num1, num2, num3, max;
scanf("%d%d%d", &num1, &num2, &num3);
max = (num1 > num2 && num1 > num3) ? num1 :
(num2 > num3) ? num2 : num3;
printf("%d", max);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of N nodes and an integer K, your task is to reverse every alternate K nodes.
In other words , you have to reverse first k nodes , then skip the next k nodes , then reverse next k nodes and so on .
NOTE: if there are not K nodes to reverse then reverse all the nodes left (See example for better understanding)<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>ReverseAlternateK()</b> that takes head node and K as parameter.
Constraints:
1 <=k<=N<=10000Return the head of the modified linked list.Sample Input:-
8 3
1 2 3 4 5 6 7 8
Sample Output:-
3 2 1 4 5 6 8 7
Explanation:
[{1 , 2 ,3 } , {4, 5 , 6} , {7 , 8}]
Reverse 1st segment.
Skip the 2nd segment.
Reverse the 3rd segment. , I have written this Solution Code: public static Node ReverseAlternateK(Node head,int k){
Node current = head;
Node next = null, prev = null;
int count = 0;
while (current != null && count < k) {
next = current.next;
current.next = prev;
prev = current;
current = next;
count++;
}
if (head != null) {
head.next = current;
}
count = 0;
while (count < k - 1 && current != null) {
current = current.next;
count++;
}
if (current != null) {
current.next = ReverseAlternateK(current.next, k);
}
return prev;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N. You need to print the sum of the first N natural numbers.The first line of input contains a single integer T, the next T lines contains a single integer N.
Constraints:
1 < = T < = 100
1 < = N < = 100Print the sum of first N natural numbers for each test case in a new line.Sample Input:
2
5
10
Sample Output:
15
55, 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());
while(t-->0){
int n=Integer.parseInt(br.readLine());
System.out.println(n*(n+1)/2);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N. You need to print the sum of the first N natural numbers.The first line of input contains a single integer T, the next T lines contains a single integer N.
Constraints:
1 < = T < = 100
1 < = N < = 100Print the sum of first N natural numbers for each test case in a new line.Sample Input:
2
5
10
Sample Output:
15
55, I have written this Solution Code: for t in range(int(input())):
n = int(input())
print(n*(n+1)//2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N. You need to print the sum of the first N natural numbers.The first line of input contains a single integer T, the next T lines contains a single integer N.
Constraints:
1 < = T < = 100
1 < = N < = 100Print the sum of first N natural numbers for each test case in a new line.Sample Input:
2
5
10
Sample Output:
15
55, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
cout<<(n*(n+1))/2<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Remove duplicates of an array and return an array of only unique elements.An array containing numbers.Space separated unique elements from the array.Sample Input:-
1 2 3 5 1 5 9 1 2 8
Sample Output:-
1 2 3 5 9 8
<b>Explanation:-</b>
Extra 1, 2, and 5 were removed since they were occurring multiple times.
Note: You only have to remove the extra occurrences i.e. each element in the final array should have a frequency equal to one., I have written this Solution Code: nan, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Remove duplicates of an array and return an array of only unique elements.An array containing numbers.Space separated unique elements from the array.Sample Input:-
1 2 3 5 1 5 9 1 2 8
Sample Output:-
1 2 3 5 9 8
<b>Explanation:-</b>
Extra 1, 2, and 5 were removed since they were occurring multiple times.
Note: You only have to remove the extra occurrences i.e. each element in the final array should have a frequency equal to one., I have written this Solution Code:
inp = eval(input(""))
new_set = []
for i in inp:
if(str(i) not in new_set):
new_set.append(str(i))
print(" ".join(new_set)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has developed a new algorithm to find sprime :
For an integer n , To find all the sprime between 1 and n , she will start from the end n , mark it as sprime, and then mark all its factors (excluding itself) as not sprime. Then she will find the next greatest unmarked number less than the current sprime number , mark it as sprime, and mark all its factors (excluding itself) as not sprime. She will continue this process till all the numbers between 1 and n has been marked either sprime or not sprime .
Your task is to calculate the the number of sprimes that are also prime between 1 and n.The first line contains T the number of test cases.
Each of the next T lines contain an integer n.
Constraint:-
1 <= T <= 100
2 <= n <= 10000000Output T lines, one for each test case, containing the required answer.Sample Input :
3
2
4
7
Sample Output :
1
1
2
Explanation:-
For test 3:- 7 and 5 are the required primes
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void sieve(boolean prime[], int n) {
int i,j;
for(i = 0; i <= n; i++)
prime[i] = true;
for(i = 2; i*i <= n; i++)
if(prime[i])
for(j = i*i; j<=n; j+=i)
prime[j] = false;
}
public static void main (String[] args) throws IOException {
int num = 10000005;
boolean prime[] = new boolean[num+1];
sieve(prime, num);
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(br.readLine().trim());
while(T --> 0) {
int n = Integer.parseInt(br.readLine().trim());
int count = 0;
for(int i=(n/2)+1; i<=n; i++)
if(prime[i])
count++;
System.out.println(count);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has developed a new algorithm to find sprime :
For an integer n , To find all the sprime between 1 and n , she will start from the end n , mark it as sprime, and then mark all its factors (excluding itself) as not sprime. Then she will find the next greatest unmarked number less than the current sprime number , mark it as sprime, and mark all its factors (excluding itself) as not sprime. She will continue this process till all the numbers between 1 and n has been marked either sprime or not sprime .
Your task is to calculate the the number of sprimes that are also prime between 1 and n.The first line contains T the number of test cases.
Each of the next T lines contain an integer n.
Constraint:-
1 <= T <= 100
2 <= n <= 10000000Output T lines, one for each test case, containing the required answer.Sample Input :
3
2
4
7
Sample Output :
1
1
2
Explanation:-
For test 3:- 7 and 5 are the required primes
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
bool a[max1];
long b[max1];
void pre(){
b[0]=0;b[1]=0;
for(int i=0;i<max1;i++){
a[i]=false;
}
long cnt=0;
for(int i=2;i<max1;i++){
if(a[i]==false){
cnt++;
for(int j=i+i;j<=max1;j=j+i){a[j]=true;}
}
b[i]=cnt;
}
}
int main(){
pre();
int t;
cin>>t;
while(t--){
long n;
cin>>n;
cout<<(b[n]-b[(n)/2])<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array and Q queries. Your task is to perform these operations:-
enqueue: this operation will add an element to your current queue.
dequeue: this operation will delete the element from the starting of the queue
displayfront: this operation will print the element presented at the frontUser task:
Since this will be a functional problem, you don't have to take input. You just have to complete the functions:
<b>enqueue()</b>:- that takes the integer to be added and the maximum size of array as parameter.
<b>dequeue()</b>:- that takes the queue as parameter.
<b>displayfront()</b> :- that takes the queue as parameter.
Constraints:
1 <= Q(Number of queries) <= 10<sup>3</sup>
<b> Custom Input:</b>
First line of input should contains two integer number of queries Q and the size of the array N. Next Q lines contains any of the given three operations:-
enqueue x
dequeue
displayfrontDuring a dequeue operation if queue is empty you need to print "Queue is empty", during enqueue operation if the maximum size of array is reached you need to print "Queue is full" and during displayfront operation you need to print the element which is at the front and if the queue is empty you need to print "Queue is empty".
Note:-Each msg or element is to be printed on a new line
Sample Input:-
8 2
displayfront
enqueue 2
displayfront
enqueue 4
displayfront
dequeue
displayfront
enqueue 5
Sample Output:-
Queue is empty
2
2
4
Queue is full
Explanation:-here size of given array is 2 so when last enqueue operation perfomed the array was already full so we display the msg "Queue is full".
Sample input:
5 5
enqueue 4
enqueue 5
displayfront
dequeue
displayfront
Sample output:-
4
5, I have written this Solution Code: public static void enqueue(int x,int k)
{
if (rear >= k) {
System.out.println("Queue is full");
}
else {
a[rear] = x;
rear++;
}
}
public static void dequeue()
{
if (rear <= front) {
System.out.println("Queue is empty");
}
else {
front++;
}
}
public static void displayfront()
{
if (rear<=front) {
System.out.println("Queue is empty");
}
else {
int x = a[front];
System.out.println(x);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr with N distinct integers from 1 to N. If you are currently at index i then in a single jump you move to the index Arr[i]. You start at standing at index 1. Find the index you will be after K jumps.First line of input contains two integers N and K.
Second line of input contains N integers, denoting Arr.
Constraints:
1 <= N <= 100000
1 <= K <= 1000000000000
1 <= Arr[i] <= NPrint the index you will be on after K jumps.Sample Input 1
5 3
3 4 2 5 1
Sample Output 1
4
Explanation:
You start at index 1
After first jump you reach index 3
After second jump you reach index 2
After third jump you reach index 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer str = new StringTokenizer(read.readLine());
int N = Integer.parseInt(str.nextToken());
long K = Long.parseLong(str.nextToken());
int[] arr = new int[N];
StringTokenizer newStr = new StringTokenizer(read.readLine());
for(int i=0; i<N; i++) {
arr[i] = Integer.parseInt(newStr.nextToken());
}
long count = 0;
int ele = 1;
long copyOfK = K;
while(K > 0) {
ele = arr[ele - 1];
count++;
K--;
if(ele == 1) {
K = copyOfK % count;
}
}
System.out.print(ele);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr with N distinct integers from 1 to N. If you are currently at index i then in a single jump you move to the index Arr[i]. You start at standing at index 1. Find the index you will be after K jumps.First line of input contains two integers N and K.
Second line of input contains N integers, denoting Arr.
