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For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
long z=0,x=0,y=0;
int choice;
Scanner in = new Scanner(System.in);
choice = in.nextInt();
String s="";
int f = 1;
while(f<=choice){
x = in.nextLong();
y = in.nextLong();
z = in.nextLong();
System.out.println((long)(Math.max((z-x),(z-y))));
f++;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: n = int(input())
for i in range(n):
l = list(map(int,input().split()))
print(l[2]-min(l)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, 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 (int 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 = 1e9 + 7;
const ll mod2 = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
signed main()
{
read(t);
assert(1 <= t && t <= ll(1e4));
while (t--)
{
readc(x, y, z);
assert(1 <= x && x <= ll(1e15));
assert(1 <= y && y <= ll(1e15));
assert(max(x, y) < z && z <= ll(1e15));
int r = 2*z - x - y - 1;
int l = z - max(x, y);
print(r - l + 1);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
int t;
cin>>t;
while(t--)
{
int x, y, z;
cin>>x>>y>>z;
cout<<max(z - y, z- x)<<endl;
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#endif
solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given the number N. You need to convert it to 1 in the minimum number of operations.
The operations allowed are as follows:
1. If N is even then divide the number by 2.
2. If N is odd then you can either add 1 to it or subtract 1 from it.
Using the above operations, find the minimum number of operations required to convert N to 1.The first line of input contains an integer T denoting the number of test cases. T test cases follow. Each test case contains 1 line of input containing integer N.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 10<sup>7</sup>
For each testcase, in a new line, print the minimum number of steps required.Sample Input:
4
1
2
3
4
Sample Output:
0
1
2
2
<b>Explanation:</b>
1 can be converted into 1 in 0 steps.
2 can be converted into 1 in 1 step: 2/2=1
3 can be converted into 1 in 3 steps: 3-1= 2 then 2/2=1
4 can be converted into 1 in 2 steps: 4/2=2 then 2/2=1, I have written this Solution Code: def stepCount(n):
count = 0
while n > 1:
if n % 2 == 0:
n = n // 2
elif n == 3 or n % 4 == 1:
n = n - 1
else:
n = n + 1
count += 1
return count
t=int(input())
for _ in range(t):
n=int(input())
print(stepCount(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given the number N. You need to convert it to 1 in the minimum number of operations.
The operations allowed are as follows:
1. If N is even then divide the number by 2.
2. If N is odd then you can either add 1 to it or subtract 1 from it.
Using the above operations, find the minimum number of operations required to convert N to 1.The first line of input contains an integer T denoting the number of test cases. T test cases follow. Each test case contains 1 line of input containing integer N.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 10<sup>7</sup>
For each testcase, in a new line, print the minimum number of steps required.Sample Input:
4
1
2
3
4
Sample Output:
0
1
2
2
<b>Explanation:</b>
1 can be converted into 1 in 0 steps.
2 can be converted into 1 in 1 step: 2/2=1
3 can be converted into 1 in 3 steps: 3-1= 2 then 2/2=1
4 can be converted into 1 in 2 steps: 4/2=2 then 2/2=1, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
#define f(i,n) for(int i=0;i<(n);++i)
#define fA(i,a,n) for(int i=a;i<=(n);++i)
#define fD(i,a,n) for(int i=a;i>=(n);--i)
#define tc int t;cin>>t;f(testcase,t)
#define pii pair<int,int>
void c_p_c() {
ios_base::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
freopen("inputf.in", "r", stdin);
freopen("outputf.in", "w", stdout);
#endif
}
int dp[10000001];
int minsteps(int n , int dp[]) {
if (n == 1)
return 0;
if (dp[n] != 0)
return dp[n];
int op1, op2, op3;
op1 = op2 = op3 = INT_MAX;
if (n % 2 == 0) {
op1 = 1 + minsteps(n / 2, dp);
}
else {
op2 = 1 + minsteps(n - 1, dp);
op3 = 1 + minsteps(n + 1, dp);
}
int ans = min(op1, min(op2, op3));
dp[n] = ans;
return dp[n];
}
int32_t main() {
memset(dp, 0, sizeof(dp));
tc {
int n;
cin >> n;
cout << minsteps(n, dp) << "\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given the number N. You need to convert it to 1 in the minimum number of operations.
The operations allowed are as follows:
1. If N is even then divide the number by 2.
2. If N is odd then you can either add 1 to it or subtract 1 from it.
Using the above operations, find the minimum number of operations required to convert N to 1.The first line of input contains an integer T denoting the number of test cases. T test cases follow. Each test case contains 1 line of input containing integer N.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 10<sup>7</sup>
For each testcase, in a new line, print the minimum number of steps required.Sample Input:
4
1
2
3
4
Sample Output:
0
1
2
2
<b>Explanation:</b>
1 can be converted into 1 in 0 steps.
2 can be converted into 1 in 1 step: 2/2=1
3 can be converted into 1 in 3 steps: 3-1= 2 then 2/2=1
4 can be converted into 1 in 2 steps: 4/2=2 then 2/2=1, I have written this Solution Code: import java.lang.*;
import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
int testcases=sc.nextInt();
while(testcases-->0)
{
long n=sc.nextLong();
System.out.println(NS.minOperations(n));
}
}
}
class NS
{
public static long minOperations(long n)
{
if(n==1)
return 0; //since 1 is already 1
if(n==2)
return 1; //convert 2 to 1. 1 step
if(n==3)
return 2; //convert 3 to 2. Then 2 to 1. 2 steps
long total=0; //save total steps
if(n%2!=0) //if odd
{
total=1+Math.min(minOperations(n-1),minOperations(n+1));
//convert n to n-1 or n+1 then minimum of those conversions
}
else
total=1+minOperations(n/2); //convert n to n/2 then count operations required for n/2 to 1
return total; //returning total at the end
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some messages. Here, we will start with the famous <b>"Hello World"</b> message. Now, here you are given a function to complete. <i>Don't worry about the ins and outs of functions, <b>just add the printing command to print "Hello World", </b></i>.your task is to just print "Hello World", without the quotes.Hello WorldHello World must be printed., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
System.out.print("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some messages. Here, we will start with the famous <b>"Hello World"</b> message. Now, here you are given a function to complete. <i>Don't worry about the ins and outs of functions, <b>just add the printing command to print "Hello World", </b></i>.your task is to just print "Hello World", without the quotes.Hello WorldHello World must be printed., I have written this Solution Code: def print_fun():
print ("Hello World")
def main():
print_fun()
if __name__ == '__main__':
main(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two non-negative integers, A and B. You are required to print the smallest non-negative integer such that it is not equal to both A and B.The input consists of two space separated integers A and B.
Constraints:
0 ≤ A ≤ 10
0 ≤ B ≤ 10
A ≠ B
Print a single integer denoting the answer.Sample Input 1:
0 1
Sample Output 1:
2
Sample Input 2:
4 5
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 Exception
{
BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb=new StringBuilder();
Set<Integer> set=new HashSet<>();
String s[]=bu.readLine().split(" ");
for(String x:s) set.add(Integer.parseInt(x));
int mex=0;
while(set.contains(mex)) mex++;
System.out.println(mex);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two non-negative integers, A and B. You are required to print the smallest non-negative integer such that it is not equal to both A and B.The input consists of two space separated integers A and B.
