Instruction
stringlengths 261
35k
| Response
stringclasses 1
value |
|---|---|
For this Question: Numbers are awesome, larger numbers are more awesome!
Given an array A of size N, you need to find the maximum sum that can be obtained from the elements of the array (the selected elements need not be contiguous). You may even decide to take no element to get a sum of 0.The first line of the input contains the size of the array.
The next line contains N (white-space separated) integers denoting the elements of the array.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>4</sup>
-10<sup>7</sup> ≤ A[i] ≤10<sup>7</sup>For each test case, output one integer representing the maximum value of the sum that can be obtained using the various elements of the array.Input 1:
5
1 2 1 -1 1
output 1:
5
input 2:
5
0 0 -1 0 0
output 2:
0
<b>Explanation 1:</b>
In order to maximize the sum among [ 1 2 1 -1 1], we need to only consider [ 1 2 1 1] and neglect the [-1]., I have written this Solution Code: n=int(input())
li = list(map(int,input().strip().split()))
sum=0
for i in li:
if i>0:
sum+=i
print(sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Numbers are awesome, larger numbers are more awesome!
Given an array A of size N, you need to find the maximum sum that can be obtained from the elements of the array (the selected elements need not be contiguous). You may even decide to take no element to get a sum of 0.The first line of the input contains the size of the array.
The next line contains N (white-space separated) integers denoting the elements of the array.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>4</sup>
-10<sup>7</sup> ≤ A[i] ≤10<sup>7</sup>For each test case, output one integer representing the maximum value of the sum that can be obtained using the various elements of the array.Input 1:
5
1 2 1 -1 1
output 1:
5
input 2:
5
0 0 -1 0 0
output 2:
0
<b>Explanation 1:</b>
In order to maximize the sum among [ 1 2 1 -1 1], we need to only consider [ 1 2 1 1] and neglect the [-1]., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
int n = Integer.parseInt(br.readLine());
String str[] = br.readLine().split(" ");
long arr[] = new long[n];
long sum=0;
for(int i=0;i<n;i++){
arr[i] = Integer.parseInt(str[i]);
if(arr[i]>0){
sum+=arr[i];
}
}
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary sorted non-increasing array arr of size <b>N</b>. You need to print the count of <b>1's</b> in the binary array.
Try to solve the problem using binary searchThe input line contains T, denotes the number of testcases.
Each test case contains two lines. The first line contains N (size of binary array). The second line contains N elements of binary array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^6
arr[i] = 0,1
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase in new line, print the count 1's in binary array.Input:
2
8
1 1 1 1 1 0 0 0
8
1 1 0 0 0 0 0 0
Output:
5
2
Explanation:
Testcase 1: Number of 1's in given binary array : 1 1 1 1 1 0 0 0 is 5.
Testcase 2: Number of 1's in given binary array : 1 1 0 0 0 0 0 0 is 2., 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 t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
int[] a=new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int count=search(a,0,n-1);
System.out.println(count);
}
}
public static int search(int[] a,int l,int h){
while(l<=h){
int mid=l+(h-l)/2;
if ((mid==h||a[mid+1]==0)&&(a[mid]==1))
return mid+1;
if (a[mid]==1)
l=mid+1;
else h=mid-1;
}
return 0;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary sorted non-increasing array arr of size <b>N</b>. You need to print the count of <b>1's</b> in the binary array.
Try to solve the problem using binary searchThe input line contains T, denotes the number of testcases.
Each test case contains two lines. The first line contains N (size of binary array). The second line contains N elements of binary array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^6
arr[i] = 0,1
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase in new line, print the count 1's in binary array.Input:
2
8
1 1 1 1 1 0 0 0
8
1 1 0 0 0 0 0 0
Output:
5
2
Explanation:
Testcase 1: Number of 1's in given binary array : 1 1 1 1 1 0 0 0 is 5.
Testcase 2: Number of 1's in given binary array : 1 1 0 0 0 0 0 0 is 2., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
int l = 0, h = n+1;
while(l+1 < h){
int m = (l + h) >> 1;
if(a[m] == 1)
l = m;
else
h = m;
}
cout << l << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary sorted non-increasing array arr of size <b>N</b>. You need to print the count of <b>1's</b> in the binary array.
Try to solve the problem using binary searchThe input line contains T, denotes the number of testcases.
Each test case contains two lines. The first line contains N (size of binary array). The second line contains N elements of binary array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^6
arr[i] = 0,1
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase in new line, print the count 1's in binary array.Input:
2
8
1 1 1 1 1 0 0 0
8
1 1 0 0 0 0 0 0
Output:
5
2
Explanation:
Testcase 1: Number of 1's in given binary array : 1 1 1 1 1 0 0 0 is 5.
Testcase 2: Number of 1's in given binary array : 1 1 0 0 0 0 0 0 is 2., I have written this Solution Code: c=int(input())
for x in range(c):
size=int(input())
s=input()
print(s.count('1')), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: static int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: def mythOrFact(N):
prime=1
cnt=N+1
for i in range (2,N):
if N%i==0:
prime=0
cnt=cnt+i
x = 3*N
x=x//2
if(cnt <= x and prime==1) or (cnt>x and prime==0):
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a contest containing 2 problems A and B. 2 strong participants P and Q participated in the contest and solved both the problems. P made AC submissions on problems A and B at time instants P<sub>A</sub> and P<sub>B</sub> respectively while Q made AC submissions on problems A and B at time instants Q<sub>A</sub> and Q<sub>B</sub>. It is given that the time penalty is the minimum time instant at which a participant has solved both the problems. Also the participant with the lower time penalty will have a better rank. Determine which participant got the better rank or if there is a TIE.The first line will contain T, number of test cases. Then the test cases follow. Each test case contains a single line of input, four integers P<sub>A</sub>, P<sub>B</sub>, Q<sub>A</sub>, Q<sub>B</sub>.
<b>Constraints</b>
1 ≤ T ≤ 1000
1 ≤ P<sub>A</sub>, P<sub>B</sub>, Q<sub>A</sub>, Q<sub>B</sub> ≤ 100For each test case, output P if P got a better rank, Q if Q got a better rank, TIE otherwise.Sample Input :
4
5 10 2 12
10 30 15 15
20 8 4 20
6 6 6 6
Sample Output :
P
Q
TIE
TIE
Explanation :
<ul><li>Time penalty incurred by participant P = 10. Time penalty incurred by participant Q = 12. Since 10 < 12, P gets a better rank. </li><li>Time penalty incurred by participant P = 30. Time penalty incurred by participant Q = 15. Since 15 < 30, Q gets a better rank. </li><li>Time penalty incurred by participant P = 20. Time penalty incurred by participant Q = 20. Since 20 = 20, P and Q gets a same rank (TIE). </li><li>Time penalty incurred by participant P = 6. Time penalty incurred by participant Q = 6. Since 6 = 6, P and Q gets a same rank (TIE). </li></ul>, I have written this Solution Code: #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t random_long(long long l = LLONG_MIN, long long r = LLONG_MAX) {
uniform_int_distribution<int64_t> generator(l, r);
return generator(rng);
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
auto start = std::chrono::high_resolution_clock::now();
int t;
cin >> t;
while (t--) {
int pa, pb, qa, qb;
cin >> pa >> pb >> qa >> qb;
int A = max(pa, pb), B = max(qa, qb);
if (A < B)
printf("P\n");
else if (A > B)
printf("Q\n");
else
printf("TIE\n");
}
auto stop = std::chrono::high_resolution_clock::now();
auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);
cerr << "Time taken : " << ((long double)duration.count()) / ((long double)1e9) << "s\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After saving the planet of Xander in 2014 from Ronan, the guardians of the galaxy decided to play a game on it. Between the vast furious oceans, they discover a row of buildings. The buildings are of different heights arranged at each index along a number line. Gamora starts at building 0 and a height of 0. Groot gets the task to <b>determine the minimum thrust that Gamora needs at the start of the game so that she can jump to the top of each building without her thrust going below zero</b>. Write a code to help Groot with his task.
The units of height relate directly to units of thrust. Gamora’s thrust level is calculated as follows:
• If gamora's thrust is less than the height of the building, her new thrust = gamora's thrust – (height – gamora's thrust)
• If gamora's thrust is greater than equal to the height of the building, her new thrust = gamora's thrust + (gamora's thrust - height)The first line contains an integer n, the number of buildings.
The second line n space-separated integers, arr[1], arr[2]…arr[n], the heights of the building.
<b>Constraints:-</b>
1 ≤ n ≤ 10<sup>5</sup>
1 ≤ arr[i] ≤ 10<sup>5</sup>The output contains a single integer denoting the minimum thrust Gamora needs at the start of the game.Sample Input:-
5
3 4 3 2 4
Sample Output:-
4
Explanation:-
If we take 3 then:-
at index 1:- 3 + 3-3 = 3
at index 2:- 3 - (4-3) = 2
at index 3:- 2 - (3-2) = 1
at index 4:- 1 - (2-1) = 0
Sample Input:-
3
4 4 4
Sample Output:-
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
FastReader sc = new FastReader();
int n = sc.nextInt();
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i]= sc.nextLong();
}
long l = 0;
long r= 1000000000;
long x=0;
while (l!=r){
x= (l+r)/2;
if(checkThrust(arr,x)) {
r=x;
}
else {
l=x+1;
}
}
System.out.print(l);
}
static boolean checkThrust(long[] arr,long r){
long thrust = r;
for (int i = 0; i < arr.length; i++) {
thrust = 2*thrust-arr[i];
if(thrust>=1000000000000L) return true;
if(thrust<0) return false;
}
return true;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After saving the planet of Xander in 2014 from Ronan, the guardians of the galaxy decided to play a game on it. Between the vast furious oceans, they discover a row of buildings. The buildings are of different heights arranged at each index along a number line. Gamora starts at building 0 and a height of 0. Groot gets the task to <b>determine the minimum thrust that Gamora needs at the start of the game so that she can jump to the top of each building without her thrust going below zero</b>. Write a code to help Groot with his task.
The units of height relate directly to units of thrust. Gamora’s thrust level is calculated as follows:
• If gamora's thrust is less than the height of the building, her new thrust = gamora's thrust – (height – gamora's thrust)
• If gamora's thrust is greater than equal to the height of the building, her new thrust = gamora's thrust + (gamora's thrust - height)The first line contains an integer n, the number of buildings.
The second line n space-separated integers, arr[1], arr[2]…arr[n], the heights of the building.
<b>Constraints:-</b>
1 ≤ n ≤ 10<sup>5</sup>
1 ≤ arr[i] ≤ 10<sup>5</sup>The output contains a single integer denoting the minimum thrust Gamora needs at the start of the game.Sample Input:-
5
3 4 3 2 4
Sample Output:-
4
Explanation:-
If we take 3 then:-
at index 1:- 3 + 3-3 = 3
at index 2:- 3 - (4-3) = 2
at index 3:- 2 - (3-2) = 1
at index 4:- 1 - (2-1) = 0
Sample Input:-
3
4 4 4
Sample Output:-
4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
const long long linf = 0x3f3f3f3f3f3f3f3fLL;
const int N = 111111;
int n;
int h[N];
int check(long long x) {
long long energy = x;
for(int i = 1; i <= n; i++) {
energy += energy - h[i];
if(energy >= linf) return 1;
if(energy < 0) return 0;
}
return 1;
}
int main() {
cin >> n;
for(int i = 1; i <= n; i++) {
cin >> h[i];
}
long long L = 0, R = linf;
long long ans=0;
while(L < R) {
long long M = (L + R) / 2;
if(check(M)) {
R = M;
ans=M;
} else {
L = M + 1;
}
}
cout << ans << endl;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is even or odd using switch case.First Line of the input contains the number n.
<b>Constraints</b>
1 <= n <= 1e9If the number is even print "Even" otherwise "Odd"Sample Input :
23
Sample Output :
Odd
Sample Input :
24
Sample Output :
Even, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t= Integer.parseInt(br.readLine());
if(t%2==0)
System.out.println("Even");
else
System.out.println("Odd");
}
catch (Exception e){
return;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is even or odd using switch case.First Line of the input contains the number n.