Constraints:
1 <= N <= 100000
1 <= K <= 1000000000000
1 <= Arr[i] <= NPrint the index you will be on after K jumps.Sample Input 1
5 3
3 4 2 5 1
Sample Output 1
4
Explanation:
You start at index 1
After first jump you reach index 3
After second jump you reach index 2
After third jump you reach index 4, I have written this Solution Code: n,k = map(int,input().split())
arr = list(map(int,input().split()))
visitedElements = {}
i = 0
curr = 0
for i in range(k):
currentIndex = arr[curr]
visitedElements[currentIndex] = visitedElements.get(currentIndex,0)+1
if visitedElements[currentIndex]>1:
break
curr = currentIndex -1
print(list(visitedElements.keys())[(k-1)%len(list(visitedElements.keys()))])
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr with N distinct integers from 1 to N. If you are currently at index i then in a single jump you move to the index Arr[i]. You start at standing at index 1. Find the index you will be after K jumps.First line of input contains two integers N and K.
Second line of input contains N integers, denoting Arr.
Constraints:
1 <= N <= 100000
1 <= K <= 1000000000000
1 <= Arr[i] <= NPrint the index you will be on after K jumps.Sample Input 1
5 3
3 4 2 5 1
Sample Output 1
4
Explanation:
You start at index 1
After first jump you reach index 3
After second jump you reach index 2
After third jump you reach index 4, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
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;
cin>>n;
int a[n+1];
int k;
cin>>k;
for(int i=1;i<=n;++i)
cin>>a[i];
int vis[n+1]={};
int x=1;
int c=0;
while(vis[x]==0){
vis[x]=1;
++c;
x=a[x];
}
k=k%c;
x=1;
while(k){
--k;
x=a[x];
}
cout<<x;
#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: You have to complete <code>printName</code> function
<ol>
<li>Takes 2 arguments which are EventEmitter object , event name </li>
<li>Using the <code>EventEmitter</code> object you need to register an eventListener for <code>personEvent</code> event which takes a callback function. </li>
<li>The callback function itself takes an argument (which would be emitted by test runner) which is of type string. Using that argument print to console.
`My name is ${argument}`.
</li>
</ol>Function will take two arguments
1) 1st argument will be an object of EventEmitter class
2) 2nd argument will be the event name (string)
Function returns a registered Event using an object of EventEmitter class (which then is used to print the name)<pre>
<code>
const emitter=EventEmitter object
const personEvent="event"
function printName(emitter,personEvent)//registers event with event Name "event"
emitter.emit("event","Dev")//emits the event and give the output "My name is Dev"
</code>
</pre>
, I have written this Solution Code: function printName(emitter,personEvent) {
emitter.on(personEvent,(arg)=>{
console.log('My name is ',arg);
});
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given an array of several arrays that each contain integers and your goal is to write a function that
will sum up all the numbers in all the arrays. For example, if the input is [[3, 2], [1], [4, 12]] then your
program should output 22 because 3 + 2 + 1 + 4 + 12 = 22An array containing arrays which can contain any number of elements.Sum of all the elements in all of the arrays.Sample input:-
[[3, 2], [1], [4, 12]]
Sample output:-
22
Explanation:-
3 + 2 + 1 + 4 + 12 = 22, I have written this Solution Code: function sum_array(arr) {
// store our final answer
var sum = 0;
// loop through entire array
for (var i = 0; i < arr.length; i++) {
// loop through each inner array
for (var j = 0; j < arr[i].length; j++) {
// add this number to the current final sum
sum += arr[i][j];
}
}
console.log(sum);
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N which is to be reduced to 1 by performing the given operation:-
In one operation you can subtract any divisor of N other than N itself from N.
Your task is to find the minimum number to reduce N to 1.The input contains a single integer N.
<b>Constraints:-</b>
1 <= N <= 1000Print the minimum number of operations need to convert N to 1.Sample Input 1:-
5
Sample Output 1:-
3
<b>Explanation:-</b>
5 - > 4 - > 2 - > 1
Sample Input 2:-
8
Sample Output 2:-
3
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int dp[1001];
int solve(int n){
if(n<=1){
return 0;
}
if(dp[n]==-1){
int ans = INT_MAX;
int x = sqrt(n);
ans=min(ans,solve(n-1));
for(int i=2;i<=x;i++){
if(n%i==0){
ans=min(ans,solve(n-i));
ans=min(ans,solve(n-(n/i)));
}}
dp[n]=ans+1;
}
return dp[n];
}
int main(){
for(int i=0;i<1001;i++){
dp[i]=-1;
}
int n;
cin>>n;
cout<<solve(n);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N which is to be reduced to 1 by performing the given operation:-
In one operation you can subtract any divisor of N other than N itself from N.
Your task is to find the minimum number to reduce N to 1.The input contains a single integer N.
<b>Constraints:-</b>
1 <= N <= 1000Print the minimum number of operations need to convert N to 1.Sample Input 1:-
5
Sample Output 1:-
3
<b>Explanation:-</b>
5 - > 4 - > 2 - > 1
Sample Input 2:-
8
Sample Output 2:-
3
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static short min = 127;
public static void main (String[] args)throws IOException {
BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(sc.readLine());
int[] dp = new int[n+1];
Arrays.fill(dp, 1000);
dp[0] = 0;
dp[1] = 0;
for(int i=2; i<=n; i++){
for(int j=2; j*j <= i; j++){
if(i%j == 0){
dp[i] = Math.min(dp[i], dp[i-j]+1);
dp[i] = Math.min(dp[i], dp[i-(i/j)]+1);
}
}
dp[i] = Math.min(dp[i], dp[i-1]+1);
}
System.out.println(dp[n]);
}
public static void check(short n, short count){
if(n == 4 || n == 3){
count +=2;
min = (short)Math.min(min, count);
return;
}
if(n ==2 || n==1){
count += (n-1);
min = (short)Math.min(min, count);
return;
}
HashSet<Short> ar = div(n);
for(short i: ar){
check((short)(n-i), (short)(count+1));
}
}
public static HashSet<Short> div(short n){
HashSet<Short> ar = new HashSet<>();
ar.add((short)1);
for(short i=2; i*i<=n; i++){
if(n%i == 0){
ar.add(i);
ar.add((short)(n/i));
}
}
return ar;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N balloons numbered from 1 to N. Every balloon has an another integer value Arr[i] assigned to it where i varies from 1 to N, and i represents the number of balloon.
A balloon at an odd position (i = 1, 3, 5. ) is happy if Arr[i] is odd.
A balloon at an even position (i = 2, 4, 6. ) is happy if Arr[i] is even.
Find the number of happy balloons.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, Arr[1], Arr[2], Arr[3],. , Arr[N].
Constrains
1 <= N <= 200000
1 <= Arr[i] <= 1000000Output a single integer, the number of happy balloons.Sample Input
5
1 3 4 6 7
Sample Output
3
Explanation
Happy balloons are balloons numbered 1, 4, 5.
Sample Input
5
1 2 3 4 5
Sample Output
5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
br.readLine();
String[] line = br.readLine().split(" ");
int happyBalloons = 0;
for(int i=1;i<=line.length;++i){
int num = Integer.parseInt(line[i-1]);
if(num%2 == i%2 ){
happyBalloons++;
}
}
System.out.println(happyBalloons);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N balloons numbered from 1 to N. Every balloon has an another integer value Arr[i] assigned to it where i varies from 1 to N, and i represents the number of balloon.
A balloon at an odd position (i = 1, 3, 5. ) is happy if Arr[i] is odd.
A balloon at an even position (i = 2, 4, 6. ) is happy if Arr[i] is even.
Find the number of happy balloons.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, Arr[1], Arr[2], Arr[3],. , Arr[N].
Constrains
1 <= N <= 200000
1 <= Arr[i] <= 1000000Output a single integer, the number of happy balloons.Sample Input
5
1 3 4 6 7
Sample Output
3
Explanation
Happy balloons are balloons numbered 1, 4, 5.
Sample Input
5
1 2 3 4 5
Sample Output
5, I have written this Solution Code: x=int(input())
arr=input().split()
for i in range(0,x):
arr[i]=int(arr[i])
count=0
for i in range(0,x):
if(arr[i]%2==0 and (i+1)%2==0):
count+=1
elif (arr[i]%2!=0 and (i+1)%2!=0):
count+=1
print (count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N balloons numbered from 1 to N. Every balloon has an another integer value Arr[i] assigned to it where i varies from 1 to N, and i represents the number of balloon.
A balloon at an odd position (i = 1, 3, 5. ) is happy if Arr[i] is odd.
A balloon at an even position (i = 2, 4, 6. ) is happy if Arr[i] is even.
Find the number of happy balloons.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, Arr[1], Arr[2], Arr[3],. , Arr[N].
Constrains
1 <= N <= 200000
1 <= Arr[i] <= 1000000Output a single integer, the number of happy balloons.Sample Input
5
1 3 4 6 7
Sample Output
3
Explanation
Happy balloons are balloons numbered 1, 4, 5.
Sample Input
5
1 2 3 4 5
Sample Output
5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
int ans = 0;
For(i, 1, n+1){
int a; cin>>a;
if(i%2 == a%2)
ans++;
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array <b>A[]</b> of size <b>N</b>. A triplet (i, j, k) is said to be special if they satisfy below two conditions:
<b>1. i < j < k</b>
<b>2. A<sub>i</sub> < A<sub>j</sub> < A<sub>k</sub></b>
<b>V<sub>(A<sub>i</sub>,A<sub>j</sub>,A<sub>k</sub>)</sub>: A<sub>i</sub> + (A<sub>j</sub> * A<sub>k</sub>)</b>
You need to find the maximum possible value V<sub>(A<sub>i</sub>,A<sub>j</sub>,A<sub>k</sub>)</sub> which can be obtained from possible triplets satisfying above two conditions.The input line contains T, denoting the number of testcases. Each testcase contains Two lines. First line contains size of array. Second line contains elements of array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
Sum of N for all test cases is less than 10^5For each testcase you need find the maximum possible value in new line otherwise print "<b>-1</b>" if no such triplet is found.Sample Input:
2
4
5 20 11 19
4
1 2 2 2
Sample Output:
214
-1
Explanation:
Testcase 1: 5 11 19 forms special triplet fulfilling above conditions, thus giving maximum value 5 + (11*19) = 214.