Constraints:
0 ≤ A ≤ 10
0 ≤ B ≤ 10
A ≠ B
Print a single integer denoting the answer.Sample Input 1:
0 1
Sample Output 1:
2
Sample Input 2:
4 5
Sample Output 2:
0, I have written this Solution Code: p=input()
q=list(p.split(" "))
for i in range(0,int(q[1])+2):
if(int(q[0])!=i and int(q[1])!=i):
print(i)
break, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two non-negative integers, A and B. You are required to print the smallest non-negative integer such that it is not equal to both A and B.The input consists of two space separated integers A and B.
Constraints:
0 ≤ A ≤ 10
0 ≤ B ≤ 10
A ≠ B
Print a single integer denoting the answer.Sample Input 1:
0 1
Sample Output 1:
2
Sample Input 2:
4 5
Sample Output 2:
0, I have written this Solution Code: //Author: Xzirium
//Time and Date: 15:00:01 23 April 2022
#include <bits/stdc++.h>
using namespace std;
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
int A,B;
cin>>A>>B;
for(int i=0 ; i<=10 ; i++)
{
if(i!=A && i!=B)
{
cout<<i<<endl;
break;
}
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a series and a number N, your task is to print the Nth number of the given series.
Series:- 24, 37, 50, 63, 76,. .. .<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>NthNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the Nth number.Sample Input:-
1
Sample Output:-
24
Sample Input:-
3
Sample Output:-
50, I have written this Solution Code: int NthNumber(int N){
return 24+(N-1)*13;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a series and a number N, your task is to print the Nth number of the given series.
Series:- 24, 37, 50, 63, 76,. .. .<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>NthNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the Nth number.Sample Input:-
1
Sample Output:-
24
Sample Input:-
3
Sample Output:-
50, I have written this Solution Code: int NthNumber(int N){
return 24+(N-1)*13;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a series and a number N, your task is to print the Nth number of the given series.
Series:- 24, 37, 50, 63, 76,. .. .<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>NthNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the Nth number.Sample Input:-
1
Sample Output:-
24
Sample Input:-
3
Sample Output:-
50, I have written this Solution Code: def NthNumber(N):
return 24+(N-1)*13
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a series and a number N, your task is to print the Nth number of the given series.
Series:- 24, 37, 50, 63, 76,. .. .<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>NthNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the Nth number.Sample Input:-
1
Sample Output:-
24
Sample Input:-
3
Sample Output:-
50, I have written this Solution Code: static int NthNumber(int N){
return 24+(N-1)*13;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>arr[]</b> of <b>N</b> distinct integers, check if this array is Sorted and Rotated clockwise. A sorted array is not considered sorted and rotated, i.e., there should be at least one rotation.
<b>Note:-</b>
The array can be sorted both increasingly and decreasinglyThe first line of input contains the number of test cases T. Each test case contains 2 lines, the first line contains N, the number of elements in the array, and the second line contains N space-separated elements of the array.
<b>Constraints:</b>
1 <= T <= 50
3 <= N <= 10^3
1 <= A[i] <= 10^4
Print "<b>Yes</b>" if the given array is sorted and rotated, else Print "<b>No</b>", without Inverted commas.Sample Input:
2
4
3 4 1 2
3
1 3 2
Sample Output:
Yes
Yes
<b>Explanation:</b>
Testcase 1: The array is sorted (1, 2, 3, 4) and rotated twice (3, 4, 1, 2).
Testcase 2: The array is sorted (3, 2, 1) and rotated once (1, 3, 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];
bool check_increasing(int n){
int id = 0;
for(int i = 2; i <= n; i++){
if(a[i] < a[i-1]){
id = i;
break;
}
}
if(id == 0) return 0;
int cur = id+1;
while(cur != id+n){
if(a[cur] <= a[cur-1])
return 0;
cur++;
}
return 1;
}
bool check_decreasing(int n){
int id = 0;
for(int i = 2; i <= n; i++){
if(a[i] > a[i-1]){
id = i;
break;
}
}
if(id == 0) return 0;
int cur = id+1;
while(cur != id+n){
if(a[cur] >= a[cur-1])
return 0;
cur++;
}
return 1;
}
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i], a[i+n] = a[i];
if(check_increasing(n))
cout << "Yes" << endl;
else if(check_decreasing(n))
cout << "Yes" << endl;
else
cout << "No" << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>arr[]</b> of <b>N</b> distinct integers, check if this array is Sorted and Rotated clockwise. A sorted array is not considered sorted and rotated, i.e., there should be at least one rotation.
<b>Note:-</b>
The array can be sorted both increasingly and decreasinglyThe first line of input contains the number of test cases T. Each test case contains 2 lines, the first line contains N, the number of elements in the array, and the second line contains N space-separated elements of the array.
<b>Constraints:</b>
1 <= T <= 50
3 <= N <= 10^3
1 <= A[i] <= 10^4
Print "<b>Yes</b>" if the given array is sorted and rotated, else Print "<b>No</b>", without Inverted commas.Sample Input:
2
4
3 4 1 2
3
1 3 2
Sample Output:
Yes
Yes
<b>Explanation:</b>
Testcase 1: The array is sorted (1, 2, 3, 4) and rotated twice (3, 4, 1, 2).
Testcase 2: The array is sorted (3, 2, 1) and rotated once (1, 3, 2)., I have written this Solution Code:
cases = int(input())
for _ in range(cases):
size = int(input())
orig = list(map(int,input().split()))
givenList = orig.copy()
givenList.sort()
flag = False
for _ in range(size-1):
ele = givenList.pop()
givenList.insert(0,ele)
if givenList == orig:
print("Yes")
flag = True
break
if not flag:
givenList.sort()
givenList.reverse()
for _ in range(size-1):
ele = givenList.pop()
givenList.insert(0,ele)
if givenList == orig:
print("Yes")
flag = True
break
if not flag:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>arr[]</b> of <b>N</b> distinct integers, check if this array is Sorted and Rotated clockwise. A sorted array is not considered sorted and rotated, i.e., there should be at least one rotation.
<b>Note:-</b>
The array can be sorted both increasingly and decreasinglyThe first line of input contains the number of test cases T. Each test case contains 2 lines, the first line contains N, the number of elements in the array, and the second line contains N space-separated elements of the array.
<b>Constraints:</b>
1 <= T <= 50
3 <= N <= 10^3
1 <= A[i] <= 10^4
Print "<b>Yes</b>" if the given array is sorted and rotated, else Print "<b>No</b>", without Inverted commas.Sample Input:
2
4
3 4 1 2
3
1 3 2
Sample Output:
Yes
Yes
<b>Explanation:</b>
Testcase 1: The array is sorted (1, 2, 3, 4) and rotated twice (3, 4, 1, 2).