<b>Constraints</b>
1 <= n <= 1e9If the number is even print "Even" otherwise "Odd"Sample Input :
23
Sample Output :
Odd
Sample Input :
24
Sample Output :
Even, I have written this Solution Code: n = int(input())
if n % 2 == 0:
print("Even")
else:
print("Odd"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is even or odd using switch case.First Line of the input contains the number n.
<b>Constraints</b>
1 <= n <= 1e9If the number is even print "Even" otherwise "Odd"Sample Input :
23
Sample Output :
Odd
Sample Input :
24
Sample Output :
Even, I have written this Solution Code: #include <stdio.h>
int main()
{
int num;
scanf("%d", &num);
switch(num % 2)
{
case 0:
printf("Even");
break;
/* Else if n%2 == 1 */
case 1:
printf("Odd");
break;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a grid having N*M elements. Find the index of the row (1-base indexing) which has the maximum sum of elements.
Note: if more than two rows have the same sum of elements, then return the smallest index.The first line of the input contains two integers N and M.
The next N lines each contain M space separated integers.
<b>Constraints:</b>
1 <= N, M <= 10<sup>3</sup>
1 <= A<sub>i, j</sub> <= 10<sup>9Print the index of the row (1-base indexing) which has the maximum sum of elements.Sample Input:
4 3
3 4 2
5 1 7
2 8 1
2 3 3
Sample Output:
2
<b>Explaination:</b>
Row number 2 has sum = 5 + 1 + 7 = 13 which is the maximum among all rows., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) {
FastReader sc=new FastReader();
int n=sc.nextInt();
int m=sc.nextInt();
int arr[][]=new int[n][m];
for(int i=0; i<n; i++)
{
for(int j=0; j<m; j++)
{
arr[i][j]= sc.nextInt();
}
}
long maxsum=0;
int index=0;
for(int i=0; i<n; i++)
{
long sum=0;
for(int j=0; j<m; j++)
{
sum += arr[i][j];
}
if(sum > maxsum)
{
maxsum=sum;
index=i+1;
}
}
System.out.print(index);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a grid having N*M elements. Find the index of the row (1-base indexing) which has the maximum sum of elements.
Note: if more than two rows have the same sum of elements, then return the smallest index.The first line of the input contains two integers N and M.
The next N lines each contain M space separated integers.
<b>Constraints:</b>
1 <= N, M <= 10<sup>3</sup>
1 <= A<sub>i, j</sub> <= 10<sup>9Print the index of the row (1-base indexing) which has the maximum sum of elements.Sample Input:
4 3
3 4 2
5 1 7
2 8 1
2 3 3
Sample Output:
2
<b>Explaination:</b>
Row number 2 has sum = 5 + 1 + 7 = 13 which is the maximum among all rows., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n, m;
cin >> n >> m;
vector<vector<int>> a(n, vector<int> (m));
vector<int> sum(n);
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
cin >> a[i][j];
sum[i] += a[i][j];
}
}
int mx = 0, ans = 0;
for(int i = 0; i < n; i++){
if(mx < sum[i]){
mx = sum[i];
ans = i;
}
}
cout << ans + 1;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
String [] str=br.readLine().trim().split(" ");
int a[]=new int[n];
for(int i=0;i<n;i++){
a[i]=Integer.parseInt(str[i]);
}
Arrays.sort(a);
int size=a[n-1]+1;
int c[]=new int[size];
for(int i=0;i<size;i++) c[i]=0;
for(int i=0;i<n;i++) c[a[i]]++;
int max=0,freq=c[1];
for(int i=2;i<size;i++){
if(freq<=c[i]){
freq=c[i];
max=i;
}
}
System.out.println(max);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=2, I have written this Solution Code: n = int(input())
a = [int(x) for x in input().split()]
freq = {}
for x in a:
if x not in freq:
freq[x] = 1
else:
freq[x] += 1
mx = max(freq.values())
rf = sorted(freq)
for i in range(len(rf) - 1, -1, -1):
if freq[rf[i]] == mx:
print(rf[i])
break, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=2, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int n; cin >> n;
for(int i = 1; i <= n; i++){
int p; cin >> p;
a[p]++;
}
int mx = 0, id = -1;
for(int i = 1; i <= 100; i++){
if(a[i] >= mx)
mx = a[i], id = i;
}
cout << id;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is learning bits manipulation today and she is stuck on a problem and asks for your help.
Problem:-
Given an integer N. Your task is to change the integer to 0 by performing the following operations:-
1. Change the rightmost digit(0th index). i. e you can change the rightmost digit to 1 if it is 0 or 0 if it is 1.
2. Change the ith bit if (i-1)th bit is 1 and all bits right to (i-1) is 0. For eg:- you can change the left most bit of these numbers:- 1100, 110, 110000.
Your task is to find the minimum operation to make the number 0.The input contains a single integer N.
Constraints:-
0 <= N <= 1000000000Print the minimum number of operations requiredSample Input:-
3
Sample Output:-
2
Explanation:-
3- >11(binary representation)
01;- Change the leftmost bit using the second operation
00:- Change the rightmost bit using the first operation
Sample Input:-
14
Sample Output:-
11
Explanation:-
14 - > 1110
1111 - > using 1st operation
1101 - > using 2nd operation
1100 - > using 1st operation
0100 - > using 2nd operation
0101 - > using 1st operation
0111 - > using 2nd operation
0110 - > using 1st operation
0010 - > using 2nd operation
0011 - > using 1st operation
0001 - > using 2nd operation
0000 - > using 1st operation, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static int minimumOneBitOperations(int n, int res)
{
if (n == 0)
return res;
int b = 1;
while ((b << 1) <= n)
b = b << 1;
return minimumOneBitOperations((b >> 1) ^ b ^ n, res + b);
}
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
System.out.println(minimumOneBitOperations(N, 0));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is learning bits manipulation today and she is stuck on a problem and asks for your help.
Problem:-
Given an integer N. Your task is to change the integer to 0 by performing the following operations:-
1. Change the rightmost digit(0th index). i. e you can change the rightmost digit to 1 if it is 0 or 0 if it is 1.
2. Change the ith bit if (i-1)th bit is 1 and all bits right to (i-1) is 0. For eg:- you can change the left most bit of these numbers:- 1100, 110, 110000.
Your task is to find the minimum operation to make the number 0.The input contains a single integer N.
Constraints:-
0 <= N <= 1000000000Print the minimum number of operations requiredSample Input:-
3
Sample Output:-
2
Explanation:-
3- >11(binary representation)
01;- Change the leftmost bit using the second operation
00:- Change the rightmost bit using the first operation
Sample Input:-
14
Sample Output:-
11
Explanation:-
14 - > 1110
1111 - > using 1st operation
1101 - > using 2nd operation
1100 - > using 1st operation
0100 - > using 2nd operation
0101 - > using 1st operation
0111 - > using 2nd operation
0110 - > using 1st operation
0010 - > using 2nd operation
0011 - > using 1st operation
0001 - > using 2nd operation
0000 - > using 1st operation, I have written this Solution Code: def mini(n, res):
if n == 0:
return res
b = 1
while (b << 1) <= n:
b = b << 1
return mini((b >> 1) ^ b ^ n, res + b)
n=int(input())
print(mini(n, 0)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is learning bits manipulation today and she is stuck on a problem and asks for your help.
Problem:-
Given an integer N. Your task is to change the integer to 0 by performing the following operations:-
1. Change the rightmost digit(0th index). i. e you can change the rightmost digit to 1 if it is 0 or 0 if it is 1.
2. Change the ith bit if (i-1)th bit is 1 and all bits right to (i-1) is 0. For eg:- you can change the left most bit of these numbers:- 1100, 110, 110000.
Your task is to find the minimum operation to make the number 0.The input contains a single integer N.
Constraints:-
0 <= N <= 1000000000Print the minimum number of operations requiredSample Input:-
3
Sample Output:-
2
Explanation:-
3- >11(binary representation)
01;- Change the leftmost bit using the second operation
00:- Change the rightmost bit using the first operation
Sample Input:-
14
Sample Output:-
11
Explanation:-
14 - > 1110
1111 - > using 1st operation
1101 - > using 2nd operation
1100 - > using 1st operation
0100 - > using 2nd operation
0101 - > using 1st operation
0111 - > using 2nd operation
0110 - > using 1st operation
0010 - > using 2nd operation
0011 - > using 1st operation
0001 - > using 2nd operation
0000 - > using 1st operation, 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 101
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int solve(int n){
if(n==1){
return 1;
}
if(n==0){
return 0;
}
int ans=1;
int cnt=0;
while(ans<n){
ans*=2;
cnt++;
}
ans/=2;
ans+=ans/2;
return solve(ans^n)+pow(2,cnt-1);
}
signed main(){
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 a singly linked list and an element K, your task is to insert the element at the tail of the linked list.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>addElement()</b> that takes head node and the integer K as a parameter.
Constraints:
1 <=N<= 1000
1 <=K, value<= 1000Return the head of the modified linked listSample Input:-
5 2
1 2 3 4 5
Sample Output:
1 2 3 4 5 2
, I have written this Solution Code: a,b=[int(x) for x in input().split()]
c=[int(x) for x in input().split()]
for i in c:
print(i,end=" ")
print(b), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a singly linked list and an element K, your task is to insert the element at the tail of the linked list.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>addElement()</b> that takes head node and the integer K as a parameter.
Constraints:
1 <=N<= 1000
1 <=K, value<= 1000Return the head of the modified linked listSample Input:-
5 2
1 2 3 4 5
Sample Output:
1 2 3 4 5 2
, I have written this Solution Code: public static Node addElement(Node head,int k) {
Node temp=head;
while(temp.next!=null){
temp=temp.next;}
Node x= new Node(k);
temp.next = x;
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the first 2 terms A and B of an Arithmetic Series, tell the Nth term of the series.<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>NthAP()</b> that takes the integer A, B, and N as a parameter.
<b>Constraints:</b>
-10<sup>3</sup> ≤ A ≤ 10<sup>3</sup>
-10<sup>3</sup> ≤ B ≤ 10<sup>3</sup>
1 ≤ N ≤ 10<sup>4</sup>Return the Nth term of AP series.Sample Input 1:
2 3 4
Sample Output 1:
5
Sample Input 2:
1 2 10
Sample output 2:
10, I have written this Solution Code: class Solution {
public static int NthAP(int a, int b, int n){
return a+(n-1)*(b-a);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. Find the number of pairs of indices (i, j) in the array A such that i < j and A<sub>i</sub> - i = A<sub>j</sub> - j.The first line of the input contains a single integer N.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= A<sub>i</sub> <= 10<sup>5</sup>Print the number of pairs of indices (i, j) in the given array A such that i < j and A<sub>i</sub> - i = A<sub>j</sub> - j.Sample Input:
4
1 3 3 4
Sample Output:
3
Explaination:
The three pairs of indices are:
(1, 3) -> A[1] - 1 = A[3] - 3 -> 1 - 1 = 3 - 3 -> 0 = 0
(1, 4) -> A[1] - 1 = A[4] - 4 -> 1 - 1 = 4 - 4 -> 0 = 0
(3, 4) -> A[3] - 3 = A[4] - 4 -> 3 - 3 = 4 - 4 -> 0 = 0, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
signed main() {
int n;
cin >> n;
map<int, int> mp;
int ans = 0;
for(int i = 1; i <= n; ++i){
int x;
cin >> x;
mp[x - i]++;
}
for(auto i:mp)
ans += i.second * (i.second - 1)/2;
cout << ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.First line consists of a single integer denoting n.
Constraint:-
1<=size of n<=100Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, 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();
int i=str.length()-1;
if(i==0){
int number=Integer.parseInt(str);
System.out.println(number);
}else{
while(str.charAt(i)=='0'){
i--;
}
for(int j=i;j>=0;j--){
System.out.print(str.charAt(j));
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.First line consists of a single integer denoting n.
Constraint:-
1<=size of n<=100Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, I have written this Solution Code: n=int(input())
def reverse(n):
return int(str(n)[::-1])
print(reverse(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.First line consists of a single integer denoting n.