Testcase 2: None of triplets satisfy the given conditions thus, maximum value is -1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int[] maxLeft(int[] a){
int N = a.length;
int[] res = new int[N];
TreeSet<Integer> ts = new TreeSet<>();
for(int i=0; i<N; i++){
int curr = a[i];
Integer maxLeftVal = ts.lower(curr);
if(maxLeftVal==null){
res[i] = -1;
}else{
res[i] = maxLeftVal;
}
ts.add(curr);
}
return res;
}
static int[] maxRight(int[] a){
int N = a.length;
int[] res = new int[N];
res[N-1] = -1;
int maxSeenIndexTillNow = N-1;
for(int i=N-2; i>=0; i--){
int curr = a[i];
if(curr<a[maxSeenIndexTillNow]){
res[i] = maxSeenIndexTillNow;
}else{
res[i] = -1;
maxSeenIndexTillNow = i;
}
}return res;
}
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(br.readLine());
for(int t=0; t<T; t++){
int N = Integer.parseInt(br.readLine());
int[] arr = new int[N];
String[] strArr = br.readLine().split(" ");
for(int i=0; i<N; i++){
arr[i] = Integer.parseInt(strArr[i]);
}
int[] valOfA = maxLeft(arr);
int[] valOfK = maxRight(arr);
long vMax = -1;
for(int i=0; i<N; i++){
int vA = valOfA[i];
int vK = valOfK[i];
if(vA==-1 || vK==-1){
}else{
long valA = vA;
long valK = arr[vK];
long valJ = arr[i];
long currVmax = vA + (valJ*valK);
vMax = Math.max(vMax, currVmax);
}
}System.out.println(vMax);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array <b>A[]</b> of size <b>N</b>. A triplet (i, j, k) is said to be special if they satisfy below two conditions:
<b>1. i < j < k</b>
<b>2. A<sub>i</sub> < A<sub>j</sub> < A<sub>k</sub></b>
<b>V<sub>(A<sub>i</sub>,A<sub>j</sub>,A<sub>k</sub>)</sub>: A<sub>i</sub> + (A<sub>j</sub> * A<sub>k</sub>)</b>
You need to find the maximum possible value V<sub>(A<sub>i</sub>,A<sub>j</sub>,A<sub>k</sub>)</sub> which can be obtained from possible triplets satisfying above two conditions.The input line contains T, denoting the number of testcases. Each testcase contains Two lines. First line contains size of array. Second line contains elements of array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
Sum of N for all test cases is less than 10^5For each testcase you need find the maximum possible value in new line otherwise print "<b>-1</b>" if no such triplet is found.Sample Input:
2
4
5 20 11 19
4
1 2 2 2
Sample Output:
214
-1
Explanation:
Testcase 1: 5 11 19 forms special triplet fulfilling above conditions, thus giving maximum value 5 + (11*19) = 214.
Testcase 2: None of triplets satisfy the given conditions thus, maximum value is -1, I have written this Solution Code: #include <bits/stdc++.h>
typedef long long int ll;
using namespace std;
ll getLowValue(set<ll>& lowValue, ll& n)
{
auto it = lowValue.lower_bound(n);
--it;
return (*it);
}
ll maxTriplet(ll arr[], ll n)
{
ll maxSuffArr[n + 1];
maxSuffArr[n] = 0;
for (int i = n - 1; i >= 0; --i)
maxSuffArr[i] = max(maxSuffArr[i + 1], arr[i]);
ll ans = -1;
set<ll> lowValue;
lowValue.insert(INT_MIN);
for (int i = 0; i < n - 1; ++i) {
if (maxSuffArr[i + 1] > arr[i]) {
ans = max(ans, getLowValue(lowValue,arr[i]) + arr[i] * maxSuffArr[i + 1]);
lowValue.insert(arr[i]);
}
}
return ans;
}
int main()
{
int t;
cin>>t;
while(t--){
ll n;
cin>>n;
ll arr[n];
for(int i=0;i<n;i++)
cin>>arr[i];
cout << maxTriplet(arr, n)<<endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In this question, you will be adding a new method called <code>squareAndSort</code> to the <code>Array prototype</code>. This means that any array created in the future will have this method available to it. This method will be called on the <code>array with number and string data type elements</code>.
The method should <code>filter</code> the <code>number datatype</code> from the <code>string datatype</code> elements and <code>return</code> a <code>sorted squared array in ascending order</code> of the remaining number element array.The squareAndSort method will be called on an array with elements of number and string datatypes.The squareAndSort method should return an array with elements of number datatype.const numbers = [5,2,1,"four",4];
console. log(numbers. squareAndSort()); // [ 1, 4, 16, 25 ], I have written this Solution Code: Array.prototype.squareAndSort = function() {
const filteredArray = this.filter(element => typeof element === "number");
const squaredArray = filteredArray.map(number => number ** 2);
const sortedArray = squaredArray.sort((a, b) => a - b);
return sortedArray;
};, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sky (the blue ranger) wants to replace Jack (the red ranger) from his position. For this, he needs to conquer the entire Base.
The Base can be represented as an M*N grid, and Sky needs to conquer it cell by cell. Sky starts conquering the Base from the cell (1, 1). In each move, he conquers the cell, and moves to an adjacent non- conquered cell (he cannot move if there is no adjacent non- conquered cell). Now, there is a catch, the last cell he needs to conquer is (M, N) so as to complete the quest for the red ranger tag!
Please let us know if Sky can replace Jack by conquering all the cells in the Base.
Note: The diagonal cells are not considered as adjacent cells.The first and the only line of input contains two integers M and N.
Constraints
1 <= M, N <= 1000Output "YES" (without quotes) if Sky can conquer the entire Base to replace Jack, else output "NO" (without quotes).Sample Input
2 2
Sample Output
NO
Explanation
The possible journeys of Sky ending at (2, 2) can be:
(1, 1) - > (1, 2) - > (2, 2)
(1, 1) - > (2, 1) - > (2, 2)
Since, in each of the path that Sky takes, the total cells covered are not 4, hence Sky cannot conquer the entire base.
Sample Input
3 3
Sample Output
YES, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s[]=br.readLine().split(" ");
int m=Integer.parseInt(s[0]);
int n=Integer.parseInt(s[1]);
if(m%2==0 && n%2==0)
System.out.println("NO");
else
System.out.println("YES");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sky (the blue ranger) wants to replace Jack (the red ranger) from his position. For this, he needs to conquer the entire Base.
The Base can be represented as an M*N grid, and Sky needs to conquer it cell by cell. Sky starts conquering the Base from the cell (1, 1). In each move, he conquers the cell, and moves to an adjacent non- conquered cell (he cannot move if there is no adjacent non- conquered cell). Now, there is a catch, the last cell he needs to conquer is (M, N) so as to complete the quest for the red ranger tag!
Please let us know if Sky can replace Jack by conquering all the cells in the Base.
Note: The diagonal cells are not considered as adjacent cells.The first and the only line of input contains two integers M and N.
Constraints
1 <= M, N <= 1000Output "YES" (without quotes) if Sky can conquer the entire Base to replace Jack, else output "NO" (without quotes).Sample Input
2 2
Sample Output
NO
Explanation
The possible journeys of Sky ending at (2, 2) can be:
(1, 1) - > (1, 2) - > (2, 2)
(1, 1) - > (2, 1) - > (2, 2)
Since, in each of the path that Sky takes, the total cells covered are not 4, hence Sky cannot conquer the entire base.
Sample Input
3 3
Sample Output
YES, I have written this Solution Code: m,n=map(int, input().split())
if(m%2 or n%2):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sky (the blue ranger) wants to replace Jack (the red ranger) from his position. For this, he needs to conquer the entire Base.
The Base can be represented as an M*N grid, and Sky needs to conquer it cell by cell. Sky starts conquering the Base from the cell (1, 1). In each move, he conquers the cell, and moves to an adjacent non- conquered cell (he cannot move if there is no adjacent non- conquered cell). Now, there is a catch, the last cell he needs to conquer is (M, N) so as to complete the quest for the red ranger tag!
Please let us know if Sky can replace Jack by conquering all the cells in the Base.
Note: The diagonal cells are not considered as adjacent cells.The first and the only line of input contains two integers M and N.
Constraints
1 <= M, N <= 1000Output "YES" (without quotes) if Sky can conquer the entire Base to replace Jack, else output "NO" (without quotes).Sample Input
2 2
Sample Output
NO
Explanation
The possible journeys of Sky ending at (2, 2) can be:
(1, 1) - > (1, 2) - > (2, 2)
(1, 1) - > (2, 1) - > (2, 2)
Since, in each of the path that Sky takes, the total cells covered are not 4, hence Sky cannot conquer the entire base.
Sample Input
3 3
Sample Output
YES, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n, m; cin>>n>>m;
if(n%2 || m%2){
cout<<"YES";
}
else{
cout<<"NO";
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements and an integer D. Your task is to rotate the array D times in a circular manner from the right to left direction. Consider the examples for better understanding:-
Try to do without creating another arrayUser task:
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>rotate()</b> that takes the array, size of the array, and the integer d as a parameter.