Testcase 2: The array is sorted (3, 2, 1) and rotated once (1, 3, 2)., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim()); //Inputting the testcases
while(t-->0){
long n = Long.parseLong(br.readLine());
int arr[] = new int[(int)n];
String inputLine[] = br.readLine().trim().split("\\s+");
for(long i=0; i<n; i++){
arr[(int)i] = Integer.parseInt(inputLine[(int)i]);
}
long mini = Integer.MAX_VALUE, maxi = Integer.MIN_VALUE;
long max_index = 0, min_index = 0;
for(long i=0; i<n; i++){
if(maxi < arr[(int)i]){
maxi = arr[(int)i];
max_index = i;
}
if(mini > arr[(int)i]){
mini = arr[(int)i];
min_index = i;
}
}
int flag = 0;
if(max_index == min_index -1)
flag = 1;
else if(min_index == max_index - 1)
flag = -1;
if(flag == 1){
for(long i = 1; flag==1 && i<=max_index; ++i){
if(arr[(int)i-1] >= arr[(int)i])
flag = 0;
}
for(long i = min_index+1; flag==1 && i<n; ++i){
if(arr[(int)i-1] >= arr[(int)i])
flag = 0;
}
if(arr[0]<=arr[(int)n-1])
flag = 0;
} else if(flag == -1){
for(long i = 1; flag ==-1 && i<=min_index; ++i){
if(arr[(int)i-1] <= arr[(int)i])
flag = 0;
}
for(long i = max_index+1; flag==-1 && i<n; ++i){
if(arr[(int)i-1] <= arr[(int)i])
flag = 0;
}
if(arr[0]>=arr[(int)n-1])
flag = 0;
}
if(flag == 0)
System.out.println("No");
else
System.out.println("Yes");
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: n=int(input())
for i in range(1,n+1):
if i%3==0 and i%5==0:
print("NewtonSchool",end=" ")
elif i%3==0:
print("Newton",end=" ")
elif i%5==0:
print("School",end=" ")
else:
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static void NewtonSchool(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("NewtonSchool ");}
else if(i%5==0){System.out.print("School ");}
else if(i%3==0){System.out.print("Newton ");}
else{System.out.print(i+" ");}
}
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
NewtonSchool(x);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 0s and 1s are super cool.
You are given a binary string (string consisting of only zeros and ones). We need to modify the string such that no 0 is followed by a 1. For achieving this, we will find the leftmost occurrence of "01" substring in the string and remove it from the string. We will repeat this operation until there is no substring of the form "01" in the string.
For example, if the initial string is "011010", it will transform in the following manner:
<b>01</b>1010 -> 1<b>01</b>0 -> 10
Find the final remaining string. If the length of remaining string is 0, print -1 instead.The first and the only line of input contains the initial string, S.
Constraints
1 <= |S| <= 300000Output the remaining string. If the length of remaining string is 0, output -1.Sample Input
011010
Sample Output
10
Explanation: Available in the question text.
Sample Input
001101
Sample Output
-1
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
StringTokenizer st;
st = new StringTokenizer(br.readLine());
String s = st.nextToken();
int curzeroes = 0;
StringBuilder sb = new StringBuilder();
int len = s.length();
for(int i = 0;i<len;i++){
if(s.charAt(i) == '1'){
if(curzeroes == 0){
sb.append("1");
}
else{
curzeroes--;
}
}
else{
curzeroes++;
}
}
for(int i = 0;i<curzeroes;i++){
sb.append("0");
}
if(sb.length() == 0 && curzeroes == 0){
bw.write("-1\n");
}
else{
bw.write(sb.toString()+"\n");
}
bw.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 0s and 1s are super cool.
You are given a binary string (string consisting of only zeros and ones). We need to modify the string such that no 0 is followed by a 1. For achieving this, we will find the leftmost occurrence of "01" substring in the string and remove it from the string. We will repeat this operation until there is no substring of the form "01" in the string.
For example, if the initial string is "011010", it will transform in the following manner:
<b>01</b>1010 -> 1<b>01</b>0 -> 10
Find the final remaining string. If the length of remaining string is 0, print -1 instead.The first and the only line of input contains the initial string, S.
Constraints
1 <= |S| <= 300000Output the remaining string. If the length of remaining string is 0, output -1.Sample Input
011010
Sample Output
10
Explanation: Available in the question text.
Sample Input
001101
Sample Output
-1
, I have written this Solution Code: arr = input()
c = 0
res = ""
n =len(arr)
for i in range(n):
if arr[i]=='0':
c+=1
else:
if c==0:
res+='1'
else:
c-=1
for i in range(c):
res+='0'
if len(res)==0:
print(-1)
else:
print(res), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 0s and 1s are super cool.
You are given a binary string (string consisting of only zeros and ones). We need to modify the string such that no 0 is followed by a 1. For achieving this, we will find the leftmost occurrence of "01" substring in the string and remove it from the string. We will repeat this operation until there is no substring of the form "01" in the string.
For example, if the initial string is "011010", it will transform in the following manner:
<b>01</b>1010 -> 1<b>01</b>0 -> 10
Find the final remaining string. If the length of remaining string is 0, print -1 instead.The first and the only line of input contains the initial string, S.
Constraints
1 <= |S| <= 300000Output the remaining string. If the length of remaining string is 0, output -1.Sample Input
011010
Sample Output
10
Explanation: Available in the question text.
Sample Input
001101
Sample Output
-1
, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
string s; cin >> s;
stack<int> st;
for(int i = 0; i < (int)s.length(); i++){
if(!st.empty() && s[st.top()] == '0' && s[i] == '1'){
st.pop();
}
else
st.push(i);
}
string res = "";
while(!st.empty()){
res += s[st.top()];
st.pop();
}
reverse(res.begin(), res.end());
if(res == "") res = "-1";
cout << res;
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 matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: t=int(input())
while t!=0:
m,n=input().split()
m,n=int(m),int(n)
for i in range(m):
arr=input().strip()
if '1' in arr:
arr='1 '*n
else:
arr='0 '*n
print(arr)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
int a[N][N];
// Driver code
int main()
{
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
bool b[n];
for(int i=0;i<n;i++){
b[i]=false;
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>a[i][j];
if(a[i][j]==1){
b[i]=true;
}
}
}
for(int i=0;i<n;i++){
if(b[i]){
for(int j=0;j<m;j++){
cout<<1<<" ";
}}
else{
for(int j=0;j<m;j++){
cout<<0<<" ";
}
}
cout<<endl;
}
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws Exception{
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bf = new BufferedReader(isr);
int t = Integer.parseInt(bf.readLine());
while (t-- > 0){
String inputs[] = bf.readLine().split(" ");
int m = Integer.parseInt(inputs[0]);
int n = Integer.parseInt(inputs[1]);
String[] matrix = new String[m];
for(int i=0; i<m; i++){
matrix[i] = bf.readLine();
}
StringBuffer ones = new StringBuffer("");
StringBuffer zeros = new StringBuffer("");
for(int i=0; i<n; i++){
ones.append("1 ");
zeros.append("0 ");
}
for(int i=0; i<m; i++){
if(matrix[i].contains("1")){
System.out.println(ones);
}else{
System.out.println(zeros);
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is building a tower with N disks of sizes 1 to N.