Constraint:-
1<=size of n<=100Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
cin>>s;
reverse(s.begin(),s.end());
int I;
for( I=0;I<s.length();I++){
if(s[I]!='0'){break;}
}
if(I==s.length()){cout<<0;return 0;}
for(int j=I;j<s.length();j++){
cout<<s[j];}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For an integer N, your task is to calculate sum of first N natural numbers.<b>User Task:</b>
Since this will be a functional problem, you don't have to worry about input. You just have to complete the function <b>sum()</b> which takes the integer N as a parameter.
Constraints:
1 <= N < = 100000000Print the sum of first N natural numbers.Sample Input:-
5
Sample Output:-
15
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code:
static void sum(int N){
long x=N;
x=x*(x+1);
x=x/2;
System.out.print(x);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N piles where the i<sup>th</sup> pile consists of A<sub>i</sub> stones. Chef and Chefina are playing a game taking alternate turns with Chef starting first. In his/her turn, a player can choose any non-empty pile and remove exactly 1 stone from it. The game ends when exactly 1 pile becomes empty. The player who made the last move wins. Determine the winner if both players play optimally.The first line of input will contain a single integer T, denoting the number of test cases. Each test case consists of multiple lines of input. The first line of each test case contains a single integer N denoting the number of piles. The next line contains N space-separated integers A<sub>1</sub>, A<sub>2</sub>,...., A<sub>N</sub> - denoting the number of stones in each pile.
<b>Constraints</b>
1 ≤ T ≤ 1000
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>9</sup>
The Sum of N over all test cases does not exceed 2. 10<sup>5</sup>.For each test case, output CHEF if Chef wins the game, otherwise output CHEFINA. Sample Input :
3
2
2 2
1
10
3
1 5 6
Sample Output :
CHEFINA
CHEFINA
CHEF
Explanation :
No matter which stone the Chef removes, Chefina will remove the remaining stone from the same pile and make it empty.
Since there is only 1 pile containing 1010 stones, the game will last exactly 10 moves with last move made by Chefina.
The chef can remove 1 stone from the first pile and thus make it empty., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
const int MX = 1e5 + 5;
typedef long long LL;
LL a[MX], T, n, ans;
int main () {
ios::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
cin >> T;
while (T --) {
bool flag = 0; ans = 0;
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = 0; i < n; i++)
if (a[i] == 1) flag = 1;
if (flag == 1) {
cout << "CHEF\n";
continue;
}
for (int i = 0; i < n; i++) ans += abs(a[i] - 2);
if (ans % 2 == 1) cout << "CHEF\n";
else cout << "CHEFINA\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andrew is now bored with usual prime numbers and decides to find prime numbers which has exactly three factors. He sits back and keeps on thinking random numbers, and asks you to tell him if the number he has chosen had exactly 3 factors. Now he has given <b>N</b> random numbers, you need to tell him "<b>Yes</b>" if those numbers had exactly 3 factors otherwise "<b>No</b>".
<b>Note:</b> You can consider the number to be factor of its own.First-line contains N, the number of random numbers he thinks of. Second-line contains the N numbers separated by space.
<b>Constraints:</b>
1 <= N <= 10^5
1 <= numbers <= 10^12Print N lines containing "Yes" or "No".Sample Input 1
3
2 4 6
Sample Output 1
No
Yes
No
Explanation:
2 has 2 factors: 1, 2
4 has 3 factors: 1,2,4
6 has 4 factors: 1,2,3,6
Sample Input 2
3
3 5 7
Sample Output 2
No
No
No, I have written this Solution Code: from math import sqrt
def isPrime(n):
if (n <= 1):
return False
if (n <= 3):
return True
if (n % 2 == 0 or n % 3 == 0):
return False
k= int(sqrt(n))+1
for i in range(5,k,6):
if (n % i == 0 or n % (i + 2) == 0):
return False
return True
def isThreeDisctFactors(n):
sq = int(sqrt(n))
if (1 * sq * sq != n):
return False
if (isPrime(sq)):
return True
else:
return False
N = int(input())
numbers = input().split()
for x in numbers:
if isThreeDisctFactors(int(x)):
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andrew is now bored with usual prime numbers and decides to find prime numbers which has exactly three factors. He sits back and keeps on thinking random numbers, and asks you to tell him if the number he has chosen had exactly 3 factors. Now he has given <b>N</b> random numbers, you need to tell him "<b>Yes</b>" if those numbers had exactly 3 factors otherwise "<b>No</b>".
<b>Note:</b> You can consider the number to be factor of its own.First-line contains N, the number of random numbers he thinks of. Second-line contains the N numbers separated by space.
<b>Constraints:</b>
1 <= N <= 10^5
1 <= numbers <= 10^12Print N lines containing "Yes" or "No".Sample Input 1
3
2 4 6
Sample Output 1
No
Yes
No
Explanation:
2 has 2 factors: 1, 2
4 has 3 factors: 1,2,4
6 has 4 factors: 1,2,3,6
Sample Input 2
3
3 5 7
Sample Output 2
No
No
No, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
static int MAX = 1000005;
static long spf[] = new long[MAX+5];
public static void main (String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(read.readLine());
SOF();
String str[] = read.readLine().trim().split(" ");
for(int i = 0; i < N; i++)
{
long x = Long.parseLong(str[i]);
long g = (long)Math.sqrt(x);
if((g*g)!=x)
{
System.out.println("No");
}
else
{
if(spf[(int)g]==g)
{
System.out.println("Yes");
}
else
System.out.println("No");
}
}
}
public static void SOF()
{
spf[1] = 0;
for (int i=2; i<MAX; i++)
// marking smallest prime factor for every
// number to be itself.
spf[i] = i;
// separately marking spf for every even
// number as 2
for (int i=4; i<MAX; i+=2)
spf[i] = 2;
for (int i=3; i*i<MAX; i++)
{
// checking if i is prime
if (spf[i] == i)
{
// marking SPF for all numbers divisitedible by i
for (int j=i*i; j<MAX; j+=i)
// marking spf[j] if it is not
// previously marked
if (spf[j]==j)
spf[j] = i;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andrew is now bored with usual prime numbers and decides to find prime numbers which has exactly three factors. He sits back and keeps on thinking random numbers, and asks you to tell him if the number he has chosen had exactly 3 factors. Now he has given <b>N</b> random numbers, you need to tell him "<b>Yes</b>" if those numbers had exactly 3 factors otherwise "<b>No</b>".
<b>Note:</b> You can consider the number to be factor of its own.First-line contains N, the number of random numbers he thinks of. Second-line contains the N numbers separated by space.
<b>Constraints:</b>
1 <= N <= 10^5
1 <= numbers <= 10^12Print N lines containing "Yes" or "No".Sample Input 1
3
2 4 6
Sample Output 1
No
Yes
No
Explanation:
2 has 2 factors: 1, 2
4 has 3 factors: 1,2,4
6 has 4 factors: 1,2,3,6
Sample Input 2
3
3 5 7
Sample Output 2
No
No
No, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 1000001
bool a[max1];
signed main()
{
string s="abcdefghijklmnopqrstuvwxyz";
for(int i=0;i<=1000000;i++){
a[i]=false;
}
for(int i=2;i<1000000;i++){
if(a[i]==false){
for(int j=i+i;j<=1000000;j+=i){
a[j]=true;
}
}
}
int n;
cin>>n;
long long x;
long long y;
for(int i=0;i<n;i++){
cin>>x;
long long y=sqrt(x);
if(y*y==x){
if(a[y]==false){cout<<"Yes"<<endl;}
else{cout<<"No"<<endl;}
}
else{
cout<<"No"<<endl;
}
}}
, 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: You are given N intervals, the i<sup>th</sup> of them being [A<sub>i</sub>, B<sub>i</sub>], where A<sub>i</sub> and B<sub>i</sub> are positive integers. Let the union of all these intervals be S. It is easy to see that S can be uniquely represented as an union of disjoint closed intervals. Your task is to find the sum of the lengths of the disjoint closed intervals that comprises S.
For example, if you are given the intervals: [1, 3], [2, 4], [5, 7] and [7, 8], then S can be uniquely represented as the union of disjoint intervals [1, 4] and [5, 8]. In this case, the answer will be 6, as (4 - 1) + (8 - 5) = 6.The first line of the input consists of a single integer N – the number of intervals.
Then N lines follow, the i<sup>th</sup> line containing two space-separated integers A<sub>i</sub> and B<sub>i</sub>.
<b> Constraints: </b>
1 ≤ N ≤ 10<sup>4</sup>
1 ≤ A<sub>i</sub> < B<sub>i</sub> ≤ 2×10<sup>5</sup>Print a single integer, the sum of the lengths of the disjoint intervals of S.Sample Input 1:
3
1 3
3 4
6 7
Sample Output 1:
4
Sample Explanation 1:
Here, S can be uniquely written as union of disjoint intervals [1, 4] and [6, 7]. Thus, answer is (4 - 1) + (7 - 6) = 4.
Sample Input 2:
4
9 12
8 10
5 6
13 15
Sample Output 2:
7
Sample Explanation 2:
Here, S can be uniquely written as union of disjoint intervals [5, 6], [8, 12] and [13, 15]. Thus, answer is (6 - 5) + (12 - 8) + (15 - 13) = 7.
, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << (a) << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (auto i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 998244353; // 1e9 + 7;
const ll N = 4e5 + 100;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int arr[N];
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
read(n);
FOR (i, 1, n)
{
readb(a, b);
a *= 2, b *= 2;
arr[a]++;
arr[b + 1]--;
}
int ans = 0, sum = 0, cur = 0;
FOR (i, 0, N - 2)
{
sum += arr[i];
if (sum) cur++;
else ans += cur/2, cur = 0;
}
print(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, I have written this Solution Code: m,n=map(int ,input().split())
matrix=[]
for i in range(m):
l1=[eval(x) for x in input().split()]
matrix.append(l1)
l2=[]
for coloumn in range(n):
sum1=0
for row in range(m):
sum1+= matrix[row][coloumn]
l2.append(sum1)
print(max(l2))
'''for row in range(n):
sum2=0
for col in range(m):
sum2 += matrix[row][col]
print(sum2)''', In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, I have written this Solution Code: // mat is the matrix/ 2d array
// the dimensions of array are:- a rows, b columns
function colMaxSum(mat,a,b) {
// write code here
// do not console.log
// return the answer as a number
let idx = -1;
// Variable to store max sum
let maxSum = Number.MIN_VALUE;
// Traverse matrix column wise
for (let i = 0; i < b; i++) {
let sum = 0;
// calculate sum of column
for (let j = 0; j < a; j++) {
sum += mat[j][i];
}
// Update maxSum if it is
// less than current sum
if (sum > maxSum) {
maxSum = sum;
// store index
idx = i;
}
}
let res;
res = [idx, maxSum];
// return result
return maxSum;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main(){
int n,m;
cin>>n>>m;
int a[m];
for(int i=0;i<m;i++){
a[i]=0;
}
int x;
int sum=0;
FOR(i,n){
FOR(j,m){
cin>>x;
a[j]+=x;
sum=max(sum,a[j]);
}
}
out(sum);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, 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 m = sc.nextInt();
int n = sc.nextInt();
int a[][] = new int[m][n];
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
a[i][j]=sc.nextInt();
}
}
int sum=0;
int ans=0;
for(int i=0;i<n;i++){
sum=0;
for(int j=0;j<m;j++){
sum+=a[j][i];
}
if(sum>ans){ans=sum;}
}
System.out.print(ans);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: This is a time of conflict in Westeros as Viserys Targaryen, the king of all seven kingdoms, rejected the hand of Lady Laena Velaryon. So, the Velaryon soldiers are not deemed trustworthy anymore. The Targaryen soldiers have to keep an eye on them.
You are given a sequential order of N soldiers standing in a line. The order is provided as a binary string, with 0 representing Velaryon soldiers, and 1 representing the Targaryen soldiers. Viserys wants each contiguous segment of N/2 soldiers to contain an even number of Targaryen soldiers.
Formally you are given a binary string of length N, where N is an even natural number. Each character of the string is either '0' or '1'. You want to rearrange the elements of the string in such a way that the final string contains an even number of 1s in each contiguous subsegment of length N/2.
Your task is to find out whether there exists a rearrangement of the soldiers that satisfies the above conditions.The first line contains an integer T, the number of test cases. Then, T test cases follow.