Constraints:
1 <= T <= 25
2 <= N <= 10^4
1<=D<=10^5
1 <= A[i] <= 10^5For each test case, you just need to rotate the array by D times. The driver code will prin the rotated array in a new line.Sample Input:
2
8
4
1 2 3 4 5 6 7 8
10
3
1 2 3 4 5 6 7 8 9 10
Sample Output:
5 6 7 8 1 2 3 4
4 5 6 7 8 9 10 1 2 3
Explanation(might now be the optimal solution):
Testcase 1:
Follow the below steps:-
After the first rotation, the array becomes 2 3 4 5 6 7 8 1
After the second rotation, the array becomes 3 4 5 6 7 8 1 2
After the third rotation, the array becomes 4 5 6 7 8 1 2 3
After the fourth rotation, the array becomes 5 6 7 8 1 2 3 4
Hence the final result: 5 6 7 8 1 2 3 4, I have written this Solution Code: public static int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
public static void rotate(int arr[], int n, int d){
d = d % n;
int g_c_d = gcd(d, n);
for (int i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
int temp = arr[i];
int j = i;
boolean win=true;
while (win) {
int k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find the smallest number which has exactly P divisors.
Note:- It is guaranteed that the number to found is less than 1000.<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>MinimumNumber()</b> that takes the integer P as a parameter.
</b>Constraints:</b>
1 <= P <= 32
<b>Note:</b>
<i>But there is a catch here given user function has already code in it which may or may not be correct, now you need to figure out these and correct if it is required</i>Return the smallest number which has exactly P divisors.Sample Input:-
4
Sample Output:-
6
Sample Input:-
1
Sample Output:-
1, I have written this Solution Code: static int solve(int n){
int cnt=0;
for(int i=1;i<=n;i++){
if(n%i==0){cnt++;}
}
return cnt;
}
static int MinimumNumber(int P){
int i=1;
for( i=1; i<=1000 && solve(i)!=P;i++){
}
return i;
}, 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 a String <b>S</b>, you need to typecast this String to Integer. If the typecasting is done successfully then we will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>checkConvertion()</b>, which contains S as parameter.
Constraints:-
1 <= |S| <= 8
The string will contain only numeric digits(1-9)You need to return the typecasted integer value. The driver code will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".Sample Input:
548
Sample Output:
Nice Job
Sample Input:
6
Sample Output:
Nice Job, I have written this Solution Code: a=input()
try:
b=int(a)
print("Nice Job")
except:
print("Wrong answer"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a String <b>S</b>, you need to typecast this String to Integer. If the typecasting is done successfully then we will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>checkConvertion()</b>, which contains S as parameter.
Constraints:-
1 <= |S| <= 8
The string will contain only numeric digits(1-9)You need to return the typecasted integer value. The driver code will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".Sample Input:
548
Sample Output:
Nice Job
Sample Input:
6
Sample Output:
Nice Job, I have written this Solution Code: static int checkConevrtion(String S)
{
return Integer.parseInt(S);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of characters letters that is sorted in non-decreasing order, and a character target. There are at least two different characters in letters.
Return the smallest character in letters that is lexicographically greater than target. If such a character does not exist, return the first character in letters.<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>nextGreatestLetter()</b> that takes char array "letters" and char "target" as parameters.
Note- In sample first char array is given in the form of string and in next line target char is given.
<b>Constraints:</b>
2 ≤ letters.length ≤ 10<sup>4</sup>
letters[i] is a lowercase English letter.
letters is sorted in non-decreasing order.
letters contains at least two different characters.
target is a lowercase English letter.Return true if num is a perfect square or false otherwise.Sample 1:
Input:
cfj
a
Output:
"c"
Explanation: The smallest character that is lexicographically greater than 'a' in letters is 'c'.
Sample 2:
Input:
cfj
c
Output:
"f"
Explanation: The smallest character that is lexicographically greater than 'c' in letters is 'f'.
, I have written this Solution Code:
class Solution {
public char nextGreatestLetter(char[] letters, char target) {
int start=0;
int end=letters.length-1;
while(start<=end)
{
int mid=start+(end-start)/2;
if(target<letters[mid])
{
end=mid-1;
}
else
{
start=mid+1;
}
}
return letters[start%letters.length];
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted linked list of size s1 and s2(sizes may or may not be same), your task is to merge them such that resultant list is also sorted.<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>Merge()</b> that takes the head node of both the linked list as the parameter.
Use <b>insert()</b> function for inserting nodes in the linked list.
<b>Constraints:</b>
1 < = s1, s2 < = 1000
1 < = val < = 10000Return the head of the merged linked list, printing will be done by the driver codeSample Input:
5 6
1 2 3 4 5
3 4 6 8 9 10
Sample Output:
1 2 3 3 4 4 5 6 8 9 10, I have written this Solution Code: public static Node Merge (Node head1, Node head2){
Node head =null;
while(head1!=null && head2!=null){
if(head1.val<head2.val){
head=insert(head,head1.val);
head1=head1.next;
}
else{
head=insert(head,head2.val);
head2=head2.next;
}
}
while(head1!=null){
head=insert(head,head1.val);
head1=head1.next;
}
while(head2!=null){
head=insert(head,head2.val);
head2=head2.next;
}
return head;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The DNA sequence is composed of a series of nucleotides abbreviated as 'A', 'C', 'G', and 'T'.
For example, "ACGAATTCCG" is a DNA sequence.
When studying DNA, it is useful to identify repeated sequences within the DNA.
Given a string s that represents a DNA sequence, return all the 10-letter long sequences (substrings) that occur more than once in a DNA molecule. You may return the answer in any order. This is a functional problem you have to complete the solve function.The first line of the input contains the string s.
<b>Constraints</b>
1 <= s. length() <= 1e5
s[i] is either 'A', 'C', 'G', or 'T'Complete the function and return the relevant strings.Sample Input:-
GGGGGGGGGGGGG
Sample Output:-
GGGGGGGGGG, I have written this Solution Code: from collections import defaultdict
dna=input()
d=defaultdict(int)
for i in range(0,len(dna)-9):
d[dna[i:i+10]]+=1
ans=[]
for i in d:
if(d[i]>1):
ans.append(i)
ans.sort()
for i in ans:
print(i), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The DNA sequence is composed of a series of nucleotides abbreviated as 'A', 'C', 'G', and 'T'.
For example, "ACGAATTCCG" is a DNA sequence.
When studying DNA, it is useful to identify repeated sequences within the DNA.
Given a string s that represents a DNA sequence, return all the 10-letter long sequences (substrings) that occur more than once in a DNA molecule. You may return the answer in any order. This is a functional problem you have to complete the solve function.The first line of the input contains the string s.
<b>Constraints</b>
1 <= s. length() <= 1e5
s[i] is either 'A', 'C', 'G', or 'T'Complete the function and return the relevant strings.Sample Input:-
GGGGGGGGGGGGG
Sample Output:-
GGGGGGGGGG, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void DNASequence(String input){
int n = input.length();
HashSet<String> set = new HashSet<>();
HashSet<String> res = new HashSet<>();
ArrayList<String> list = new ArrayList<>();
for(int i=0; i<=n-10 ; i++){
String result = input.substring(i, i+10);
if(set.contains(result)){
res.add(result);
}
set.add(result);
}
for(String r : res){
list.add(r);
}
Collections.sort(list);
for(String r : list){
System.out.println(r);
}
}
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
String input = sc.next();
DNASequence(input);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The DNA sequence is composed of a series of nucleotides abbreviated as 'A', 'C', 'G', and 'T'.
For example, "ACGAATTCCG" is a DNA sequence.
When studying DNA, it is useful to identify repeated sequences within the DNA.
Given a string s that represents a DNA sequence, return all the 10-letter long sequences (substrings) that occur more than once in a DNA molecule. You may return the answer in any order. This is a functional problem you have to complete the solve function.The first line of the input contains the string s.
<b>Constraints</b>
1 <= s. length() <= 1e5
s[i] is either 'A', 'C', 'G', or 'T'Complete the function and return the relevant strings.Sample Input:-
GGGGGGGGGGGGG
Sample Output:-
GGGGGGGGGG, I have written this Solution Code: vector<string> solve(string &s) {
string res = "";
for (int i = 0; i < 10; i++) {
res += s[i];
}
unordered_map<string, int> freq;
freq[res] += 1;
for (int i = 10; i < s.length(); i++) {
res.erase(res.begin());
res += s[i];
freq[res] += 1;
}
vector<string> result;
for (auto &it : freq) {
if (it.second > 1) {
result.push_back(it.first);
}
}
return result;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N. The array contains positive and negative elements both. You need to count number of subsequences that has product of its elements as positive. The single positive element of the array will also contribute in the count.
<b>Note:</b> Since the answer can be huge you need to take count as the modulo of (10^9 + 7).The first line of each test case contains N size of array. Second-line contains elements of array separated by space.
<b>Constraints:</b>
1 <= N <= 10^5
-10^6 <= A[i] <= 10^6For each test case you need to print the count of subsequences which has a product of elements as positive.Sample Input 1
4
-1 -2 0 -1
Sample Output 1
3
Sample Input 2
1
5
Sample Output 2
1
Explanation:
Testcase1: The positive product subsequences are {-1, -2}, {-2, -1}, {-1, -1} so a total of 3., I have written this Solution Code: import math
m=int(math.pow(10,9))+7
def power(n):
ans=1
for i in range(0,n):
ans=(ans*2)%m
return ans
def cntSubSeq(arr, n):
pos_count = 0;
neg_count = 0
for i in range (n):
if (arr[i] > 0) :
pos_count += 1
if (arr[i] < 0):
neg_count += 1
result = power(pos_count)
if (neg_count > 0):
result = (result *power(neg_count-1))%m
result -= 1
return result
n=int(input())
arr=input().split()
for i in range(0,n):
arr[i]=int(arr[i])
print (int(cntSubSeq(arr,n))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N. The array contains positive and negative elements both. You need to count number of subsequences that has product of its elements as positive. The single positive element of the array will also contribute in the count.
<b>Note:</b> Since the answer can be huge you need to take count as the modulo of (10^9 + 7).The first line of each test case contains N size of array. Second-line contains elements of array separated by space.