The rules for building the tower are simple:-
*Every day you are provided with one disk having distinct size.
*The disk with larger sizes should be placed at the bottom of the tower.
*The disk with smaller sizes should be placed at the top of the tower.
*You cannot put a new disk on the top of the tower until all the larger disks that are given to you get placed.
Your task is to print N lines denoting the disk sizes that can be put on the tower on the i<sup>th</sup> day.The first line of input contains a single integer N. The next line of input contains N space separated integers depicting the size of the discs.
Constraints:-
1 <= N <= 100000
1 <= Size <= NPrint N lines. In the ith line, print the size of disks that can be placed on the top of the tower in descending order of the disk sizes.
If on the ith day no disks can be placed, then leave that line empty.Sample Input:-
5
4 5 1 2 3
Sample Output:-
5 4
3 2 1
Explanation
On the first day, the disk of size 4 is given. But you cannot put the disk on the bottom of the tower as a disk of size 5 is still remaining.
On the second day, the disk of size 5 will be given so now the disk of sizes 5 and 4 can be placed on the tower.
On the third and fourth days, disks cannot be placed on the tower as the disk of 3 needs to be given yet. Therefore, these lines are empty.
On the fifth day, all the disks of sizes 3, 2, and 1 can be placed on the top of the tower.
Sample Input:-
3
3 2 1
Sample Output:-
3
2
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String args[])throws IOException
{
int n;
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
n=Integer.parseInt(br.readLine());
int a[]=new int[n];
StringTokenizer st=new StringTokenizer(br.readLine());
for(int i=0;i<n;i++)
a[i]=Integer.parseInt(st.nextToken());
boolean vis[]=new boolean[n+1];
int max=n;
StringBuilder sb=new StringBuilder();
for(int i=0;i<n;i++)
{
vis[a[i]]=true;
if(vis[max])
{
while(vis[max])
{
sb.append(max).append(" ");
max--;
}
}
sb.append("\n");
}
System.out.println(sb);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is building a tower with N disks of sizes 1 to N.
The rules for building the tower are simple:-
*Every day you are provided with one disk having distinct size.
*The disk with larger sizes should be placed at the bottom of the tower.
*The disk with smaller sizes should be placed at the top of the tower.
*You cannot put a new disk on the top of the tower until all the larger disks that are given to you get placed.
Your task is to print N lines denoting the disk sizes that can be put on the tower on the i<sup>th</sup> day.The first line of input contains a single integer N. The next line of input contains N space separated integers depicting the size of the discs.
Constraints:-
1 <= N <= 100000
1 <= Size <= NPrint N lines. In the ith line, print the size of disks that can be placed on the top of the tower in descending order of the disk sizes.
If on the ith day no disks can be placed, then leave that line empty.Sample Input:-
5
4 5 1 2 3
Sample Output:-
5 4
3 2 1
Explanation
On the first day, the disk of size 4 is given. But you cannot put the disk on the bottom of the tower as a disk of size 5 is still remaining.
On the second day, the disk of size 5 will be given so now the disk of sizes 5 and 4 can be placed on the tower.
On the third and fourth days, disks cannot be placed on the tower as the disk of 3 needs to be given yet. Therefore, these lines are empty.
On the fifth day, all the disks of sizes 3, 2, and 1 can be placed on the top of the tower.
Sample Input:-
3
3 2 1
Sample Output:-
3
2
1, I have written this Solution Code: def Solve(arr):
size = len(arr)
queue = {}
for i in arr:
queue[i] = True
placed = []
if i == size:
placed.append(i)
size -= 1
while size in queue:
placed.append(size)
size -= 1
yield placed
N = int(input())
arr = list(map(int, input().split()))
out_ = Solve(arr)
for i_out_ in out_:
print(' '.join(map(str, i_out_))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is building a tower with N disks of sizes 1 to N.
The rules for building the tower are simple:-
*Every day you are provided with one disk having distinct size.
*The disk with larger sizes should be placed at the bottom of the tower.
*The disk with smaller sizes should be placed at the top of the tower.
*You cannot put a new disk on the top of the tower until all the larger disks that are given to you get placed.
Your task is to print N lines denoting the disk sizes that can be put on the tower on the i<sup>th</sup> day.The first line of input contains a single integer N. The next line of input contains N space separated integers depicting the size of the discs.
Constraints:-
1 <= N <= 100000
1 <= Size <= NPrint N lines. In the ith line, print the size of disks that can be placed on the top of the tower in descending order of the disk sizes.
If on the ith day no disks can be placed, then leave that line empty.Sample Input:-
5
4 5 1 2 3
Sample Output:-
5 4
3 2 1
Explanation
On the first day, the disk of size 4 is given. But you cannot put the disk on the bottom of the tower as a disk of size 5 is still remaining.
On the second day, the disk of size 5 will be given so now the disk of sizes 5 and 4 can be placed on the tower.
On the third and fourth days, disks cannot be placed on the tower as the disk of 3 needs to be given yet. Therefore, these lines are empty.
On the fifth day, all the disks of sizes 3, 2, and 1 can be placed on the top of the tower.
Sample Input:-
3
3 2 1
Sample Output:-
3
2
1, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
vector<vector<int> > SolveTower (int N, vector<int> a) {
// Write your code here
vector<vector<int> > ans;
int done = N;
priority_queue<int> pq;
for(auto x: a){
pq.push(x);
vector<int> aux;
while(!pq.empty() && pq.top() == done){
aux.push_back(done);
pq.pop();
done--;
}
ans.push_back(aux);
}
sort(a.begin(), a.end());
a.resize(unique(a.begin(), a.end()) - a.begin());
assert(a.size() == N);
return ans;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int N;
cin >> N;
assert(1 <= N && N <= 1e6);
vector<int> a(N);
for(int i_a = 0; i_a < N; i_a++)
{
cin >> a[i_a];
}
vector<vector<int> > out_;
out_ = SolveTower(N, a);
for(int i = 0; i < out_.size(); i++){
for(int j = 0; j < out_[i].size(); j++){
cout << out_[i][j] << " ";
}
cout << "\n";
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the Armstrong numbers which are present between 1 to N.
<b>A number is said to Armstrong if it is equal to sum of cube of its digits. </b>The input contains a single integer N.
Constraints:-
1 < = N < = 1000Print all the number which are armstrong numbers less than equal to N.Sample Input:-
2
Sample Output:-
1
Sample input:-
4
Sample Output:
1, I have written this Solution Code: import math
n= int(input())
for i in range (1,n+1):
arm=i
summ=0
while(arm!=0):
rem=math.pow(arm%10,3)
summ=summ+rem
arm=math.floor(arm/10)
if summ == i :
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the Armstrong numbers which are present between 1 to N.
<b>A number is said to Armstrong if it is equal to sum of cube of its digits. </b>The input contains a single integer N.