The first line of each test case contains an even positive integer N, the length of the line.
The second line of each test case contains a binary string of length N, representing the current arrangement of soldiers.
<b> Constraints: </b>
1 ≤ T ≤ 10
2 ≤ N ≤ 10<sup>4</sup>
N is evenPrint a single character in a new line for each test case. Print '1' (without quotes) if a required rearrangement exists, and '0' (without quotes) otherwise.Sample Input:
3
2
10
2
00
4
0011
Sample Output:
0
1
0
(In the last case, no matter how you rearrange the string, there will always be a single one in at least one subsegment of length 2 of the string), I have written this Solution Code: import java.io.*;
import java.util.*;
class Main{
public static void main(String[] args)throws IOException
{
StringBuilder out=new StringBuilder();
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int test=Integer.parseInt(br.readLine());
while(test-->0)
{
int n=Integer.parseInt(br.readLine());
String s=br.readLine();
int c1=0;
for(int i=0;i<s.length();i++)
{
if(s.charAt(i)=='1') c1++;
}
if(c1%4==0) out.append("1\n");
else if(c1==s.length() && (c1/2)%2==0) out.append("1\n");
else
out.append("0\n");
}
System.out.print(out);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: This is a time of conflict in Westeros as Viserys Targaryen, the king of all seven kingdoms, rejected the hand of Lady Laena Velaryon. So, the Velaryon soldiers are not deemed trustworthy anymore. The Targaryen soldiers have to keep an eye on them.
You are given a sequential order of N soldiers standing in a line. The order is provided as a binary string, with 0 representing Velaryon soldiers, and 1 representing the Targaryen soldiers. Viserys wants each contiguous segment of N/2 soldiers to contain an even number of Targaryen soldiers.
Formally you are given a binary string of length N, where N is an even natural number. Each character of the string is either '0' or '1'. You want to rearrange the elements of the string in such a way that the final string contains an even number of 1s in each contiguous subsegment of length N/2.
Your task is to find out whether there exists a rearrangement of the soldiers that satisfies the above conditions.The first line contains an integer T, the number of test cases. Then, T test cases follow.
The first line of each test case contains an even positive integer N, the length of the line.
The second line of each test case contains a binary string of length N, representing the current arrangement of soldiers.
<b> Constraints: </b>
1 ≤ T ≤ 10
2 ≤ N ≤ 10<sup>4</sup>
N is evenPrint a single character in a new line for each test case. Print '1' (without quotes) if a required rearrangement exists, and '0' (without quotes) otherwise.Sample Input:
3
2
10
2
00
4
0011
Sample Output:
0
1
0
(In the last case, no matter how you rearrange the string, there will always be a single one in at least one subsegment of length 2 of the string), I have written this Solution Code: T=int(input())
for i in range(T):
n=int(input())
a=input()
count_1=0
for i in a:
if i=='1':
count_1+=1
if count_1%2==0 and ((count_1)//2)%2==0:
print('1')
else:
print('0'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: This is a time of conflict in Westeros as Viserys Targaryen, the king of all seven kingdoms, rejected the hand of Lady Laena Velaryon. So, the Velaryon soldiers are not deemed trustworthy anymore. The Targaryen soldiers have to keep an eye on them.
You are given a sequential order of N soldiers standing in a line. The order is provided as a binary string, with 0 representing Velaryon soldiers, and 1 representing the Targaryen soldiers. Viserys wants each contiguous segment of N/2 soldiers to contain an even number of Targaryen soldiers.
Formally you are given a binary string of length N, where N is an even natural number. Each character of the string is either '0' or '1'. You want to rearrange the elements of the string in such a way that the final string contains an even number of 1s in each contiguous subsegment of length N/2.
Your task is to find out whether there exists a rearrangement of the soldiers that satisfies the above conditions.The first line contains an integer T, the number of test cases. Then, T test cases follow.
The first line of each test case contains an even positive integer N, the length of the line.
The second line of each test case contains a binary string of length N, representing the current arrangement of soldiers.
<b> Constraints: </b>
1 ≤ T ≤ 10
2 ≤ N ≤ 10<sup>4</sup>
N is evenPrint a single character in a new line for each test case. Print '1' (without quotes) if a required rearrangement exists, and '0' (without quotes) otherwise.Sample Input:
3
2
10
2
00
4
0011
Sample Output:
0
1
0
(In the last case, no matter how you rearrange the string, there will always be a single one in at least one subsegment of length 2 of the string), I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
signed main() {
int t;
cin >> t;
for(int i=0; i<t; i++) {
int n;
cin >> n;
string s;
cin >> s;
int cnt = 0;
for(int j=0; j<n; j++) {
if(s[j] == '1') cnt++;
}
if(cnt % 4 == 0) cout << 1 << "\n";
else cout << 0 << "\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 integers, an integer X, and Q queries. For each query, you will be given an integer K. Your task is to find the Kth occurrence of the number X in the array. If the Kth occurrence does not exist print -1.The first line of input contains three integers N, X, and Q. The second line of input contains N space- separated integers. The Next Q lines of input contain a single integer each representing K.
Constraints:-
1 <= Q, K, N <= 100000
1 <= X, Arr[] <= 100000For each query in a new line print the index of the Kth occurrence. If the Kth occurrence does not exist print -1.Sample Input:-
10 3 4
1 2 3 4 2 3 4 5 2 3
1
2
3
4
Sample Output:_
3
6
10
-1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader inp = new BufferedReader(new InputStreamReader(System.in));
String[] NXQ = inp.readLine().split("\\s+");
int N = Integer.parseInt(NXQ[0]);
int X = Integer.parseInt(NXQ[1]);
int Q = Integer.parseInt(NXQ[2]);
String[] StrArr = inp.readLine().split("\\s+");
int[] arr = new int[N];
ArrayList<Integer> posX = new ArrayList<>();
int j = 0;
for (int i = 0; i < N; i++) {
arr[i] = Integer.parseInt(StrArr[i]);
if (arr[i] == X) {
posX.add(i + 1);
}
}
for (int i = 0; i < Q; i++) {
int query = Integer.parseInt(inp.readLine());
if (query > posX.size()) {
System.out.println("-1");
}
else {
System.out.println(posX.get(query - 1));
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, an integer X, and Q queries. For each query, you will be given an integer K. Your task is to find the Kth occurrence of the number X in the array. If the Kth occurrence does not exist print -1.The first line of input contains three integers N, X, and Q. The second line of input contains N space- separated integers. The Next Q lines of input contain a single integer each representing K.
Constraints:-
1 <= Q, K, N <= 100000
1 <= X, Arr[] <= 100000For each query in a new line print the index of the Kth occurrence. If the Kth occurrence does not exist print -1.Sample Input:-
10 3 4
1 2 3 4 2 3 4 5 2 3
1
2
3
4
Sample Output:_
3
6
10
-1, I have written this Solution Code: N, X , Q = map(int, input().split())
L = list(map(int, input().split()))
Queries = []
while Q > 0:
Queries.append(int(input()))
Q -= 1
indices = [i for i,x in enumerate(L) if x == X]
for query in Queries:
if len(indices)< query:
print(-1)
else:
print(indices[query-1]+1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, an integer X, and Q queries. For each query, you will be given an integer K. Your task is to find the Kth occurrence of the number X in the array. If the Kth occurrence does not exist print -1.The first line of input contains three integers N, X, and Q. The second line of input contains N space- separated integers. The Next Q lines of input contain a single integer each representing K.
Constraints:-
1 <= Q, K, N <= 100000
1 <= X, Arr[] <= 100000For each query in a new line print the index of the Kth occurrence. If the Kth occurrence does not exist print -1.Sample Input:-
10 3 4
1 2 3 4 2 3 4 5 2 3
1
2
3
4
Sample Output:_
3
6
10
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n,x,q;
cin>>n>>x>>q;
vector<int> a(n),v;
for(int i=0;i<n;i++){
cin>>a[i];
if(a[i]==x){
v.push_back(i+1);
}
}
while(q--){
cin>>x;
if(x<=v.size()){
cout<<v[x-1]<<'\n';
}
else{
cout<<-1<<'\n';
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers containing only 0 or 1. You can do the following operations on the array:
<ul><li>swap elements at two indices</li><li>choose one index and change its value from 0 to 1 or vice-versa.</li></ul>
You have to do the minimum number of the above operations such that the final array is non-decreasing.
<b>Note</b> Consider 1 based Array-indexingThe first line of input contains a single integer N.
The second line of input contains N space-separated integers denoting the array.
Constraints:
1 ≤ N ≤ 100000
elements of the array are 0 or 1.Minimum number of moves required such that the final array is non- decreasing.Sample Input 1
5
1 1 0 0 1
Sample Output 1
2
Explanation:
Swap indices (1, 3)
Swap indices (2, 4)
Sample Input 2
5
0 0 1 1 1
Sample Output 2
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));
int n = Integer.parseInt(br.readLine().trim());
String array[] = br.readLine().trim().split(" ");
boolean decreasingOrder = false;
int[] arr = new int[n];
int totalZeroCount = 0,
totalOneCount = 0;
for(int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(array[i]);
if(i != 0 && arr[i] < arr[i - 1])
decreasingOrder = true;
if(arr[i] % 2 == 0)
++totalZeroCount;
else
++totalOneCount;
}
if(!decreasingOrder) {
System.out.println("0");
} else {
int oneCount = 0;
for(int i = 0; i < totalZeroCount; i++) {
if(arr[i] == 1)
++oneCount;
}
System.out.println(oneCount);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers containing only 0 or 1. You can do the following operations on the array:
<ul><li>swap elements at two indices</li><li>choose one index and change its value from 0 to 1 or vice-versa.</li></ul>
You have to do the minimum number of the above operations such that the final array is non-decreasing.
<b>Note</b> Consider 1 based Array-indexingThe first line of input contains a single integer N.
The second line of input contains N space-separated integers denoting the array.
Constraints:
1 ≤ N ≤ 100000
elements of the array are 0 or 1.Minimum number of moves required such that the final array is non- decreasing.Sample Input 1
5
1 1 0 0 1
Sample Output 1
2
Explanation:
Swap indices (1, 3)
Swap indices (2, 4)
Sample Input 2
5
0 0 1 1 1
Sample Output 2
0, 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];
for(int i=0;i<n;++i){
cin>>a[i];
}
int cnt = 0;
for (int i = 0; i < n; i++) {
if (a[i]==0) cnt++;
}
int ans = 0;
for (int i = 0; i < cnt; i++) if (a[i] == 1) ans++;
cout<<ans;
#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: Given an array of N integers containing only 0 or 1. You can do the following operations on the array:
<ul><li>swap elements at two indices</li><li>choose one index and change its value from 0 to 1 or vice-versa.</li></ul>
You have to do the minimum number of the above operations such that the final array is non-decreasing.
<b>Note</b> Consider 1 based Array-indexingThe first line of input contains a single integer N.
The second line of input contains N space-separated integers denoting the array.
Constraints:
1 ≤ N ≤ 100000
elements of the array are 0 or 1.Minimum number of moves required such that the final array is non- decreasing.Sample Input 1
5
1 1 0 0 1
Sample Output 1
2
Explanation:
Swap indices (1, 3)
Swap indices (2, 4)
Sample Input 2
5
0 0 1 1 1
Sample Output 2
0, I have written this Solution Code: n=int(input())
l=list(map(int,input().split()))
x=l.count(0)
c=0
for i in range(0,x):
if(l[i]==1):
c+=1
print(c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A child is running up a staircase with n steps and can hop either 1 step, 2 steps or 3 steps at a time. Implement a method to count how many possible ways the child can run-up to the stairs. You need to return all possible number of ways.
The time complexity of your code should be O(n).
Print answer modulo 10^9 + 7.The input line contains T, denoting the number of testcases. Each testcase contains single line containing N, i.e. stairs having number of steps.
Constraints
1 <= T <= 10
1 <= N <= 10^5Find the total number of ways. Since the answer can be very large, thus print them as modulo 10^9 + 7.Sample Input
2
4
3
Sample output
7
4
Explanation:
Testcase 1:
In this case we have stairs with 4 steps. So we can figure out how many ways are there to reach at the top.