<b>Constraints:</b>
1 <= N <= 10^5
-10^6 <= A[i] <= 10^6For each test case you need to print the count of subsequences which has a product of elements as positive.Sample Input 1
4
-1 -2 0 -1
Sample Output 1
3
Sample Input 2
1
5
Sample Output 2
1
Explanation:
Testcase1: The positive product subsequences are {-1, -2}, {-2, -1}, {-1, -1} so a total of 3., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
static final int MOD = 1000000007;
static long getCount(int arr[], int n) {
long cntPrevPos = 0, cntPrevNeg = 0;
if(arr[0] < 0) {
cntPrevNeg = 1;
}
else if(arr[0] > 0) {
cntPrevPos = 1;
}
for(int i=1; i<n; i++) {
long cntCurrPos = 0, cntCurrNeg = 0;
if(arr[i] < 0) {
cntCurrNeg = 1 + cntPrevPos + cntPrevNeg;
cntCurrNeg %= MOD;
cntCurrPos = cntPrevNeg + cntPrevPos;
cntCurrPos %= MOD;
}
else if(arr[i] > 0) {
cntCurrNeg = cntPrevNeg + cntPrevNeg;
cntCurrNeg %= MOD;
cntCurrPos = 1 + cntPrevPos + cntPrevPos;
cntCurrPos %= MOD;
}
else
continue;
cntPrevNeg = cntCurrNeg;
cntPrevPos = cntCurrPos;
}
return cntPrevPos;
}
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(" ");
int arr[] = new int[n];
for(int i=0; i<n; i++)
arr[i] = Integer.parseInt(str[i]);
System.out.println(getCount(arr, n));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: void magicTrick(int a, int b){
cout<<a+b/2;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: static void magicTrick(int a, int b){
System.out.println(a + b/2);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: A,B = map(int,input().split(' '))
C = A+B//2
print(C)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, 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);
long n = sc.nextLong();
int p=(int)Math.sqrt(n);
for(int i=2;i<=p;i++){
if(n%i==0){System.out.print("NO");return;}
}
System.out.print("YES");
}
}
, 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 check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, I have written this Solution Code: import math
def isprime(A):
if A == 1:
return False
sqrt = int(math.sqrt(A))
for i in range(2,sqrt+1):
if A%i == 0:
return False
return True
inp = int(input())
if isprime(inp):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long n;
cin>>n;
long x = sqrt(n);
for(int i=2;i<=x;i++){
if(n%i==0){cout<<"NO";return 0;}
}
cout<<"YES";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Siddharth, a great entrepreneur, chooses a person as an intern if he/she good at maths. Thus Sid gives a simple maths problem to the candidate based on which he decides to who is gonna get the offer. Below is the description of the problem asked by Sid
Given arr <b>A[]</b> containing <b>N</b> positive integers. The task is to check whether it is possible to divide the given array into an <b>odd number of non-empty subarrays each of odd length such that each sub-array starts and ends with an odd number</b>. If it is possible to divide the array then print "<b>YES</b>" otherwise "<b>NO</b>".
<b>Note:</b> Don't forget that the complete array is also considered as a subarrayThe first line of each test case contains N size of array. Second-line contains N space-separated elements.
<b>Constraints:</b>
1 <= N <= 999999
0 <= A[i] <= 999999999Print "YES", if it is possible to divide the array. Otherwise, print "NO". Without quotes.Sample Input 1:
5
9 0 3 1 5
Sample Output 1:
YES
Sample Input 2:
3
1 2 0
Sample Output 2:
NO
Explanation:
Testcase 1: Array {9, 0, 3, 1, 5} can be divided as {9, 0, 3}, {1}, {5}., 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 n=sc.nextInt();
int[] a=new int[n];
int counter=0;
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
if(a[i]%2!=0) counter++;
}
String ans="NO";
if((a[0]%2!=0)&&(a[n-1]%2!=0)&&(n%2!=0))
ans="YES";
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Siddharth, a great entrepreneur, chooses a person as an intern if he/she good at maths. Thus Sid gives a simple maths problem to the candidate based on which he decides to who is gonna get the offer. Below is the description of the problem asked by Sid
Given arr <b>A[]</b> containing <b>N</b> positive integers. The task is to check whether it is possible to divide the given array into an <b>odd number of non-empty subarrays each of odd length such that each sub-array starts and ends with an odd number</b>. If it is possible to divide the array then print "<b>YES</b>" otherwise "<b>NO</b>".
<b>Note:</b> Don't forget that the complete array is also considered as a subarrayThe first line of each test case contains N size of array. Second-line contains N space-separated elements.
<b>Constraints:</b>
1 <= N <= 999999
0 <= A[i] <= 999999999Print "YES", if it is possible to divide the array. Otherwise, print "NO". Without quotes.Sample Input 1:
5
9 0 3 1 5
Sample Output 1:
YES
Sample Input 2:
3
1 2 0
Sample Output 2:
NO
Explanation:
Testcase 1: Array {9, 0, 3, 1, 5} can be divided as {9, 0, 3}, {1}, {5}., I have written this Solution Code:
def fun(n,arr):
if n%2==0:
if arr[0]%2==1 and arr[n-1]%2==1:
for i in range(1,n):
if arr[i]%2==1 and arr[i-1]%2==1:
return "YES"
else:
return "NO"
else:
if arr[0]%2==1 and arr[n-1]%2==1:
return "YES"
else:
return "NO"
n=int(input())
arr=[int(i) for i in input().split()]
print(fun(n,arr)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Siddharth, a great entrepreneur, chooses a person as an intern if he/she good at maths. Thus Sid gives a simple maths problem to the candidate based on which he decides to who is gonna get the offer. Below is the description of the problem asked by Sid
Given arr <b>A[]</b> containing <b>N</b> positive integers. The task is to check whether it is possible to divide the given array into an <b>odd number of non-empty subarrays each of odd length such that each sub-array starts and ends with an odd number</b>. If it is possible to divide the array then print "<b>YES</b>" otherwise "<b>NO</b>".
<b>Note:</b> Don't forget that the complete array is also considered as a subarrayThe first line of each test case contains N size of array. Second-line contains N space-separated elements.
<b>Constraints:</b>
1 <= N <= 999999
0 <= A[i] <= 999999999Print "YES", if it is possible to divide the array. Otherwise, print "NO". Without quotes.Sample Input 1:
5
9 0 3 1 5
Sample Output 1:
YES
Sample Input 2:
3
1 2 0
Sample Output 2:
NO
Explanation:
Testcase 1: Array {9, 0, 3, 1, 5} can be divided as {9, 0, 3}, {1}, {5}., I have written this Solution Code: 'use strict'
function main(input)
{
const inputList = input.split('\n');
const arrSize = Number(inputList[0].split(' ')[0])
const arr = inputList[1].split(' ').
map(x => Number(x))
if(arrSize %2 == 0 || Number(arr[0]) %2 == 0 ||
Number(arr[arrSize-1])%2 == 0)
console.log("NO")
else console.log("YES")
}
main(require("fs").readFileSync("/dev/stdin", "utf8"));, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted linked list of size s1 and s2(sizes may or may not be same), your task is to merge them such that resultant list is also sorted.<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>Merge()</b> that takes the head node of both the linked list as the parameter.
Use <b>insert()</b> function for inserting nodes in the linked list.
<b>Constraints:</b>
1 < = s1, s2 < = 1000
1 < = val < = 10000Return the head of the merged linked list, printing will be done by the driver codeSample Input:
5 6
1 2 3 4 5
3 4 6 8 9 10
Sample Output:
1 2 3 3 4 4 5 6 8 9 10, I have written this Solution Code: public static Node Merge (Node head1, Node head2){
Node head =null;
while(head1!=null && head2!=null){
if(head1.val<head2.val){
head=insert(head,head1.val);
head1=head1.next;
}
else{
head=insert(head,head2.val);
head2=head2.next;
}
}
while(head1!=null){
head=insert(head,head1.val);
head1=head1.next;
}
while(head2!=null){
head=insert(head,head2.val);
head2=head2.next;
}
return head;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an sorted array <b>Arr[]</b> of size <b>N</b>, containing both <b>negative</b> and <b>positive</b> integers, you need to print the squared sorted output.
<b>Note</b> Try using two pointer approachThe first line of input contains T, denoting the number of test cases. Each testcase contains 2 lines. The first line contains the N size of the array. The second line contains elements of an array separated by space.
Constraints:
1 ≤ T ≤ 100
1 ≤ N ≤ 10000
-10000 ≤ A[i] ≤ 10000
The Sum of N over all test cases does not exceed 10^6For each test case you need to print the sorted squared output in new lineInput:
1
5
-7 -2 3 4 6
Output:
4 9 16 36 49, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main
{
public static void main(String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine());
while (t-- > 0) {
int n = Integer.parseInt(read.readLine());
int[] arr = new int[n];
String str[] = read.readLine().trim().split(" ");
for(int i = 0; i < n; i++)
arr[i] = Integer.parseInt(str[i]);
arr = sortedSquares(arr);
for(int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
System.out.println();
}
}
public static int[] sortedSquares(int[] A) {
int[] nums = new int[A.length];
int k=A.length-1;
int i=0, j=A.length-1;
while(i<=j){
if(Math.abs(A[i]) <= Math.abs(A[j])){
nums[k--] = A[j]*A[j];
j--;
}
else{
nums[k--] = A[i]*A[i];
i++;
}
}
return nums;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an sorted array <b>Arr[]</b> of size <b>N</b>, containing both <b>negative</b> and <b>positive</b> integers, you need to print the squared sorted output.
<b>Note</b> Try using two pointer approachThe first line of input contains T, denoting the number of test cases. Each testcase contains 2 lines. The first line contains the N size of the array. The second line contains elements of an array separated by space.