Constraints:-
1 < = N < = 1000Print all the number which are armstrong numbers less than equal to N.Sample Input:-
2
Sample Output:-
1
Sample input:-
4
Sample Output:
1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
bool checkArmstrong(int n)
{
int temp = n, sum = 0;
while(n > 0)
{
int d = n%10;
sum = sum + d*d*d;
n = n/10;
}
if(sum == temp)
return true;
return false;
}
int main()
{
int n;
cin>>n;
for(int i = 1; i <= n; i++)
{
if(checkArmstrong(i) == true)
cout << i << " ";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the Armstrong numbers which are present between 1 to N.
<b>A number is said to Armstrong if it is equal to sum of cube of its digits. </b>The input contains a single integer N.
Constraints:-
1 < = N < = 1000Print all the number which are armstrong numbers less than equal to N.Sample Input:-
2
Sample Output:-
1
Sample input:-
4
Sample Output:
1, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int digitsum,num,digit;
for(int i=1;i<=n;i++){
digitsum=0;
num=i;
while(num>0){
digit=num%10;
digitsum+=digit*digit*digit;
num/=10;
}
if(digitsum==i){System.out.print(i+" ");}
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.io.*;
public class Main {
static final int MOD = 1000000007;
public static void main(String args[]) throws IOException {
BufferedReader br
= new BufferedReader(new InputStreamReader(System.in));
String str[] = br.readLine().trim().split(" ");
int a = Integer.parseInt(str[0]);
int b = Integer.parseInt(str[1]);
int c = Integer.parseInt(str[2]);
int d = Integer.parseInt(str[3]);
int e = Integer.parseInt(str[4]);
System.out.println(grades(a, b, c, d, e));
}
static char grades(int a, int b, int c, int d, int e)
{
int sum = a+b+c+d+e;
int per = sum/5;
if(per >= 80)
return 'A';
else if(per >= 60)
return 'B';
else if(per >= 40)
return 'C';
else
return 'D';
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: li = list(map(int,input().strip().split()))
avg=0
for i in li:
avg+=i
avg=avg/5
if(avg>=80):
print("A")
elif(avg>=60 and avg<80):
print("B")
elif(avg>=40 and avg<60):
print("C")
else:
print("D"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has an array A of length N consisting of integers. The elements of the array may be positive, negative or zero. Further, Alice has a special integer K and she can choose any subarray of A and make all its elements equal to K. In other words, Alice can choose two integers i and j such that 1 ≤ i ≤ j ≤ N, and make A[i] = A[i + 1] = ... = A[j] = K.
Note that Alice can do this operation at most once, and possibly not do it at all. Find the maximum possible sum of the array.The first line consists of two space-separated integers N and K.
The second line consists of N space-separated integers – A[1], A[2], ... A[N].
<b>Constraints:</b>
1 ≤ N ≤ 1000
-1000 ≤ K ≤ 1000
-1000 ≤ A[i] ≤ 1000Print a single integer denoting the maximum possible sum of the array after the operation.Sample Input 1:
3 1
-1 5 -2
Sample Output 1:
5
Sample Explanation 1:
It is optimal to assign the subarray [3, 3]. The sum then becomes -1 + 5 + 1 = 5.
Sample Input 2:
3 0
-1 -2 3
Sample Output 2:
3
Sample Explanation 2:
It is optimal to assign the subarray [1, 2]. The sum then becomes 0 + 0 + 3 = 3.
Sample Input 3:
2 0
2 4
Sample Output 3:
6
Sample Explanation 3:
It is optimal to not assign any subarrays., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
try {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] tokens = br.readLine().split(" ");
int n = Integer.parseInt(tokens[0]), k = Integer.parseInt(tokens[1]);
int[] arr = new int[n];
tokens = br.readLine().split(" ");
for (int i = 0; i < n; i++)
arr[i] = Integer.parseInt(tokens[i]);
int[] sums = new int[n];
sums[0] = arr[0];
for (int i = 1; i < n; i++)
sums[i] = arr[i] + sums[i - 1];
int maxSum = sums[n - 1];
for (int i = 0; i < n; i++) {
int sum = sums[n - 1] - sums[i] + k * (i + 1);
if (sum > maxSum)
maxSum = sum;
}
for (int i = 1; i < n; i++) {
for (int j = i; j < n; j++) {
int sum = sums[n - 1] - (sums[j] - sums[i - 1]) + k * (j - i + 1);
if (sum > maxSum)
maxSum = sum;
}
}
System.out.println(maxSum);
} catch (Exception e) {
System.out.println("Exception caught");
return;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has an array A of length N consisting of integers. The elements of the array may be positive, negative or zero. Further, Alice has a special integer K and she can choose any subarray of A and make all its elements equal to K. In other words, Alice can choose two integers i and j such that 1 ≤ i ≤ j ≤ N, and make A[i] = A[i + 1] = ... = A[j] = K.
Note that Alice can do this operation at most once, and possibly not do it at all. Find the maximum possible sum of the array.The first line consists of two space-separated integers N and K.
The second line consists of N space-separated integers – A[1], A[2], ... A[N].
<b>Constraints:</b>
1 ≤ N ≤ 1000
-1000 ≤ K ≤ 1000
-1000 ≤ A[i] ≤ 1000Print a single integer denoting the maximum possible sum of the array after the operation.Sample Input 1:
3 1
-1 5 -2
Sample Output 1:
5
Sample Explanation 1:
It is optimal to assign the subarray [3, 3]. The sum then becomes -1 + 5 + 1 = 5.
Sample Input 2:
3 0
-1 -2 3
Sample Output 2:
3
Sample Explanation 2:
It is optimal to assign the subarray [1, 2]. The sum then becomes 0 + 0 + 3 = 3.
Sample Input 3:
2 0
2 4
Sample Output 3:
6
Sample Explanation 3:
It is optimal to not assign any subarrays., I have written this Solution Code: N, K = [int(x) for x in input().split()]
arr = [int(x) for x in input().split()]
initialsum = sum(arr)
maxtamp = 0
temp = 0
for i in range(N):
temp += (K - arr[i])
maxtamp = max(maxtamp, temp)
if temp < 0:
temp = 0
print(initialsum + maxtamp), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has an array A of length N consisting of integers. The elements of the array may be positive, negative or zero. Further, Alice has a special integer K and she can choose any subarray of A and make all its elements equal to K. In other words, Alice can choose two integers i and j such that 1 ≤ i ≤ j ≤ N, and make A[i] = A[i + 1] = ... = A[j] = K.
Note that Alice can do this operation at most once, and possibly not do it at all. Find the maximum possible sum of the array.The first line consists of two space-separated integers N and K.
The second line consists of N space-separated integers – A[1], A[2], ... A[N].
<b>Constraints:</b>
1 ≤ N ≤ 1000
-1000 ≤ K ≤ 1000
-1000 ≤ A[i] ≤ 1000Print a single integer denoting the maximum possible sum of the array after the operation.Sample Input 1:
3 1
-1 5 -2
Sample Output 1:
5
Sample Explanation 1:
It is optimal to assign the subarray [3, 3]. The sum then becomes -1 + 5 + 1 = 5.