1st: (1, 1, 1, 1)
2nd: (1, 1, 2)
3rd: (1, 2, 1)
4th: (1, 3)
5th: (2, 1, 1)
6th: (2, 2)
7th: (3, 1), I have written this Solution Code:
dp = [1,2,4]
n = int(input())
for i in range(n):
k=int(input())
one = 1
two = 2
three = 4
four = 0
if k<4:
print(dp[k-1])
continue
for i in range(3,k):
four = one + two + three
one = two
two = three
three = four
print(four%(10**9+7)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A child is running up a staircase with n steps and can hop either 1 step, 2 steps or 3 steps at a time. Implement a method to count how many possible ways the child can run-up to the stairs. You need to return all possible number of ways.
The time complexity of your code should be O(n).
Print answer modulo 10^9 + 7.The input line contains T, denoting the number of testcases. Each testcase contains single line containing N, i.e. stairs having number of steps.
Constraints
1 <= T <= 10
1 <= N <= 10^5Find the total number of ways. Since the answer can be very large, thus print them as modulo 10^9 + 7.Sample Input
2
4
3
Sample output
7
4
Explanation:
Testcase 1:
In this case we have stairs with 4 steps. So we can figure out how many ways are there to reach at the top.
1st: (1, 1, 1, 1)
2nd: (1, 1, 2)
3rd: (1, 2, 1)
4th: (1, 3)
5th: (2, 1, 1)
6th: (2, 2)
7th: (3, 1), I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static long waysBottomUp(int n, long dp[]){
for(int i=4; i<=n; i++){
dp[i]= ((dp[i-1]%1000000007) + (dp[i-2]%1000000007) + (dp[i-3]%1000000007))%1000000007;
}
return dp[n];
}
public static void main (String[] args)throws IOException {
BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
long dp[] = new long[100001];
int t = Integer.parseInt(rd.readLine());
while(t-->0){
int n = Integer.parseInt(rd.readLine());
dp[0] =1;
dp[1] = 1;
dp[2] = 2;
dp[3] = 4;
System.out.println(waysBottomUp(n, dp)%1000000007);}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A child is running up a staircase with n steps and can hop either 1 step, 2 steps or 3 steps at a time. Implement a method to count how many possible ways the child can run-up to the stairs. You need to return all possible number of ways.
The time complexity of your code should be O(n).
Print answer modulo 10^9 + 7.The input line contains T, denoting the number of testcases. Each testcase contains single line containing N, i.e. stairs having number of steps.
Constraints
1 <= T <= 10
1 <= N <= 10^5Find the total number of ways. Since the answer can be very large, thus print them as modulo 10^9 + 7.Sample Input
2
4
3
Sample output
7
4
Explanation:
Testcase 1:
In this case we have stairs with 4 steps. So we can figure out how many ways are there to reach at the top.
1st: (1, 1, 1, 1)
2nd: (1, 1, 2)
3rd: (1, 2, 1)
4th: (1, 3)
5th: (2, 1, 1)
6th: (2, 2)
7th: (3, 1), I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
a[0] = 1;
for(int i = 1; i < N; i++){
if(i >= 1) a[i] += a[i-1];
if(i >= 2) a[i] += a[i-2];
if(i >= 3) a[i] += a[i-3];
a[i] %= mod;
}
while(t--){
int n; cin >> n;
cout << a[n] << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You must be familiar with various types of operators. One of the most commonly used operators in any language are increment and decrement operators. Given two numbers X and Y. Your task is to print the value of X decremented by 1 and value of Y after incremented by 1.The first line of the input contains two integers X and Y.
<b>Constraints:</b>
1 ≤ X, Y ≤ 10000You need to perform the task as mentioned in the question, and finally, print the result separated by a space.Sample Input 1:
4 5
Sample Output 1:
3 6
Sample Input 2:
5 6
Sample Output 2:
4 7, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
int y = sc.nextInt();
x--;
y++;
System.out.print(x);
System.out.print(" ");
System.out.print(y);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You must be familiar with various types of operators. One of the most commonly used operators in any language are increment and decrement operators. Given two numbers X and Y. Your task is to print the value of X decremented by 1 and value of Y after incremented by 1.The first line of the input contains two integers X and Y.
<b>Constraints:</b>
1 ≤ X, Y ≤ 10000You need to perform the task as mentioned in the question, and finally, print the result separated by a space.Sample Input 1:
4 5
Sample Output 1:
3 6
Sample Input 2:
5 6
Sample Output 2:
4 7, I have written this Solution Code: def incremental_decremental(x, y):
x -= 1
y += 1
print(x, y, end =' ')
def main():
input1 = input().split()
x = int(input1[0])
y = int(input1[1])
#z = int(input1[2])
incremental_decremental(x, y)
if __name__ == '__main__':
main(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array of size N and an integer K. You have to divide the given array into K non-empty subarrays such that each array index is in exactly one of the subarrays.
Divide the array such that the following value is minimized:
summation (max(i) - min(i)) for i from 1 to K
where max(i) is the maximum value in the subarray i.
and min(i) is the minimum value in the subarray i.
For example, if we have an array [1, 2, 3, 4, 5, 6] and we split it as [1, 2] [3, 4, 5] [6] then the value we get will be (2 - 1) + (5 - 3) + (6 - 6) = 3.The first line of the input contains two integers N and K.
The next line contain N integers.
Constraints
1 <= K <= N <= 100000
1 <= Arr[i] <= 10^9Print the minimum value you can obtain after dividing the array.Sample Input
6 3
4 8 15 16 23 42
Sample Output
12
Explanation: we split the array as [4, 8, 15, 16], [23], [42]
Sample Input:-
6 3
1 2 3 4 5 6
Sample Output:-
3
Sample Input:-
5 5
1 2 3 4 5
Sample Output:-
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 input[] = br.readLine().split(" ");
int n = Integer.parseInt(input[0]);
int k = Integer.parseInt(input[1]);
String[] stringArr = br.readLine().split(" ");
int[] arr = new int[n];
for(int i=0; i<n; i++){
arr[i] = Integer.parseInt(stringArr[i]);
}
int[] diff = new int[n-1];
int j=0;
for(int i=1; i<n; i++){
diff[j] = arr[i]-arr[i-1];
j++;
}
Arrays.sort(diff);
int sum = 0;
for(j = 0; j <= n-k-1; j++) {
sum += diff[j];
}
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array of size N and an integer K. You have to divide the given array into K non-empty subarrays such that each array index is in exactly one of the subarrays.
Divide the array such that the following value is minimized:
summation (max(i) - min(i)) for i from 1 to K
where max(i) is the maximum value in the subarray i.
and min(i) is the minimum value in the subarray i.
For example, if we have an array [1, 2, 3, 4, 5, 6] and we split it as [1, 2] [3, 4, 5] [6] then the value we get will be (2 - 1) + (5 - 3) + (6 - 6) = 3.The first line of the input contains two integers N and K.
The next line contain N integers.
Constraints
1 <= K <= N <= 100000
1 <= Arr[i] <= 10^9Print the minimum value you can obtain after dividing the array.Sample Input
6 3
4 8 15 16 23 42
Sample Output
12
Explanation: we split the array as [4, 8, 15, 16], [23], [42]
Sample Input:-
6 3
1 2 3 4 5 6
Sample Output:-
3
Sample Input:-
5 5
1 2 3 4 5
Sample Output:-
0, 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 N = (3e5);
int n, k;
int a[N];
signed main(){
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
cin >> n >> k;
for(int i = 0; i < n; ++i){
cin>>a[i];
}
vector <int> v;
for(int i = 1; i < n; ++i) v.push_back(a[i - 1] - a[i]);
sort(v.begin(), v.end());
int res = a[n - 1] - a[0];
for(int i = 0; i < k - 1; ++i) res += v[i];
cout << res;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a singly linked list and an element K, your task is to insert the element at the tail of the linked list.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>addElement()</b> that takes head node and the integer K as a parameter.
Constraints:
1 <=N<= 1000
1 <=K, value<= 1000Return the head of the modified linked listSample Input:-
5 2
1 2 3 4 5
Sample Output:
1 2 3 4 5 2
, I have written this Solution Code: a,b=[int(x) for x in input().split()]
c=[int(x) for x in input().split()]
for i in c:
print(i,end=" ")
print(b), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a singly linked list and an element K, your task is to insert the element at the tail of the linked list.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>addElement()</b> that takes head node and the integer K as a parameter.
Constraints:
1 <=N<= 1000
1 <=K, value<= 1000Return the head of the modified linked listSample Input:-
5 2
1 2 3 4 5
Sample Output:
1 2 3 4 5 2
, I have written this Solution Code: public static Node addElement(Node head,int k) {
Node temp=head;
while(temp.next!=null){
temp=temp.next;}
Node x= new Node(k);
temp.next = x;
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: static int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: def mythOrFact(N):
prime=1
cnt=N+1
for i in range (2,N):
if N%i==0:
prime=0
cnt=cnt+i
x = 3*N
x=x//2
if(cnt <= x and prime==1) or (cnt>x and prime==0):
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.First line consists of a single integer denoting n.
Constraint:-
1<=size of n<=100Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, 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();
int i=str.length()-1;
if(i==0){
int number=Integer.parseInt(str);
System.out.println(number);
}else{
while(str.charAt(i)=='0'){
i--;
}
for(int j=i;j>=0;j--){
System.out.print(str.charAt(j));
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.First line consists of a single integer denoting n.
Constraint:-
1<=size of n<=100Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, I have written this Solution Code: n=int(input())
def reverse(n):
return int(str(n)[::-1])
print(reverse(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.First line consists of a single integer denoting n.
Constraint:-
1<=size of n<=100Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
cin>>s;
reverse(s.begin(),s.end());
int I;
for( I=0;I<s.length();I++){
if(s[I]!='0'){break;}
}
if(I==s.length()){cout<<0;return 0;}
for(int j=I;j<s.length();j++){
cout<<s[j];}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rick has got a huge candy bag. The bag contains a total of O orange candies, A apple candies, and U candies with an unknown flavor (can be orange or apple as well).
Now Morty wants to know the maximum number of candies that can be drawn from the bag blindfolded such that no more than X orange and Y apple candies are drawn out.
Can you solve the problem for Morty?The first line of the input contains three integers O, A, and U, representing the number of orange, apple, and unknown flavored candies respectively.
The second line of the input contains two integers X and Y representing the maximum number of orange and apple candies that can be drawn of the bag.
Constraints
0 <= O, A, U <= 10<sup>9</sup>
0 <= X, Y <= 10<sup>9</sup>Output a single integer, the maximum number of candies that can be drawn out of Rick's bag.Sample Input 1
4 2 7
2 7
Sample Output 1
2
Explanation: If we draw out more than 2 candies, there is a possible chance for 3 candies to be orange.
Sample Input 2
1 2 3
8 4
Sample Output 2
4, I have written this Solution Code: [o,a,u]=[int(i) for i in input().split()]
s=o+a+u
[x,y]=[int(i) for i in input().split()]
o+=u
a+=u
if o<=x and a<=y:
print(s)
elif o<=x:
print(y)
elif a<=y:
print(x)
else:
print(min(x,y)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rick has got a huge candy bag. The bag contains a total of O orange candies, A apple candies, and U candies with an unknown flavor (can be orange or apple as well).
Now Morty wants to know the maximum number of candies that can be drawn from the bag blindfolded such that no more than X orange and Y apple candies are drawn out.
Can you solve the problem for Morty?The first line of the input contains three integers O, A, and U, representing the number of orange, apple, and unknown flavored candies respectively.
The second line of the input contains two integers X and Y representing the maximum number of orange and apple candies that can be drawn of the bag.
Constraints
0 <= O, A, U <= 10<sup>9</sup>
0 <= X, Y <= 10<sup>9</sup>Output a single integer, the maximum number of candies that can be drawn out of Rick's bag.Sample Input 1
4 2 7
2 7
Sample Output 1
2
Explanation: If we draw out more than 2 candies, there is a possible chance for 3 candies to be orange.