Constraints:
1 ≤ T ≤ 100
1 ≤ N ≤ 10000
-10000 ≤ A[i] ≤ 10000
The Sum of N over all test cases does not exceed 10^6For each test case you need to print the sorted squared output in new lineInput:
1
5
-7 -2 3 4 6
Output:
4 9 16 36 49, I have written this Solution Code: t = int(input())
for i in range(t):
n = int(input())
for i in sorted(map(lambda j:int(j)**2,input().split())):
print(i,end=' ')
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have bought N cakes and the ith cake has A[i] level of sweetness. Now you want to display the cakes in exhibition but there is a condition that you cannot display two cakes with same sweetness level. Find the maximum number of cakes that you can display.The first line of input contains a single integer N.
Second line contains N space seperated integers A[1], A[2] ..... A[N]
Constraints :
1 <= N <= 100000
1 <= A[i] <= 100000Output a single integer denoting the maximum number of cakes you can display.Sample input 1
5
1 2 3 4 5
Sample output 1
5
Sample input 2
4
1 2 2 3
Sample output 2
3
Explanation:-
Test Case 2:- There are 2 cakes with sweetness level 2 as per rule only one of them can be displayed so the ans is 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());
int a[]=new int[100001];
String str[]=br.readLine().split(" ");
for(int i=0;i<str.length;i++){
a[Integer.parseInt(str[i])]++;
}
long cnt=0;
for(int i=0;i<100001;i++){
if(a[i]>0)
cnt++;
}
System.out.println(cnt);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have bought N cakes and the ith cake has A[i] level of sweetness. Now you want to display the cakes in exhibition but there is a condition that you cannot display two cakes with same sweetness level. Find the maximum number of cakes that you can display.The first line of input contains a single integer N.
Second line contains N space seperated integers A[1], A[2] ..... A[N]
Constraints :
1 <= N <= 100000
1 <= A[i] <= 100000Output a single integer denoting the maximum number of cakes you can display.Sample input 1
5
1 2 3 4 5
Sample output 1
5
Sample input 2
4
1 2 2 3
Sample output 2
3
Explanation:-
Test Case 2:- There are 2 cakes with sweetness level 2 as per rule only one of them can be displayed so the ans is 3, I have written this Solution Code: a = int(input())
arr = input().split()
print(len(set(arr))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have bought N cakes and the ith cake has A[i] level of sweetness. Now you want to display the cakes in exhibition but there is a condition that you cannot display two cakes with same sweetness level. Find the maximum number of cakes that you can display.The first line of input contains a single integer N.
Second line contains N space seperated integers A[1], A[2] ..... A[N]
Constraints :
1 <= N <= 100000
1 <= A[i] <= 100000Output a single integer denoting the maximum number of cakes you can display.Sample input 1
5
1 2 3 4 5
Sample output 1
5
Sample input 2
4
1 2 2 3
Sample output 2
3
Explanation:-
Test Case 2:- There are 2 cakes with sweetness level 2 as per rule only one of them can be displayed so the ans is 3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a;
unordered_map<int,int> m;
for(int i=0;i<n;i++){
cin>>a;
m[a]++;
}
cout<<m.size();
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Chapo is the most exotic thing ever. Sid wants to give Chapo to the person who solves this little tricky question on Fibonacci numbers.
Fib(i) refers to i-th Fibonacci Number. First few Fibonacci numbers are Fib(0)=0, Fib(1)=1, Fib(2)=1, Fib(3)=2 ...
Given k=331.
Also, F(N) = (sum of Fib(i*k) where i varies from 0 to N-1).
You will be given two positive integers, m and n. You need to find the value of F(m**n)%1000000009. Here, m**n means m raised to the power n.
Can you get a chapo?The first and the only line of input contains 2 integers m and n.
Constraints
1 <= m, n <= 1000000000Print a single integer, the value of F(m**n)%1000000009.Sample Input 1
2 1
Sample Output 1
242667188
Sample Input 2
100 100
Sample Output 2
822562876
Explanation for Sample 1:
F(2**1)%1000000009 = F(2)%1000000009 = (Fib(0)+Fib(331))%1000000009 = 242667188.
Note: In 2nd test case, there will be 100**100 terms., I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define pb push_back
#define pii pair<int,int>
#define vi vector<int>
#define vii vector<pii>
#define mi map<int,int>
#define mii map<pii,int>
#define all(a) (a).begin(),(a).end()
#define x first
#define y second
#define sz(x) (int)x.size()
#define endl '\n'
#define hell 1000000009
#define rep(i,a,b) for(int i=a;i<b;i++)
using namespace std;
class matrix{
public:
int n;
vector<vi> val;
matrix(int n=2);
matrix operator+(const matrix&) const;
matrix operator-(const matrix&) const;
matrix operator*(const matrix&) const;
matrix& operator=(const matrix& b){
val=b.val;
n=b.n;
return *this;
}
void set(int x){
rep(i,0,n){
rep(j,0,n){
val[i][j]=x;
}
}
}
void print(){
rep(i,0,n){
rep(j,0,n){
cout<<val[i][j]<<" ";
}
cout<<endl;
}
}
matrix mul(int a){
matrix ans(this->n);
rep(i,0,this->n){
rep(j,0,this->n){
ans.val[i][j]=a*this->val[i][j]%hell;
}
}
return ans;
}
};
matrix::matrix(int x){
n=x;
val.resize(x);
rep(i,0,x){
val[i].resize(x);
}
}
matrix matrix::operator+(const matrix& b) const{
matrix ans(b.n);
rep(i,0,b.n){
rep(j,0,b.n){
ans.val[i][j]=this->val[i][j]+b.val[i][j];
}
}
return ans;
}
matrix matrix::operator-(const matrix& b) const{
matrix ans(b.n);
rep(i,0,b.n){
rep(j,0,b.n){
ans.val[i][j]=this->val[i][j]-b.val[i][j];
}
}
return ans;
}
matrix matrix::operator*(const matrix& b) const{
matrix ans(b.n);
rep(i,0,b.n){
rep(j,0,b.n){
rep(k,0,b.n){
ans.val[i][j]+=this->val[i][k]*b.val[k][j];
ans.val[i][j]%=hell;
}
}
}
return ans;
}
matrix identity(int N=2){
matrix ans(N);
rep(i,0,N){
ans.val[i][i]=1;
}
return ans;
}
matrix matexpo(matrix a, int n){
if(n==0){
return identity(a.n);
}
if(n==1) return a;
matrix x=matexpo(a,n/2);
matrix r=x*x;
if(n&1) r=r*a;
return r;
}
matrix GP(matrix A,matrix R,int N){
matrix T(2*R.n),B=A;
rep(i,0,2*R.n){
rep(j,0,2*R.n){
if(i<R.n and j<R.n){
T.val[i][j]=R.val[i][j];
}
else if(i<R.n){
T.val[i][j]=(i==j-R.n);
}
else if(j>=R.n){
T.val[i][j]=(i==j);
}
}
}
T=matexpo(T,N);
rep(i,0,A.n){
rep(j,0,A.n){
B.val[i][j]=T.val[i][j+A.n];
}
}
return B*A;
}
int expo(int base,int exponent,int mod){
int ans=1;
while(exponent!=0){
if(exponent&1) ans=(1LL*ans*base)%mod;
base=(1LL*base*base)%mod;
exponent>>=1;
}
return ans%mod;
}
void solve(){
int m,n,k=331;
cin>>m>>n;
int p=(expo(m,n,hell-1)-1+hell-1)%(hell-1);
matrix mat(2);
mat.val[0][0]=mat.val[0][1]=mat.val[1][0]=1;
mat.val[1][1]=0;
matrix a=matexpo(mat,k-1);
matrix r=matexpo(mat,k);
matrix res=GP(a,r,p);
cout<<res.val[0][0]<<endl;
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t=1;
// cin>>t;
while(t--){
solve();
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Abhi and Niyu play a game where they need to destroy N palaces (numbered through 1 to N) to pass a level. Abhi performs an attack with frequency p hits per second and Niyu performs an attack with frequency q hits per second i.e. each character spends fixed time to raise a weapon and then they hit (the time to raise the weapon is 1 / p seconds for Abhi and 1 / q seconds for Niyu). The i-th palace destroys if it receives m number of hits.
Abhi and Niyu wonder who makes the last hit on each Palace.
If Abhi and Niyu make the last hit at the same time, we assume that both of them have made the last hit.
Note:- Each castle is to be destroyed individually.The first line contains three integers N, p, q — the number of palaces, the frequency of Abhi's and Niyu's attack, correspondingly.
Next N lines contain a single integer m i. e the number of hits required to destroy the ith palace
Constraints:-
1 ≤ m ≤ 10<sup>9</sup>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ p, q ≤ 10<sup>6</sup>Print n lines. In the i-th line print word "Abhi", if the last hit on the i-th palace was performed by Abhi, "Niyu", if Niyu performed the last hit, or "Both", if both performed it at the same time.Sample Input:-
4 3 2
1
2
3
4
Sample Output:-
Abhi
Niyu
Abhi
Both
Explanation:-
In the first sample Abhi makes the first hit at time 1/3, Niyu makes the second hit at time 1/2, Abhi makes the third hit at time 2/3, and both boys make the fourth and fifth hit simultaneously at the time 1.