Sample Input 2:
3 0
-1 -2 3
Sample Output 2:
3
Sample Explanation 2:
It is optimal to assign the subarray [1, 2]. The sum then becomes 0 + 0 + 3 = 3.
Sample Input 3:
2 0
2 4
Sample Output 3:
6
Sample Explanation 3:
It is optimal to not assign any subarrays., I have written this Solution Code: #include <iostream>
using namespace std;
int main()
{
int n, x;
cin >> n >> x;
int a[n + 1] = {}, pre[n + 1] = {};
for (int i = 1; i <= n; i++) cin >> a[i];
for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + a[i];
int ans = pre[n];
for (int i = 1; i <= n; i++)
for (int j = i; j <= n; j++)
ans = max(ans, pre[i - 1] + (pre[n] - pre[j]) + x*(j - i + 1));
cout << ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N (N is always even), you need to find exactly (N/2) numbers where each number represents the pair-wise product of consecutive elements of the array A.
In simple terms print (A[1]*A[2]), (A[3]*A[4]), ..., (A[N-1]*A[N]).
(Use long long int to avoid overflow)For each test case, the first line of the input contains an integer N (even number) denoting the number of elements in the array. The next line contains N (white-space separated) integers.
Constraints
1 <= N <= 10^2
1 <= A[i] <= 1000000000For each test case, output N/2 elements representing the pairwise product of adjacent elements in the array.Input-1
10
2 1 4 1 6 2 2 6 4 1
Output-1
2 4 12 12 4
Input-2
8
1 23 54 2 3 6 43 2
Output-2
23 108 18 86
Explanation(might not be the optimal solution):
Testcase 1:
Follow the below steps:-
Step 1: [2 1 4 1 6 2 2 6 4 1]
Step 2: (2 1), (4 1), (6 2), (2 6) and (4 1)
Step 3: 2 4 12 12 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
String n1 = br.readLine();
int n = Integer.parseInt(n1);
long arr[] = new long[n];
String data = br.readLine();
String s[] = data.split(" ");
long mul = 1;
for(int i = 0; i < arr.length; i++){
arr[i] = Integer.parseInt(s[i]);
}
for(int i = 0; i < arr.length; i+=2){
mul = arr[i]*arr[i+1];
System.out.print(mul+" ");
mul = 1;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N (N is always even), you need to find exactly (N/2) numbers where each number represents the pair-wise product of consecutive elements of the array A.
In simple terms print (A[1]*A[2]), (A[3]*A[4]), ..., (A[N-1]*A[N]).
(Use long long int to avoid overflow)For each test case, the first line of the input contains an integer N (even number) denoting the number of elements in the array. The next line contains N (white-space separated) integers.
Constraints
1 <= N <= 10^2
1 <= A[i] <= 1000000000For each test case, output N/2 elements representing the pairwise product of adjacent elements in the array.Input-1
10
2 1 4 1 6 2 2 6 4 1
Output-1
2 4 12 12 4
Input-2
8
1 23 54 2 3 6 43 2
Output-2
23 108 18 86
Explanation(might not be the optimal solution):
Testcase 1:
Follow the below steps:-
Step 1: [2 1 4 1 6 2 2 6 4 1]
Step 2: (2 1), (4 1), (6 2), (2 6) and (4 1)
Step 3: 2 4 12 12 4, I have written this Solution Code: a=int(input())
b=list(map(int,input().split()))
for i in range(0,a-1,2):
print(b[i]*b[i+1],end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N (N is always even), you need to find exactly (N/2) numbers where each number represents the pair-wise product of consecutive elements of the array A.
In simple terms print (A[1]*A[2]), (A[3]*A[4]), ..., (A[N-1]*A[N]).
(Use long long int to avoid overflow)For each test case, the first line of the input contains an integer N (even number) denoting the number of elements in the array. The next line contains N (white-space separated) integers.
Constraints
1 <= N <= 10^2
1 <= A[i] <= 1000000000For each test case, output N/2 elements representing the pairwise product of adjacent elements in the array.Input-1
10
2 1 4 1 6 2 2 6 4 1
Output-1
2 4 12 12 4
Input-2
8
1 23 54 2 3 6 43 2
Output-2
23 108 18 86
Explanation(might not be the optimal solution):
Testcase 1:
Follow the below steps:-
Step 1: [2 1 4 1 6 2 2 6 4 1]
Step 2: (2 1), (4 1), (6 2), (2 6) and (4 1)
Step 3: 2 4 12 12 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<n;i+=2){
cout<<a[i]*a[i+1]<<" ";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: a="Hello World"
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
System.out.println("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, check if the given string contains any white space or not.Input contains a single string S.
Constraints:-
1 < = |S| < = 20
Note:- String will only contain lowercase english letters.Print "Yes" if the given string contains a whitespace else print "No"Sample Input:-
newton school
Sample Output:-
Yes
Sample Input:-
newtonschool
Sample Output:-
No, 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();
if(str.contains(" ")) {
System.out.println("Yes");
}else {
System.out.println("No");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, check if the given string contains any white space or not.Input contains a single string S.
Constraints:-
1 < = |S| < = 20
Note:- String will only contain lowercase english letters.Print "Yes" if the given string contains a whitespace else print "No"Sample Input:-
newton school
Sample Output:-
Yes
Sample Input:-
newtonschool
Sample Output:-
No, I have written this Solution Code: s = input()
if " " in s:
print("Yes")
else:
print("No")
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, check if the given string contains any white space or not.Input contains a single string S.
Constraints:-
1 < = |S| < = 20
Note:- String will only contain lowercase english letters.Print "Yes" if the given string contains a whitespace else print "No"Sample Input:-
newton school
Sample Output:-
Yes
Sample Input:-
newtonschool
Sample Output:-
No, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
string s;
getline(cin,s);
for(int i=0;i<s.length();i++){
if(s[i]==' '){cout<<"Yes";return 0;}
}
cout<<"No";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 integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, 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') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main (String[] args)throws IOException {
Reader sc = new Reader();
int N = sc.nextInt();
int[] arr = new int[N];
for(int i=0;i<N;i++){
arr[i] = sc.nextInt();
}
int max=0;
if(arr[0]<arr[N-1])
System.out.print(N-1);
else{
for(int i=0;i<N-1;i++){
int j = N-1;
while(j>i){
if(arr[i]<arr[j]){
if(max<j-i){
max = j-i;
} break;
}
j--;
}
if(i==j)
break;
if(j==N-1)
break;
}
if(max==0)
System.out.print("-1");
else
System.out.print(max);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
/* For a given array arr[],
returns the maximum j – i such that
arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
int maxDiff;
int i, j;
int *LMin = new int[(sizeof(int) * n)];
int *RMax = new int[(sizeof(int) * n)];
/* Construct LMin[] such that
LMin[i] stores the minimum value
from (arr[0], arr[1], ... arr[i]) */
LMin[0] = arr[0];
for (i = 1; i < n; ++i)
LMin[i] = min(arr[i], LMin[i - 1]);
/* Construct RMax[] such that
RMax[j] stores the maximum value from
(arr[j], arr[j+1], ..arr[n-1]) */
RMax[n - 1] = arr[n - 1];
for (j = n - 2; j >= 0; --j)
RMax[j] = max(arr[j], RMax[j + 1]);
/* Traverse both arrays from left to right
to find optimum j - i. This process is similar to
merge() of MergeSort */
i = 0, j = 0, maxDiff = -1;
while (j < n && i < n)
{
if (LMin[i] < RMax[j])
{
maxDiff = max(maxDiff, j - i);
j = j + 1;
}
else
i = i + 1;
}
return maxDiff;
}
// Driver Code
signed main()
{
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
int maxDiff = maxIndexDiff(a, n);
cout << maxDiff;
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 integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
rightMax = [0] * n
rightMax[n - 1] = arr[n - 1]
for i in range(n - 2, -1, -1):
rightMax[i] = max(rightMax[i + 1], arr[i])
maxDist = -2**31
i = 0
j = 0
while (i < n and j < n):
if (rightMax[j] >= arr[i]):
maxDist = max(maxDist, j - i)
j += 1
else:
i += 1
if maxDist==0:
maxDist=-1
print(maxDist), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rohit loves vowels and he likes a string if it contains all the vowels atleast once.