Sample Input 2
1 2 3
8 4
Sample Output 2
4, I have written this Solution Code: import java.io.*;
import java.util.*;
import static java.lang.Math.*;
public class Main implements Runnable {
private boolean console=false;
private long MOD = 1000_000_007L;
private int MAX = 1000_001;
private void solve1(){
long a=in.nl(),b=in.nl(),c= in.nl();
long x=in.nl(),y=in.nl();
long ans =0;
if(x>y){
long t = x;
x=y;
y=t;
t = a;
a=b;
b=t;
}
if(x-a-c <0){
ans = x;
}else if(y-b-c<0) {
ans = y;
}else {
ans = a+b+c;
}
out.printLn(ans);
}
private void solve() {
int testCases = 1;
while (testCases-->0){
solve1();
}
}
private void add(TreeMap<Integer, Integer> map, int key){
map.put(key,map.getOrDefault(key,0)+1);
}
private void remove(TreeMap<Integer,Integer> map,int key){
if(!map.containsKey(key))
return;
map.put(key,map.getOrDefault(key,0)-1);
if(map.get(key)==0)
map.remove(key);
}
@Override
public void run() {
long time = System.currentTimeMillis();
try {
init();
} catch (FileNotFoundException e) {
e.printStackTrace();
}
try {
solve();
out.flush();
System.err.println(System.currentTimeMillis()-time);
System.exit(0);
}catch (Exception e){
e.printStackTrace(); System.exit(1);
}
}
private FastInput in;
private FastOutput out;
public static void main(String[] args) throws Exception {
new Main().run();
}
private void init() throws FileNotFoundException {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
try {
if (!console && System.getProperty("user.name").equals("puneetkumar")) {
outputStream = new FileOutputStream("/Users/puneetkumar/output.txt");
inputStream = new FileInputStream("/Users/puneetkumar/input.txt");
}
} catch (Exception ignored) {
}
out = new FastOutput(outputStream);
in = new FastInput(inputStream);
}
private void maualAssert(int a,int b,int c){
if(a<b || a>c) throw new RuntimeException();
}
private void maualAssert(long a,long b,long c){
if(a<b || a>c) throw new RuntimeException();
}
private void sort(int[] arr) {
List<Integer> list = new ArrayList<>();
for (int object : arr) list.add(object);
Collections.sort(list);
for (int i = 0; i < list.size(); ++i) arr[i] = list.get(i);
}
private void sort(long[] arr) {
List<Long> list = new ArrayList<>();
for (long object : arr) list.add(object);
Collections.sort(list);
for (int i = 0; i < list.size(); ++i) arr[i] = list.get(i);
}
private long ModPow(long x, long y, long MOD) {
long res = 1L;
x = x % MOD;
while (y >= 1L) {
if ((y & 1L) > 0) res = (res * x) % MOD;
x = (x * x) % MOD;
y >>= 1L;
}
return res;
}
private int gcd(int a, int b) {
if (a == 0) return b;
return gcd(b % a, a);
}
private long gcd(long a, long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
private int[] arrInt(int n){
int[] arr=new int[n];for(int i=0;i<n;++i)arr[i]=in.ni();
return arr;
}
private long[] arrLong(int n){
long[] arr=new long[n];for(int i=0;i<n;++i)arr[i]=in.nl();
return arr;
}
private int arrMax(int[] arr){
int ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
private long arrMax(long[] arr){
long ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
private int arrMin(int[] arr){
int ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
private long arrMin(long[] arr){
long ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
class FastInput { InputStream obj;
public FastInput(InputStream obj) {
this.obj = obj;
}
private byte inbuffer[] = new byte[1024];
private int lenbuffer = 0, ptrbuffer = 0;
private int readByte() { if (lenbuffer == -1) throw new InputMismatchException();
if (ptrbuffer >= lenbuffer) { ptrbuffer = 0;
try { lenbuffer = obj.read(inbuffer);
} catch (IOException e) { throw new InputMismatchException(); } }
if (lenbuffer <= 0) return -1;return inbuffer[ptrbuffer++]; }
String ns() { int b = skip();StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b)))
{ sb.appendCodePoint(b);b = readByte(); }return sb.toString();}
int ni() {
int num = 0, b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); }}
long nl() { long num = 0;int b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10L + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); } }
private boolean isSpaceChar(int c) {
return (!(c >= 33 && c <= 126));
}
int skip() { int b;while ((b = readByte()) != -1 && isSpaceChar(b)) ;return b; }
float nf() {return Float.parseFloat(ns());}
double nd() {return Double.parseDouble(ns());}
char nc() {return (char) skip();}
}
class FastOutput{
private final PrintWriter writer;
public FastOutput(OutputStream outputStream) {
writer = new PrintWriter(outputStream);
}
public PrintWriter getWriter(){
return writer;
}
public void print(Object obj){
writer.print(obj);
}
public void printLn(){
writer.println();
}
public void printLn(Object obj){
writer.print(obj);
printLn();
}
public void printSp(Object obj){
writer.print(obj+" ");
}
public void printArr(int[] arr){
for(int i:arr)
printSp(i);
printLn();
}
public void printArr(long[] arr){
for(long i:arr)
printSp(i);
printLn();
}
public void flush(){
writer.flush();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rick has got a huge candy bag. The bag contains a total of O orange candies, A apple candies, and U candies with an unknown flavor (can be orange or apple as well).
Now Morty wants to know the maximum number of candies that can be drawn from the bag blindfolded such that no more than X orange and Y apple candies are drawn out.
Can you solve the problem for Morty?The first line of the input contains three integers O, A, and U, representing the number of orange, apple, and unknown flavored candies respectively.
The second line of the input contains two integers X and Y representing the maximum number of orange and apple candies that can be drawn of the bag.
Constraints
0 <= O, A, U <= 10<sup>9</sup>
0 <= X, Y <= 10<sup>9</sup>Output a single integer, the maximum number of candies that can be drawn out of Rick's bag.Sample Input 1
4 2 7
2 7
Sample Output 1
2
Explanation: If we draw out more than 2 candies, there is a possible chance for 3 candies to be orange.
Sample Input 2
1 2 3
8 4
Sample Output 2
4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int a, b, c;
cin >> a >> b >> c;
int n, m;
cin >> n >> m;
int ans = a+b+c;
if (n < a + c) ans = min(n, ans);
if (m < b + c) ans = min(m, ans);
cout << ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Olivia likes triangles. Hence, John decided to give her three integers and ask whether a triangle with edge length as given three integers is possible.
Help Olivia in answering it.The input consists of a single line containing three space-separated integers A, B, and C.
<b>Constraints </b>
1 <= A, B, C <= 100Output "Yes" if the triangle is possible otherwise, "No" (without quotes).Sample Input 1:
5 3 4
Sample Output 1:
Yes
Sample Explanation 1:
The possible triangle is a right-angled triangle with a hypotenuse of 5., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int a, b, c;
cin >> a >> b >> c;
if(a<(b+c) && b<(c+a) && c<(a+b)){
cout << "Yes\n";
}else{
cout << "No\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and K, find the lexicographically smallest string of length N using only the first K lowercase letters of the alphabet such that each letter is used at least once and no two adjacent characters are equal.
If such a string doesn't exist, print -1.The first line of input contains a single integer, T (1 <= T <= 100).
Then T lines follow, each containing two space-separated integers, N (1 <= N <= 10<sup>5</sup>) and K (1 <= K <= 26).
It is guaranteed that sum of N over all test cases does not exceed 10<sup>6</sup>For each test case, output its answer in a new line.Sample Input:
2
2 3
3 2
Sample Output:
-1
aba, 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));
BufferedWriter bo=new BufferedWriter(new OutputStreamWriter(System.out));
int t;
try{
t=Integer.parseInt(br.readLine());
}
catch(Exception e)
{
return;
}
while(t-->0)
{
String[] g=br.readLine().split(" ");
int n=Integer.parseInt(g[0]);
int k=Integer.parseInt(g[1]);
if(k>n || (k==1) || (k>26))
{
if(n==1 && k==1)
bo.write("a\n");
else
bo.write(-1+"\n");
}
else
{
int extra=k-2;
boolean check=true;
while(n>extra)
{
if(check==true)
bo.write("a");
else
bo.write("b");
if(check==true)
check=false;
else
check=true;
n--;
}
for(int i=0;i<extra;i++)
bo.write((char)(i+99));
bo.write("\n");
}
}
bo.close();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and K, find the lexicographically smallest string of length N using only the first K lowercase letters of the alphabet such that each letter is used at least once and no two adjacent characters are equal.
If such a string doesn't exist, print -1.The first line of input contains a single integer, T (1 <= T <= 100).
Then T lines follow, each containing two space-separated integers, N (1 <= N <= 10<sup>5</sup>) and K (1 <= K <= 26).
It is guaranteed that sum of N over all test cases does not exceed 10<sup>6</sup>For each test case, output its answer in a new line.Sample Input:
2
2 3
3 2
Sample Output:
-1
aba, I have written this Solution Code: t=int(input())
for tt in range(t):
n,k=map(int,input().split())
if (k==1 and n>1) or (k>n):
print(-1)
continue
s="abcdefghijklmnopqrstuvwxyz"
ss="ab"
if (n-k)%2==0:
a=ss*((n-k)//2)+s[:k]
else:
a=ss*((n-k)//2)+s[:2]+"a"+s[2:k]
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and K, find the lexicographically smallest string of length N using only the first K lowercase letters of the alphabet such that each letter is used at least once and no two adjacent characters are equal.
If such a string doesn't exist, print -1.The first line of input contains a single integer, T (1 <= T <= 100).
Then T lines follow, each containing two space-separated integers, N (1 <= N <= 10<sup>5</sup>) and K (1 <= K <= 26).
It is guaranteed that sum of N over all test cases does not exceed 10<sup>6</sup>For each test case, output its answer in a new line.Sample Input:
2
2 3
3 2
Sample Output:
-1
aba, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long long int ll;
typedef unsigned long long int ull;
const long double PI = acos(-1);
const ll mod=1e9+7;
const ll mod1=998244353;
const int inf = 1e9;
const ll INF=1e18;
void precompute(){
}
void TEST_CASE(){
int n,k;
cin >> n >> k;
if(k==1){
if(n>1){
cout << -1 << endl;
}else{
cout << 'a' << endl;
}
}else if(n<k){
cout << -1 << endl;
}else if(n==k){
string s="";
for(int i=0 ; i<k ; i++){
s+=('a'+i);
}
cout << s << endl;
}else{
string s="";
for(int i=0 ; i<(n-k+2) ; i++){
if(i%2){
s+="b";
}else{
s+="a";
}
}
for(int i=2 ; i<k ; i++){
s+=('a'+i);
}
cout << s << endl;
}
}
signed main(){
fast;
//freopen ("INPUT.txt","r",stdin);
//freopen ("OUTPUT.txt","w",stdout);
int test=1,TEST=1;
precompute();
cin >> test;
while(test--){
TEST_CASE();
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array <b>arr[]</b> of size <b>N</b> without duplicates, and given a value <b>x</b>. Find the floor of x in given array. Floor of x is defined as the largest element K in arr[] such that K is smaller than or equal to x.
Try to use binary search to solve this problem.First line of input contains number of testcases T. For each testcase, first line of input contains number of elements in the array and element whose floor is to be searched. Last line of input contains array elements.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 10^5
1 ≤ arr[i] ≤ 10^18
0 ≤ X ≤ arr[n-1]Output the index of floor of x if exists, else print -1. Use 0-indexingInput:
3
7 0
1 2 8 10 11 12 19
7 5
1 2 8 10 11 12 19
7 10
1 2 8 10 11 12 19
Output:
-1
1
3
Explanation:
Testcase 1: No element less than or equal to 0 is found. So output is "-1".
Testcase 2: Number less than or equal to 5 is 2, whose index is 1(0-based indexing).
Testcase 3: Number less than or equal to 10 is 10 and its index is 3., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static int BS(int arr[],int x,int start,int end){
if( start > end) return start-1;
int mid = start + (end - start)/2;
if(arr[mid]==x) return mid;
if(arr[mid]>x) return BS(arr,x,start,mid-1);
return BS(arr,x,mid+1,end);
}
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
String[] s;
for(;t>0;t--){
s = br.readLine().split(" ");
int n = Integer.parseInt(s[0]);
int x = Integer.parseInt(s[1]);
s = br.readLine().split(" ");
int[] arr = new int[n];
for(int i=0;i<n;i++)
arr[i] = Integer.parseInt(s[i]);
int res = BS(arr,x,0,arr.length-1);
System.out.println(res);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array <b>arr[]</b> of size <b>N</b> without duplicates, and given a value <b>x</b>. Find the floor of x in given array. Floor of x is defined as the largest element K in arr[] such that K is smaller than or equal to x.