Sample Input:-
2 1 1
1
2
Sample Output:-
Both
Both
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static String[] answer = new String[1000001];
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] line1 = br.readLine().split(" ");
int size = Integer.parseInt(line1[0]);
long abhi = Integer.parseInt(line1[1]);
long niyu = Integer.parseInt(line1[2]);
answer[0] = "Both";
long[] castle = new long[size];
long max = 0;
for(int i = 0 ; i < size ; i++) {
castle[i] = Long.parseLong(br.readLine());
max = Math.max(castle[i], max);
}
StringBuilder sb = new StringBuilder();
int last = findWinner(abhi, niyu, max);
for(int i = 0 ; i < size ; i++){
sb.append(answer[(int) (castle[i] % last)]).append("\n");
}
System.out.println(sb);
}
static int findWinner(long abhi, long niyu, long basepower){
long temp = basepower + 1;
long i = 1, j = 1; int itr = 1;
while(temp > itr){
if(i*niyu < j*abhi){
basepower--;
answer[itr] = "Abhi";
i++;
itr++;
}
else if(i*niyu > j*abhi){
answer[itr] = "Niyu";
basepower--;
j++;
itr++;
}
else{
answer[itr] = "Both";
itr++;
answer[itr] = "Both";
break;
}
}
return itr ;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Abhi and Niyu play a game where they need to destroy N palaces (numbered through 1 to N) to pass a level. Abhi performs an attack with frequency p hits per second and Niyu performs an attack with frequency q hits per second i.e. each character spends fixed time to raise a weapon and then they hit (the time to raise the weapon is 1 / p seconds for Abhi and 1 / q seconds for Niyu). The i-th palace destroys if it receives m number of hits.
Abhi and Niyu wonder who makes the last hit on each Palace.
If Abhi and Niyu make the last hit at the same time, we assume that both of them have made the last hit.
Note:- Each castle is to be destroyed individually.The first line contains three integers N, p, q — the number of palaces, the frequency of Abhi's and Niyu's attack, correspondingly.
Next N lines contain a single integer m i. e the number of hits required to destroy the ith palace
Constraints:-
1 ≤ m ≤ 10<sup>9</sup>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ p, q ≤ 10<sup>6</sup>Print n lines. In the i-th line print word "Abhi", if the last hit on the i-th palace was performed by Abhi, "Niyu", if Niyu performed the last hit, or "Both", if both performed it at the same time.Sample Input:-
4 3 2
1
2
3
4
Sample Output:-
Abhi
Niyu
Abhi
Both
Explanation:-
In the first sample Abhi makes the first hit at time 1/3, Niyu makes the second hit at time 1/2, Abhi makes the third hit at time 2/3, and both boys make the fourth and fifth hit simultaneously at the time 1.
Sample Input:-
2 1 1
1
2
Sample Output:-
Both
Both
, I have written this Solution Code: n,p,q = map(int,input().split())
l = []
for i in range(n):
ele = int(input())
l.append(ele)
li = []
size = 0
f = 0
x = 2*(p+q)
dic = {0:[1,1]}
for j in range(1,2*(p+q)+1):
pi = dic[j-1][0]
qi = dic[j-1][1]
ans = ""
if(pi/p == qi/q):
ans = "Both"
pi+=1
qi+=1
dic[j] = [pi,qi];
dic[j+1] = [pi,qi]
f = 1
elif(pi/p < qi/q):
ans = "Abhi"
pi+=1
dic[j] = [pi,qi];
elif(pi/p > qi/q):
qi+=1
ans = "Niyu"
dic[j] = [pi,qi];
li.append(ans)
size += 1
if(f == 1):
li.append(ans)
size += 1
break
for m in l:
print(li[m%size-1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Abhi and Niyu play a game where they need to destroy N palaces (numbered through 1 to N) to pass a level. Abhi performs an attack with frequency p hits per second and Niyu performs an attack with frequency q hits per second i.e. each character spends fixed time to raise a weapon and then they hit (the time to raise the weapon is 1 / p seconds for Abhi and 1 / q seconds for Niyu). The i-th palace destroys if it receives m number of hits.
Abhi and Niyu wonder who makes the last hit on each Palace.
If Abhi and Niyu make the last hit at the same time, we assume that both of them have made the last hit.
Note:- Each castle is to be destroyed individually.The first line contains three integers N, p, q — the number of palaces, the frequency of Abhi's and Niyu's attack, correspondingly.
Next N lines contain a single integer m i. e the number of hits required to destroy the ith palace
Constraints:-
1 ≤ m ≤ 10<sup>9</sup>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ p, q ≤ 10<sup>6</sup>Print n lines. In the i-th line print word "Abhi", if the last hit on the i-th palace was performed by Abhi, "Niyu", if Niyu performed the last hit, or "Both", if both performed it at the same time.Sample Input:-
4 3 2
1
2
3
4
Sample Output:-
Abhi
Niyu
Abhi
Both
Explanation:-
In the first sample Abhi makes the first hit at time 1/3, Niyu makes the second hit at time 1/2, Abhi makes the third hit at time 2/3, and both boys make the fourth and fifth hit simultaneously at the time 1.
Sample Input:-
2 1 1
1
2
Sample Output:-
Both
Both
, 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 EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 2000001
#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
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}int n,x,y,i,t,cntx,cnty;
int rez[max1];
signed main()
{
cin>>n>>x>>y;
cntx = cnty = 0;
int k=0;
while (cntx < x||cnty < y)
{
if ((cntx+1)*y >(cnty+1)*x)
{
cnty++;
rez[k]=2;
k++;
}
else
if ((cntx+1)*y < (cnty+1)*x)
{
cntx++;
rez[k]=1;
k++;
} else
{
cntx++;
cnty++;
rez[k]=3;
k++;
rez[k]=3;
k++;
}
}
for (i = 0; i < n; i++)
{ //cout<<i<<endl;
cin>>t;
t--;
int tmp = rez[t%(x+y)];
if (tmp == 1)
out("Abhi");
else if (tmp == 2)
out("Niyu");
else
out("Both");
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of characters 'A' or 'B' only, you need to find the maximum length of substring consisting of character 'A' only.The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100000
S consists of characters 'A' or 'B' only.Output a single integer, the answer to the problem.Sample Input
ABAAABBBAA
Sample Output
3
Explanation: Substring from character 3-5 is the longest consisting of As only.
Sample Input
AAAA
Sample Output
4, 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 st = br.readLine();
int len = 0;
int c=0;
for(int i=0;i<st.length();i++){
if(st.charAt(i)=='A'){
c++;
len = Math.max(len,c);
}else{
c=0;
}
}
System.out.println(len);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of characters 'A' or 'B' only, you need to find the maximum length of substring consisting of character 'A' only.The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100000
S consists of characters 'A' or 'B' only.Output a single integer, the answer to the problem.Sample Input
ABAAABBBAA
Sample Output
3
Explanation: Substring from character 3-5 is the longest consisting of As only.
Sample Input
AAAA
Sample Output
4, I have written this Solution Code: S=input()
max=0
flag=0
for i in range(0,len(S)):
if(S[i]=='A' or S[i]=='B'):
if(S[i]=='A'):
flag+=1
if(flag>max):
max=flag
else:
flag=0
print(max), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of characters 'A' or 'B' only, you need to find the maximum length of substring consisting of character 'A' only.The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100000
S consists of characters 'A' or 'B' only.Output a single integer, the answer to the problem.Sample Input
ABAAABBBAA
Sample Output
3
Explanation: Substring from character 3-5 is the longest consisting of As only.
Sample Input
AAAA
Sample Output
4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
string s; cin>>s;
int ct = 0;
int ans = 0;
for(char c: s){
if(c == 'A')
ct++;
else
ct=0;
ans = max(ans, ct);
}
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: Arpit Gupta has brought a toy for his valentine. She is playing with that toy which runs for T seconds when winded.If winded when the toy is already running, from that moment it will run for T seconds (not additional T seconds) For example if T is 10 and toy has run for 5 seconds and winded at this moment then in total it will run for 15 seconds.
Arpit's Valentine winds the toy N times.She winds the toy at t[i] seconds after the first time she winds it.How long will the toy run in total?First Line of input contains two integers N and T
Second Line of input contains N integers, list of time Arpit's Valentine has wound the toy at.
Constraints
1 <= N <= 100000
1 <= T <= 1000000000
1 <= t[i] <= 1000000000
t[0] = 0Print a single integer the total time the toy has run.Sample input 1
2 4
0 3
Sample output 1
7
Sample input 2
2 10
0 5
Sample output 2
15
Explanation:
Testcase1:-
at first the toy is winded at 0 it will go till 4 but it again winded at 3 making it go for more 4 seconds
so the total is 7, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] str=br.readLine().split(" ");
int n=Integer.parseInt(str[0]);
int t=Integer.parseInt(str[1]);
int[] arr=new int[n];
str=br.readLine().split(" ");
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(str[i]);
}
int sum=0;
for(int i=1;i<n;i++){
int dif=arr[i]-arr[i-1];
if(dif>t){
sum=sum+t;
}else{
sum=sum+dif;
}
}
sum+=t;
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Arpit Gupta has brought a toy for his valentine. She is playing with that toy which runs for T seconds when winded.If winded when the toy is already running, from that moment it will run for T seconds (not additional T seconds) For example if T is 10 and toy has run for 5 seconds and winded at this moment then in total it will run for 15 seconds.
Arpit's Valentine winds the toy N times.She winds the toy at t[i] seconds after the first time she winds it.How long will the toy run in total?First Line of input contains two integers N and T
Second Line of input contains N integers, list of time Arpit's Valentine has wound the toy at.
Constraints
1 <= N <= 100000
1 <= T <= 1000000000
1 <= t[i] <= 1000000000
t[0] = 0Print a single integer the total time the toy has run.Sample input 1
2 4
0 3
Sample output 1
7
Sample input 2
2 10
0 5
Sample output 2
15
Explanation:
Testcase1:-
at first the toy is winded at 0 it will go till 4 but it again winded at 3 making it go for more 4 seconds
so the total is 7, I have written this Solution Code: n , t = [int(x) for x in input().split() ]
l= [int(x) for x in input().split() ]
c = 0
for i in range(len(l)-1):
if l[i+1] - l[i]<=t:
c+=l[i+1] - l[i]
else:
c+=t
c+=t
print(c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Arpit Gupta has brought a toy for his valentine. She is playing with that toy which runs for T seconds when winded.If winded when the toy is already running, from that moment it will run for T seconds (not additional T seconds) For example if T is 10 and toy has run for 5 seconds and winded at this moment then in total it will run for 15 seconds.