You are given a String S of length N integers. Find minimum length substring of S which contains all vowels atleast once (it may contain other characters too).The first line contains string length and the second line contains the string.
Constraints
1 <= N <= 100000
String contains lowercase english alphabetsThe output should contain only one integer which is the length of minimum length substring containing all vowels. If no such substring exists print -1.Sample input 1:
7
aeiddou
Sample output 1:
7
Sample input 2:
7
daeioud
Sample output 2:
5
Sample input 3:
7
daeiodd
Sample output 3:
-1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static int nextI(String str,int i)
{
char c=str.charAt(i);
return 0;
}
public static boolean vowel(char c)
{
if(c=='a' || c=='e' || c=='i' || c=='o' || c=='u') return true;
else return false;
}
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();
if(n<5)
{
System.out.print(-1);
return;
}
int i=0;
while(i<n && !vowel(str.charAt(i))) ++i;
if(i==n)
{
System.out.print(-1);
return;
}
int min,max;
int len=n;
boolean found=false;
int index,l;
while(i<n)
{
max=-1;
min=n;
if(str.charAt(i)!='a')
{
index=str.indexOf('a',i+1);
if(index==-1) break;
if(index>max) max=index;
if(index<min) min=index;
}
if(str.charAt(i)!='e')
{
index=str.indexOf('e',i+1);
if(index==-1) break;
if(index>max) max=index;
if(index<min) min=index;
}
if(str.charAt(i)!='i')
{
index=str.indexOf('i',i+1);
if(index==-1) break;
if(index>max) max=index;
if(index<min) min=index;
}
if(str.charAt(i)!='o')
{
index=str.indexOf('o',i+1);
if(index==-1) break;
if(index>max) max=index;
if(index<min) min=index;
}
if(str.charAt(i)!='u')
{
index=str.indexOf('u',i+1);
if(index==-1) break;
if(index>max) max=index;
if(index<min) min=index;
}
l=max-i+1;
found=true;
if(l<len) len=l;
index=str.indexOf(str.charAt(i),i+1);
if(index==-1) break;
if(index<min) min=index;
i=min;
}
System.out.print((found)?len:-1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rohit loves vowels and he likes a string if it contains all the vowels atleast once.
You are given a String S of length N integers. Find minimum length substring of S which contains all vowels atleast once (it may contain other characters too).The first line contains string length and the second line contains the string.
Constraints
1 <= N <= 100000
String contains lowercase english alphabetsThe output should contain only one integer which is the length of minimum length substring containing all vowels. If no such substring exists print -1.Sample input 1:
7
aeiddou
Sample output 1:
7
Sample input 2:
7
daeioud
Sample output 2:
5
Sample input 3:
7
daeiodd
Sample output 3:
-1, I have written this Solution Code: t=input()
t=int(t)
x=input()
x=x[::-1]
start=0
a=0
e=0
i=0
o=0
u=0
l=[]
n=len(x)
j=0
value=100000
while(j<n):
if(x[j]=='a'):
a+=1
l.append(x[j])
elif x[j]=='e':
e+=1
l.append(x[j])
elif x[j]=='i':
i+=1
l.append(x[j])
elif x[j]=='o':
o+=1
l.append(x[j])
elif x[j]=='u':
u+=1
l.append(x[j])
while(a>=1 and e>=1 and i>=1 and o>=1 and u>=1):
if(value>(j-start+1)):
value=j-start+1
if(x[start]=='a'):
a-=1
elif(x[start]=='e'):
e-=1
elif(x[start]=='i'):
i-=1
elif(x[start]=='o'):
o-=1
elif(x[start]=='u'):
u-=1
if(l):
l.pop(0)
start=start+1
j+=1
if(value==100000):
print("-1")
else:
print (value), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rohit loves vowels and he likes a string if it contains all the vowels atleast once.
You are given a String S of length N integers. Find minimum length substring of S which contains all vowels atleast once (it may contain other characters too).The first line contains string length and the second line contains the string.
Constraints
1 <= N <= 100000
String contains lowercase english alphabetsThe output should contain only one integer which is the length of minimum length substring containing all vowels. If no such substring exists print -1.Sample input 1:
7
aeiddou
Sample output 1:
7
Sample input 2:
7
daeioud
Sample output 2:
5
Sample input 3:
7
daeiodd
Sample output 3:
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
t=1;
while(t--){
int n;
cin>>n;
string s;
cin>>s;
map<char,int> m;
m['a']=0;
m['e']=0;
m['i']=0;
m['o']=0;
m['u']=0;
int cnt=0, ans=INT_MAX,j=0;
for(int i=0;i<n;i++){
if(m.find(s[i])!=m.end()){
if(m[s[i]]==0){cnt++;}
m[s[i]]++;
}
if(cnt==5){
ans=min(ans,i-j+1);
while(1){
if(m.find(s[j])!=m.end()){m[s[j]]--;
if(m[s[j]]==0){j++;
break;}
}
j++;
ans=min(ans,i-j+1);
}
cnt--;
}
}
if(ans==INT_MAX){cout<<-1;return 0;}
cout<<ans<<endl;}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, 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 T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of even length N consisting of non-negative integers. You need to make each element of the array equal to zero, by doing some number of moves (possibly zero).
In a single move, you can choose two integers l and r such that 1 ≤ l ≤ r ≤ N. Let x = A[l] ⊕ A[l+1] ⊕ A[l+2] ... A[r], where ⊕ represents xor operator. Then, replace the whole subarray with x. In other words, assign A[i] = x for l ≤ i ≤ r.
Print the minimum number of moves required to make the entire array zero.The first line of the input contains a single integer T – the number of test cases.
Then T test cases follow, each test case in the following format:
The first line contains a single integer N – the length of array A.