Try to use binary search to solve this problem.First line of input contains number of testcases T. For each testcase, first line of input contains number of elements in the array and element whose floor is to be searched. Last line of input contains array elements.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 10^5
1 ≤ arr[i] ≤ 10^18
0 ≤ X ≤ arr[n-1]Output the index of floor of x if exists, else print -1. Use 0-indexingInput:
3
7 0
1 2 8 10 11 12 19
7 5
1 2 8 10 11 12 19
7 10
1 2 8 10 11 12 19
Output:
-1
1
3
Explanation:
Testcase 1: No element less than or equal to 0 is found. So output is "-1".
Testcase 2: Number less than or equal to 5 is 2, whose index is 1(0-based indexing).
Testcase 3: Number less than or equal to 10 is 10 and its index is 3., I have written this Solution Code: def bsearch(arr,e,l,r):
if(r>=l):
mid = l + (r - l)//2
if (arr[mid] == e):
return mid
elif arr[mid] > e:
return bsearch(arr, e,l, mid-1)
else:
return bsearch(arr,e, mid + 1, r)
else:
return r
t=int(input())
for i in range(t):
ip=list(map(int,input().split()))
l=list(map(int,input().split()))
le=ip[0]
e=ip[1]
print(bsearch(sorted(l),e,0,le-1)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array <b>arr[]</b> of size <b>N</b> without duplicates, and given a value <b>x</b>. Find the floor of x in given array. Floor of x is defined as the largest element K in arr[] such that K is smaller than or equal to x.
Try to use binary search to solve this problem.First line of input contains number of testcases T. For each testcase, first line of input contains number of elements in the array and element whose floor is to be searched. Last line of input contains array elements.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 10^5
1 ≤ arr[i] ≤ 10^18
0 ≤ X ≤ arr[n-1]Output the index of floor of x if exists, else print -1. Use 0-indexingInput:
3
7 0
1 2 8 10 11 12 19
7 5
1 2 8 10 11 12 19
7 10
1 2 8 10 11 12 19
Output:
-1
1
3
Explanation:
Testcase 1: No element less than or equal to 0 is found. So output is "-1".
Testcase 2: Number less than or equal to 5 is 2, whose index is 1(0-based indexing).
Testcase 3: Number less than or equal to 10 is 10 and its index is 3., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n, x; cin >> n >> x;
for(int i = 0; i < n; i++)
cin >> a[i];
int l = -1, h = n;
while(l+1 < h){
int m = (l + h) >> 1;
if(a[m] <= x)
l = m;
else
h = m;
}
cout << l << endl;
}
return 0;
}, In this Programming Language: C++, 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: You are given a chessboard of size N x N, where the top left square is black. Each square contains a value. Find the sum of the values of all black squares and all white squares.
Remember that in a chessboard black and white squares are alternate.The first line of input will be the N size of the matrix. Then next N lines will consist of elements of the matrix. Each row will contain N elements since it is a square matrix.
<b>Constraints:-</b>
1 ≤ N ≤ 800
1 ≤ Matrix[i][j] ≤ 100000
Print two lines, the first line containing the sum of black squares and the second line containing the sum of white squares.Input 1:
3
1 2 3
4 5 6
7 8 9
Output 1:
25
20
Sample Input 2:
4
1 2 3 4
6 8 9 10
11 12 13 14
15 16 17 18
Sample Output 2:
80
79
<b>Explanation 1</b>
The black square contains 1, 3, 5, 7, 9; sum = 25
The white square contains 2, 4, 6, 8; sum = 20, I have written this Solution Code: n=int(input())
bSum=0
wSum=0
for i in range(n):
c=(input().split())
for j in range(len(c)):
if(i%2==0):
if(j%2==0):
bSum+=int(c[j])
else:
wSum+=int(c[j])
else:
if(j%2==0):
wSum+=int(c[j])
else:
bSum+=int(c[j])
print(bSum)
print(wSum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a chessboard of size N x N, where the top left square is black. Each square contains a value. Find the sum of the values of all black squares and all white squares.
Remember that in a chessboard black and white squares are alternate.The first line of input will be the N size of the matrix. Then next N lines will consist of elements of the matrix. Each row will contain N elements since it is a square matrix.
<b>Constraints:-</b>
1 ≤ N ≤ 800
1 ≤ Matrix[i][j] ≤ 100000
Print two lines, the first line containing the sum of black squares and the second line containing the sum of white squares.Input 1:
3
1 2 3
4 5 6
7 8 9
Output 1:
25
20
Sample Input 2:
4
1 2 3 4
6 8 9 10
11 12 13 14
15 16 17 18
Sample Output 2:
80
79
<b>Explanation 1</b>
The black square contains 1, 3, 5, 7, 9; sum = 25
The white square contains 2, 4, 6, 8; sum = 20, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws IOException {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int mat[][] = new int[N][N];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
mat[i][j] = sc.nextInt();
}
alternate_Matrix_Sum(mat,N);
}
static void alternate_Matrix_Sum(int mat[][], int N)
{
long sum =0, sum1 = 0;
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
if((i+j)%2 == 0)
sum += mat[i][j];
else sum1 += mat[i][j];
}
}
System.out.println(sum);
System.out.print(sum1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
String [] str=br.readLine().trim().split(" ");
int a[]=new int[n];
for(int i=0;i<n;i++){
a[i]=Integer.parseInt(str[i]);
}
Arrays.sort(a);
int size=a[n-1]+1;
int c[]=new int[size];
for(int i=0;i<size;i++) c[i]=0;
for(int i=0;i<n;i++) c[a[i]]++;
int max=0,freq=c[1];
for(int i=2;i<size;i++){
if(freq<=c[i]){
freq=c[i];
max=i;
}
}
System.out.println(max);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=2, I have written this Solution Code: n = int(input())
a = [int(x) for x in input().split()]
freq = {}
for x in a:
if x not in freq:
freq[x] = 1
else:
freq[x] += 1
mx = max(freq.values())
rf = sorted(freq)
for i in range(len(rf) - 1, -1, -1):
if freq[rf[i]] == mx:
print(rf[i])
break, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=2, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int n; cin >> n;
for(int i = 1; i <= n; i++){
int p; cin >> p;
a[p]++;
}
int mx = 0, id = -1;
for(int i = 1; i <= 100; i++){
if(a[i] >= mx)
mx = a[i], id = i;
}
cout << id;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have a box of paint. The paint can be used to cover area of 1 square unit. You have N magic spells with magic value from 1 to N. You can use each spell at max once.
When you use a spell on the box of paint, the quantity of paint in the box gets multiplied by the magic value of that spell. For example, if currently the paint in the box can cover A square units and a spell of B magic value is cast on the box, then after the spell the paint in the box can be used to cover A*B square units of area.
You want to paint a square (completely filled) of maximum area with integer side length with the paint such that after painting the square, the paint in the box gets empty. You can apply any magic spells you want in any order before the painting starts and not during or after the painting.
Find the area of maximum side square you can paint. As the answer can be huge find the area modulo 1000000007.The first and the only line of input contains a single integer N.
Constraints:
1 <= N <= 1000000Print the area of maximum side square you can paint modulo 1000000007.Sample Input 1
3
Sample Output 1
1
Explanation: We apply no magic spell.
Sample Input 2
4
Sample Output 2
4
Explanation: We apply magic spell number 4., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define 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;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
int si[1000001];
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int ans=1;
int mo=1000000007;
for(int i=2;i<=n;++i){
if(si[i]==0){
int x=i;
int c=0;
while(x<=n){
c+=n/x;
x*=i;
}
if(c%2==0){
ans=(ans*i)%mo;
}
int j=i;
while(j<=n){
si[j]=1;
j+=i;
}
}
else{
ans=(ans*i)%mo;
}
}
cout<<ans;
#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 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: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
import java.math.BigInteger;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
BigInteger sum;
String ip1 = sc.next();
String ip2 = sc.next();
BigInteger a = new BigInteger(ip1);
BigInteger b = new BigInteger(ip2);
sum = a.add(b);
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-02-03 02:46:30
**/
#include <bits/stdc++.h>
#define NX 105
#define MX 3350
using namespace std;
const int mod = 998244353;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
typedef long long INT;
const int pb = 10;
const int base_digits = 9;
const int base = 1000000000;
const int DIV = 100000;
struct bigint {
vector<int> a;
int sign;
bigint() : sign(1) {}
bigint(INT v) { *this = v; }
bigint(const string &s) { read(s); }
void operator=(const bigint &v) { sign = v.sign, a = v.a; }
void operator=(INT v) {
sign = 1;
if (v < 0) sign = -1, v = -v;
for (; v > 0; v = v / base) a.push_back(v % base);
}
bigint operator+(const bigint &v) const {
if (sign == v.sign) {
bigint res = v;
for (int i = 0, carry = 0; i < (int)max(a.size(), v.a.size()) ||
carry;
i++) {
if (i == (int)res.a.size()) res.a.push_back(0);
res.a[i] += carry + (i < (int)a.size() ? a[i] : 0);
carry = res.a[i] >= base;
if (carry) res.a[i] -= base;
}
return res;
}
return *this - (-v);
}
bigint operator-(const bigint &v) const {
if (sign == v.sign) {
if (abs() >= v.abs()) {
bigint res = *this;
for (int i = 0, carry = 0; i < (int)v.a.size() || carry; i++) {
res.a[i] -= carry + (i < (int)v.a.size() ? v.a[i] : 0);
carry = res.a[i] < 0;
if (carry) res.a[i] += base;
}
res.trim();
return res;
}
return -(v - *this);
}
return *this + (-v);
}
void operator*=(int v) {
if (v < 0) sign = -sign, v = -v;
for (int i = 0, carry = 0; i < (int)a.size() || carry; i++) {
if (i == (int)a.size()) a.push_back(0);
INT cur = a[i] * (INT)v + carry;
carry = (int)(cur / base);
a[i] = (int)(cur % base);
}
trim();
}
bigint operator*(int v) const {
bigint res = *this;
res *= v;
return res;
}
friend pair<bigint, bigint> DIVmod(const bigint &a1, const bigint &b1) {
int norm = base / (b1.a.back() + 1);
bigint a = a1.abs() * norm;
bigint b = b1.abs() * norm;
bigint q, r;
q.a.resize(a.a.size());
for (int i = a.a.size() - 1; i >= 0; i--) {
r *= base;
r += a.a[i];
int s1 = r.a.size() <= b.a.size() ? 0 : r.a[b.a.size()];
int s2 = r.a.size() <= b.a.size() - 1 ? 0 : r.a[b.a.size() - 1];
int d = ((INT)base * s1 + s2) / b.a.back();
r -= b * d;
while (r < 0) r += b, --d;
q.a[i] = d;
}
q.sign = a1.sign * b1.sign;
r.sign = a1.sign;
q.trim();
r.trim();
return make_pair(q, r / norm);
}
bigint operator/(const bigint &v) const { return DIVmod(*this, v).first; }
bigint operator%(const bigint &v) const { return DIVmod(*this, v).second; }
void operator/=(int v) {
if (v < 0) sign = -sign, v = -v;
for (int i = (int)a.size() - 1, rem = 0; i >= 0; i--) {
INT cur = a[i] + rem * (INT)base;
a[i] = (int)(cur / v);
rem = (int)(cur % v);
}
trim();
}
bigint operator/(int v) const {
bigint res = *this;
res /= v;
return res;
}
int operator%(int v) const {
if (v < 0) v = -v;
int m = 0;
for (int i = a.size() - 1; i >= 0; --i)
m = (a[i] + m * (INT)base) % v;
return m * sign;
}
void operator+=(const bigint &v) { *this = *this + v; }
void operator-=(const bigint &v) { *this = *this - v; }
void operator*=(const bigint &v) { *this = *this * v; }
void operator/=(const bigint &v) { *this = *this / v; }
bool operator<(const bigint &v) const {
if (sign != v.sign) return sign < v.sign;
if (a.size() != v.a.size()) return a.size() * sign < v.a.size() * v.sign;
for (int i = a.size() - 1; i >= 0; i--)
if (a[i] != v.a[i]) return a[i] * sign < v.a[i] * sign;
return false;
}
bool operator>(const bigint &v) const { return v < *this; }
bool operator<=(const bigint &v) const { return !(v < *this); }
bool operator>=(const bigint &v) const { return !(*this < v); }
bool operator==(const bigint &v) const { return !(*this < v) && !(v < *this); }