Arpit's Valentine winds the toy N times.She winds the toy at t[i] seconds after the first time she winds it.How long will the toy run in total?First Line of input contains two integers N and T
Second Line of input contains N integers, list of time Arpit's Valentine has wound the toy at.
Constraints
1 <= N <= 100000
1 <= T <= 1000000000
1 <= t[i] <= 1000000000
t[0] = 0Print a single integer the total time the toy has run.Sample input 1
2 4
0 3
Sample output 1
7
Sample input 2
2 10
0 5
Sample output 2
15
Explanation:
Testcase1:-
at first the toy is winded at 0 it will go till 4 but it again winded at 3 making it go for more 4 seconds
so the total is 7, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long n,t;
cin>>n>>t;
long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
long cur=t;
long ans=t;
for(int i=1;i<n;i++){
ans+=min(a[i]-a[i-1],t);
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, 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 str1 = br.readLine();
String str2 = br.readLine();
countCommonCharacters(str1, str2);
}
static void countCommonCharacters(String s1, String s2) {
int[] arr1 = new int[26];
int[] arr2 = new int[26];
for (int i = 0; i < s1.length(); i++)
arr1[s1.codePointAt(i) - 97]++;
for (int i = 0; i < s2.length(); i++)
arr2[s2.codePointAt(i) - 97]++;
int lenToCut = 0;
for (int i = 0; i < 26; i++) {
int leastOccurrence = Math.min(arr1[i], arr2[i]);
lenToCut += (2 * leastOccurrence);
}
System.out.println(findRelation((s1.length() + s2.length() - lenToCut) % 6));
}
static String findRelation(int value) {
switch (value) {
case 1:
return "Friends";
case 2:
return "Love";
case 3:
return "Affection";
case 4:
return "Marriage";
case 5:
return "Enemy";
default:
return "Siblings";
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code: name1 = input().strip().lower()
name2 = input().strip().lower()
listA = [0]*26
listB = [0]*26
for i in name1:
listA[ord(i)-ord('a')] = listA[ord(i)-ord('a')] + 1
for i in name2:
listB[ord(i)-ord('a')] = listB[ord(i)-ord('a')] + 1
count = 0
for i in range(0,26):
count = count+abs(listA[i]-listB[i])
res = ["Siblings","Friends","Love","Affection", "Marriage", "Enemy"]
print(res[count%6]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
int main(){
string s1,s2;
cin>>s1>>s2;
string a[6];
a[1]= "Friends";
a[2]= "Love";
a[3]="Affection";
a[4]= "Marriage";
a[5]= "Enemy";
a[0]= "Siblings";
int b[26],c[26];
for(int i=0;i<26;i++){b[i]=0;c[i]=0;}
for(int i=0;i<s1.length();i++){
b[s1[i]-'a']++;
}
for(int i=0;i<s2.length();i++){
c[s2[i]-'a']++;
}
int sum=0;
for(int i=0;i<26;i++){
sum+=abs(b[i]-c[i]);
}
cout<<a[sum%6];
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings numbered from 1 to N in a straight line, the i<sup>th</sup> building having height h<sub>i</sub>. You start from building 1 and move towards building N. If the height of the next building is greater than the current, you expend twice the difference in the heights of the building's energy units. If the height of the next building is lesser than the current, you expend the difference in the heights of the building's energy units.
Find the total amount of energy you would have to expend up until you reach building N.The first line of the input contains a single integer N.
The second line of the input contains N space seperated integers.
Constraints:
2 <= N <= 10<sup>5</sup>
1 <= h<sub>i</sub> <= 10<sup>9</sup>Print the total amount of energy you would have to expend uptill you reach building N.Sample Input:
5
3 5 2 1 7
Sample Output:
20
<b>Explanation</b>
3 --> 5 = 2*2=4
5 --> 2 = 3*1=3
2 --> 1= 1*1=1
1 --> 7= 6*2=12
Total = 4+3+1+12=20, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin >> n;
vector<int> a(n);
for(auto &i : a) cin >> i;
int ans = 0;
for(int i = 1; i < n; i++){
if(a[i] > a[i - 1]){
ans += 2 * (a[i] - a[i - 1]);
}
else ans += a[i - 1] - a[i];
}
cout << ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings numbered from 1 to N in a straight line, the i<sup>th</sup> building having height h<sub>i</sub>. You start from building 1 and move towards building N. If the height of the next building is greater than the current, you expend twice the difference in the heights of the building's energy units. If the height of the next building is lesser than the current, you expend the difference in the heights of the building's energy units.
Find the total amount of energy you would have to expend up until you reach building N.The first line of the input contains a single integer N.
The second line of the input contains N space seperated integers.
Constraints:
2 <= N <= 10<sup>5</sup>
1 <= h<sub>i</sub> <= 10<sup>9</sup>Print the total amount of energy you would have to expend uptill you reach building N.Sample Input:
5
3 5 2 1 7
Sample Output:
20
<b>Explanation</b>
3 --> 5 = 2*2=4
5 --> 2 = 3*1=3
2 --> 1= 1*1=1
1 --> 7= 6*2=12
Total = 4+3+1+12=20, 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 arr[] = new int[n];
for(int i=0; i<n; i++) arr[i] = Integer.parseInt(st.nextToken());
long expend = 0;
for(int i=1; i<n; i++){
long diff = Math.abs(arr[i-1]-arr[i]);
if(arr[i-1]<arr[i]){
expend+=2*diff;
} else expend+=diff;
}
System.out.print(expend);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an unsorted array of length n and must sort it using merge sort while also printing the amount of merges that occur throughout the sorting process.The first line of input will be n, which represents the array's length, followed by the n array items in the second line.
Constraints:
0<= n <=100000First- line should be the sorted array and the second should be the number of mergers that occurs when the array is sorted using merge sort.Sample Input:
5
5 1 2 7 3
Output:
1 2 3 5 7
4, I have written this Solution Code: def mergeSort(arr,count):
if len(arr)>1:
mid= len(arr)//2
a=arr[:mid]
b=arr[mid:]
count=mergeSort(a,count)
count=mergeSort(b,count)
count+=1
i = j = k = 0
l1 = len(a)
l2 = len(b)
while i< l1 and j <l2:
if a[i]<b[j]:
arr[k]=a[i]
i +=1
else:
arr[k]=b[j]
j+=1
k+=1
while i <l1:
arr[k]=a[i]
i+=1
k+=1
while j<l2:
arr[k]=b[j]
j+=1
k+=1
return count
N=int(input())
arr=list(map(int,input().split()))
count=mergeSort(arr,0)
print(' '.join(map(str,arr)))
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an unsorted array of length n and must sort it using merge sort while also printing the amount of merges that occur throughout the sorting process.The first line of input will be n, which represents the array's length, followed by the n array items in the second line.
Constraints:
0<= n <=100000First- line should be the sorted array and the second should be the number of mergers that occurs when the array is sorted using merge sort.Sample Input:
5
5 1 2 7 3
Output:
1 2 3 5 7
4, I have written this Solution Code: import java.util.Scanner;
public class Main
{
int noOfMerge=0;
void merge(int arr[], int l, int m, int r)
{
int n1 = m - l + 1;
int n2 = r - m;
int L[] = new int [n1];
int R[] = new int [n2];
for (int i=0; i<n1; ++i)
L[i] = arr[l + i];
for (int j=0; j<n2; ++j)
R[j] = arr[m + 1+ j];
int i = 0, j = 0,k=l;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
} else {
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
void sort(int arr[], int l, int r)
{
if (l < r)
{
int m = (l+r)/2;
sort(arr, l, m);
sort(arr , m+1, r);
merge(arr, l, m, r);
noOfMerge+=1;
}
}
public static void main(String args[])
{
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] array=new int[n];
for(int i=0;i<n;i++)
array[i]=scanner.nextInt();
Main ob = new Main();
ob.sort(array, 0, n-1);
for (int i=0; i<n; ++i)
System.out.print(array[i] + " ");
System.out.println();
System.out.println(ob.noOfMerge);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Complete the constructor function <code>Account</code>. It should contain the properties <code>accountNum, password, name, balance </code>, equal to the respective parameters, and a method <code>addInterest</code> which will <code>add the interest to the existing balance</code> using the formula <code>balance * (default interest rate / 100)</code> and then return the updated <code>balance</code>. The default interest rate should be 10%.
Now create a new class <code>SavingsAccount</code> extended from the <code>Account</code> constructor function such that it accepts a new field called <code>interest</code>.
Now in this class add another <code>addInterest</code> method such that the interest is not 10% now instead it's the new interest passed to this class. The method should also <code>add the interest to the existing balance</code> using the formula <code>balance * (interest / 100)</code> and then return the updated <code>balance</code>.
In both classes, (i.e. the parent class and the child class) the <code>addInterest</code> method should be completed which takes no arguments but returns the current balance of that instance after adding the interest.The Account class constructor will take 4 arguments
<ul><li>accountNum(string), password(string), name(string), balance(number)</li></ul>
The SavingsAccount class will take 5 arguments
<ul><li> accountNum(string), password(string), name(string), balance(number), interest(number)</li></ul>The method <code>addInterest</code> will return the current balance after adding interest.
<code>
const hey = new SavingsAccount('newton1103', '*****', 'company', 1500, 20);
let bal = hey.addInterest()
console.log(bal) //1800
</code>
, I have written this Solution Code: function account (accountNum, password, name, balance){
this.accountNum = accountNum;
this.password = password
this.name = name
this.balance = balance
this.addInterest = function(){
this.balance += this.balance*0.1
return this.balance;
}
}
class savingsAccount extends account{
constructor(accountNum, password, name, balance,interest){
super(accountNum, password, name, balance)
this.interest = interest
}
addInterest = function(){
this.balance += this.balance*(this.interest/100)
return this.balance;
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
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