bool operator!=(const bigint &v) const { return *this < v || v < *this; }
void trim() {
while (!a.empty() && !a.back()) a.pop_back();
if (a.empty()) sign = 1;
}
bool isZero() const { return a.empty() || (a.size() == 1 && !a[0]); }
bigint operator-() const {
bigint res = *this;
res.sign = -sign;
return res;
}
bigint abs() const {
bigint res = *this;
res.sign *= res.sign;
return res;
}
INT longValue() const {
INT res = 0;
for (int i = a.size() - 1; i >= 0; i--) res = res * base + a[i];
return res * sign;
}
friend bigint gcd(const bigint &a, const bigint &b) { return b.isZero() ? a : gcd(b, a % b); }
friend bigint lcm(const bigint &a, const bigint &b) { return a / gcd(a, b) * b; }
void read(const string &s) {
sign = 1;
a.clear();
int pos = 0;
while (pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
if (s[pos] == '-') sign = -sign;
pos++;
}
for (int i = s.size() - 1; i >= pos; i -= base_digits) {
int x = 0;
for (int j = max(pos, i - base_digits + 1); j <= i; j++) x = x *
pb +
s[j] - '0';
a.push_back(x);
}
trim();
}
friend istream &operator>>(istream &stream, bigint &v) {
string s;
stream >> s;
v.read(s);
return stream;
}
friend ostream &operator<<(ostream &stream, const bigint &v) {
if (v.sign == -1) stream << '-';
stream << (v.a.empty() ? 0 : v.a.back());
for (int i = (int)v.a.size() - 2; i >= 0; --i) stream << setw(base_digits) << setfill('0') << v.a[i];
return stream;
}
static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) {
vector<INT> p(max(old_digits, new_digits) + 1);
p[0] = 1;
for (int i = 1; i < (int)p.size(); i++) p[i] = p[i - 1] * pb;
vector<int> res;
INT cur = 0;
int cur_digits = 0;
for (int i = 0; i < (int)a.size(); i++) {
cur += a[i] * p[cur_digits];
cur_digits += old_digits;
while (cur_digits >= new_digits) {
res.push_back(int(cur % p[new_digits]));
cur /= p[new_digits];
cur_digits -= new_digits;
}
}
res.push_back((int)cur);
while (!res.empty() && !res.back()) res.pop_back();
return res;
}
typedef vector<INT> vll;
static vll karatsubaMultiply(const vll &a, const vll &b) {
int n = a.size();
vll res(n + n);
if (n <= 32) {
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
res[i + j] += a[i] * b[j];
return res;
}
int k = n >> 1;
vll a1(a.begin(), a.begin() + k);
vll a2(a.begin() + k, a.end());
vll b1(b.begin(), b.begin() + k);
vll b2(b.begin() + k, b.end());
vll a1b1 = karatsubaMultiply(a1, b1);
vll a2b2 = karatsubaMultiply(a2, b2);
for (int i = 0; i < k; i++) a2[i] += a1[i];
for (int i = 0; i < k; i++) b2[i] += b1[i];
vll r = karatsubaMultiply(a2, b2);
for (int i = 0; i < (int)a1b1.size(); i++) r[i] -= a1b1[i];
for (int i = 0; i < (int)a2b2.size(); i++) r[i] -= a2b2[i];
for (int i = 0; i < (int)r.size(); i++) res[i + k] += r[i];
for (int i = 0; i < (int)a1b1.size(); i++) res[i] += a1b1[i];
for (int i = 0; i < (int)a2b2.size(); i++) res[i + n] += a2b2[i];
return res;
}
bigint operator*(const bigint &v) const {
vector<int> a5 = convert_base(this->a, base_digits, 5);
vector<int> b5 = convert_base(v.a, base_digits, 5);
vll a(a5.begin(), a5.end());
vll b(b5.begin(), b5.end());
while (a.size() < b.size()) a.push_back(0);
while (b.size() < a.size()) b.push_back(0);
while (a.size() & (a.size() - 1)) a.push_back(0), b.push_back(0);
vll c = karatsubaMultiply(a, b);
bigint res;
res.sign = sign * v.sign;
for (int i = 0, carry = 0; i < (int)c.size(); i++) {
INT cur = c[i] + carry;
res.a.push_back((int)(cur % DIV));
carry = (int)(cur / DIV);
}
res.a = convert_base(res.a, 5, base_digits);
res.trim();
return res;
}
inline bool isOdd() { return a[0] & 1; }
};
int main() {
bigint n, m;
cin >> n >> m;
cout << n + m << "\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b your task is to print the summation of these two numbersInput contains a single line containing two space separated integers a and b.
Constraints:-
1 <= a, b <= 10<sup>20000</sup>Print the sum of a and b.Sample Input:-
3 8
Sample Output:-
11
Sample Input:-
15 1
Sample Output:-
16, I have written this Solution Code: n,m = map(int,input().split())
print(n+m) , In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: How would you add your own method to the Array object so
the following code would work?
const arr = [1, 2, 3]
console. log(arr.average())
// 2input will be an array, run like this
const anyArray = [5,6...]
anyArray.average should return average of the arraysAverage of the given arrayconst myArray = [1,2,3,4,5]
console.log(myArray.average())
// 3, I have written this Solution Code: Array.prototype.average = function() {
// calculate sum
var sum = this.reduce(function(prev, cur) { return prev + cur; });
// return sum divided by number of elements
return sum / this.length;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The Newton City is going to participate in a Basket Ball tournament. You have been made the coordinator for the final team selection committee.
Lot of applications have been received and you are given with a two integer arrays. First array, height[], stores the height of the player while the second array, basketsScored[], has the number of Baskets scored for each player.
To find the top K players, you need to choose players according to their heights, that means greater the height, more suitable is the player. If height of two players is same then consider the player with greater number of baskets.
Print the heights and the baskets of the top K players.First line of the input takes the number of test cases, T. Further, for each test case, first line contains two space separated integers, N and K. Next two lines contains N space separated integers denoting the height of the players and basketsScored.
Constraints:
1 <= N <= 10^6
1 <= K <= N
1 <= height[i] <= 10^6
0 <= basketsScored[i] <= 10^6For each test case, print the K players on next lines and in each line print the value of height and basketsScored with space separation.Input:
5 3
1 3 3 3 5
1 2 3 1 2
Output:
5 2
3 3
3 2
Explanation: The top 3 players will be the one with greater heights, and the players with larger no. of baskets in case of same heights., I have written this Solution Code: import java.io.*;
import java.util.*;
class player {
int height, score;
}
class mycomp implements Comparator<player>{
public int compare(player a, player b){
return b.height - a.height;
}
}
class mycomp2 implements Comparator<player>{
public int compare(player a, player b){
if(a.height == b.height){
return b.score - a.score;
}
return b.height - a.height;
}
}
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(
new InputStreamReader(System.in));
StringTokenizer st
= new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());
int arr1[] = new int[n];
int arr2[] = new int[n];
StringTokenizer sc = new StringTokenizer(br.readLine());
for(int i=0;i<n;i++){
arr1[i] = Integer.parseInt(sc.nextToken());
}
StringTokenizer sg = new StringTokenizer(br.readLine());
for(int i=0;i<n;i++){
arr2[i] = Integer.parseInt(sg.nextToken());
}
topKplayers(arr1, arr2, k);
}
public static void topKplayers(int[] arr1, int[] arr2, int k){
ArrayList<player> pl = new ArrayList<>();
for(int i=0;i<arr1.length;i++){
player play = new player();
play.height = arr1[i];
play.score = arr2[i];
pl.add(play);
}
Collections.sort(pl, new mycomp());
Collections.sort(pl, new mycomp2());
for(player play : pl){
if(k-- > 0){
System.out.println(play.height + " " + play.score);
}
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The Newton City is going to participate in a Basket Ball tournament. You have been made the coordinator for the final team selection committee.
Lot of applications have been received and you are given with a two integer arrays. First array, height[], stores the height of the player while the second array, basketsScored[], has the number of Baskets scored for each player.
To find the top K players, you need to choose players according to their heights, that means greater the height, more suitable is the player. If height of two players is same then consider the player with greater number of baskets.
Print the heights and the baskets of the top K players.First line of the input takes the number of test cases, T. Further, for each test case, first line contains two space separated integers, N and K. Next two lines contains N space separated integers denoting the height of the players and basketsScored.
Constraints:
1 <= N <= 10^6
1 <= K <= N
1 <= height[i] <= 10^6
0 <= basketsScored[i] <= 10^6For each test case, print the K players on next lines and in each line print the value of height and basketsScored with space separation.Input:
5 3
1 3 3 3 5
1 2 3 1 2
Output:
5 2
3 3
3 2
Explanation: The top 3 players will be the one with greater heights, and the players with larger no. of baskets in case of same heights., I have written this Solution Code: numOfPlayers,topK = map(int,input().split())
heightArr = list(map(int,input().split()))
scoreArr = list(map(int,input().split()))
heightArrTwo = list(heightArr)
heightArrTwo.sort()
heightArrTwo.reverse()
pairArr = []
for i in zip(heightArr,scoreArr):
pairArr.append(i)
pairArr.sort()
pairArr.reverse()
con = topK
for i in pairArr:
he,sc = i
print(he,sc)
con -= 1
if con == 0:
break, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The Newton City is going to participate in a Basket Ball tournament. You have been made the coordinator for the final team selection committee.
Lot of applications have been received and you are given with a two integer arrays. First array, height[], stores the height of the player while the second array, basketsScored[], has the number of Baskets scored for each player.
To find the top K players, you need to choose players according to their heights, that means greater the height, more suitable is the player. If height of two players is same then consider the player with greater number of baskets.
Print the heights and the baskets of the top K players.First line of the input takes the number of test cases, T. Further, for each test case, first line contains two space separated integers, N and K. Next two lines contains N space separated integers denoting the height of the players and basketsScored.
Constraints:
1 <= N <= 10^6
1 <= K <= N
1 <= height[i] <= 10^6
0 <= basketsScored[i] <= 10^6For each test case, print the K players on next lines and in each line print the value of height and basketsScored with space separation.Input:
5 3
1 3 3 3 5
1 2 3 1 2
Output:
5 2
3 3
3 2
Explanation: The top 3 players will be the one with greater heights, and the players with larger no. of baskets in case of same heights., I have written this Solution Code: #include <bits/stdc++.h> // header file includes every Standard library
using namespace std;
int main() {
// Your code here
int n, k; cin>>n>>k;
int height[n], basket[n];
for(int i=0; i<n; i++){
cin>>height[i];
}
vector<pair<int,int>> vp;
for(int i=0; i<n; i++){
cin>>basket[i];
vp.push_back({height[i], basket[i]});
}
auto comp = [&](pair<int, int> p1, pair<int, int> p2){
if(p1.first == p2.first){
return p1.second > p2.second;
}
return p1.first > p2.first;
};
sort(vp.begin(), vp.end(), comp);
for(int i=0; i<k; i++){
cout<<vp[i].first<<" "<<vp[i].second<<endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
32, 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();
if(s==null){
System.exit(0);
}
StringTokenizer st = new StringTokenizer(s, " ");
int power = Integer.parseInt(st.nextToken());
int multiple = Integer.parseInt(st.nextToken());
int res = power;
for(int i = 1;i<=multiple;i++){
res = res*2;
}
System.out.println(res);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
32, I have written this Solution Code: #include <iostream>
using namespace std;
int main()
{
int x, n;
cin >> x >> n;
cout << x*(1 << n);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
32, I have written this Solution Code: x,n = map(int,input().split())
print(x*(2**n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.