The next line contains N space-separated integers representing the elements of the array A.
<b>Constraints:</b>
1≤ T ≤ 50
1 ≤ N ≤ 10<sup>3</sup>
0 ≤ A[i] ≤ 10<sup>9</sup>
N is even.Output T lines, the i<sup>th</sup> line containing a single integer – the answer to the i<sup>th</sup> test case.Sample Input:
2
2
2 2
4
7 1 2 0
Sample Output:
1
2, I have written this Solution Code: import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
public class Main {
public static void main(String[] args) throws Exception {
Thread thread = new Thread(null, new TaskAdapter(), "@debanjandhar12", 1 << 29);
thread.start();
thread.join();
}
static class TaskAdapter implements Runnable {
@Override
public void run() {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
ArrayXor solver = new ArrayXor();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
}
static class ArrayXor {
public void solve(int testNumber, InputReader in, OutputWriter out) {
int n = in.readInt();
int[] a = in.readIntArray(n);
boolean ansIs0 = true;
for (int i = 0; i < n; i++) {
if (a[i] != 0) {
ansIs0 = false;
break;
}
}
if (ansIs0) {
out.printLine(0);
return;
}
int xor = 0;
for (int i = 0; i < n; i++) {
xor ^= a[i];
}
if (xor == 0) {
out.printLine(1);
return;
}
out.printLine(2);
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int[] readIntArray(int size) {
int[] array = new int[size];
for (int i = 0; i < size; i++) {
array[i] = readInt();
}
return array;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int readInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void close() {
writer.close();
}
public void printLine(int i) {
writer.println(i);
}
}
}, 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 even length N consisting of non-negative integers. You need to make each element of the array equal to zero, by doing some number of moves (possibly zero).
In a single move, you can choose two integers l and r such that 1 ≤ l ≤ r ≤ N. Let x = A[l] ⊕ A[l+1] ⊕ A[l+2] ... A[r], where ⊕ represents xor operator. Then, replace the whole subarray with x. In other words, assign A[i] = x for l ≤ i ≤ r.
Print the minimum number of moves required to make the entire array zero.The first line of the input contains a single integer T – the number of test cases.
Then T test cases follow, each test case in the following format:
The first line contains a single integer N – the length of array A.
The next line contains N space-separated integers representing the elements of the array A.
<b>Constraints:</b>
1≤ T ≤ 50
1 ≤ N ≤ 10<sup>3</sup>
0 ≤ A[i] ≤ 10<sup>9</sup>
N is even.Output T lines, the i<sup>th</sup> line containing a single integer – the answer to the i<sup>th</sup> test case.Sample Input:
2
2
2 2
4
7 1 2 0
Sample Output:
1
2, I have written this Solution Code: for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
if a.count(0) == n:
print(0)
else:
c= 0
xor= 0
for x in a:
xor ^= x
if xor == 0:
print(1)
else:
print(2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of even length N consisting of non-negative integers. You need to make each element of the array equal to zero, by doing some number of moves (possibly zero).
In a single move, you can choose two integers l and r such that 1 ≤ l ≤ r ≤ N. Let x = A[l] ⊕ A[l+1] ⊕ A[l+2] ... A[r], where ⊕ represents xor operator. Then, replace the whole subarray with x. In other words, assign A[i] = x for l ≤ i ≤ r.
Print the minimum number of moves required to make the entire array zero.The first line of the input contains a single integer T – the number of test cases.
Then T test cases follow, each test case in the following format:
The first line contains a single integer N – the length of array A.
The next line contains N space-separated integers representing the elements of the array A.
<b>Constraints:</b>
1≤ T ≤ 50
1 ≤ N ≤ 10<sup>3</sup>
0 ≤ A[i] ≤ 10<sup>9</sup>
N is even.Output T lines, the i<sup>th</sup> line containing a single integer – the answer to the i<sup>th</sup> test case.Sample Input:
2
2
2 2
4
7 1 2 0
Sample Output:
1
2, I have written this Solution Code:
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define pi 3.141592653589793238
#define int long long
#define ll long long
#define ld long double
using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long powm(long long a, long long b,long long mod) {
long long res = 1;
while (b > 0) {
if (b & 1)
res = res * a %mod;
a = a * a %mod;
b >>= 1;
}
return res;
}
ll gcd(ll a, ll b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(0);
#ifndef ONLINE_JUDGE
if(fopen("input.txt","r"))
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
}
#endif
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
int a[n];
bool flag=true;
int ans=0;
for(int i=0;i<n;i++)
{
cin>>a[i];
ans^=a[i];
if(a[i])
flag=false;
}
if(flag)
cout<<0<<'\n';
else if(ans==0)
cout<<1<<'\n';
else
cout<<2<<'\n';
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: def partition(array, low, high):
pivot = array[high]
i = low - 1
for j in range(low, high):
if array[j] <= pivot:
i = i + 1
(array[i], array[j]) = (array[j], array[i])
(array[i + 1], array[high]) = (array[high], array[i + 1])
return i + 1
def quick_sort(array, low, high):
if low < high:
pi = partition(array, low, high)
quick_sort(array, low, pi - 1)
quick_sort(array, pi + 1, high)
t=int(input())
for i in range(t):
n=int(input())
a=input().strip().split()
a=[int(i) for i in a]
quick_sort(a, 0, n - 1)
for i in a:
print(i,end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code:
public static int[] quickSort(int arr[], int low, int high)
{
if (low < high)
{
/* pi is partitioning index, arr[pi] is
now at right place */
int pi = partition(arr, low, high);
// Recursively sort elements before
// partition and after partition
quickSort(arr, low, pi-1);
quickSort(arr, pi+1, high);
}
return arr;
}
public static int partition(int arr[], int low, int high)
{
int pivot = arr[high];
int i = (low-1); // index of smaller element
for (int j=low; j<high; j++)
{
// If current element is smaller than the pivot
if (arr[j] < pivot)
{
i++;
// swap arr[i] and arr[j]
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
// swap arr[i+1] and arr[high] (or pivot)
int temp = arr[i+1];
arr[i+1] = arr[high];
arr[high] = temp;
return i+1;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: function quickSort(arr, low, high)
{
if(low < high)
{
let pi = partition(arr, low, high);
quickSort(arr, low, pi-1);
quickSort(arr, pi+1, high);
}
return arr;
}
function partition(arr, low, high)
{
let pivot = arr[high];
let i = (low-1); // index of smaller element
for (let j=low; j<high; j++)
{
// If current element is smaller than the pivot
if (arr[j] < pivot)
{
i++;
// swap arr[i] and arr[j]
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
// swap arr[i+1] and arr[high] (or pivot)
let temp = arr[i+1];
arr[i+1] = arr[high];
arr[high] = temp;
return i+1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given are integers a, b, c, and d. If x and y are integers and a≤x≤b and c≤y≤d hold, what is the maximum possible value of x*y?The input consists of four space-separated integers.
a b c d
<b>Constraints</b>
−10^9≤a≤b≤10^9
−10^9≤c≤d≤10^9
All values in the input are integers.Print the maximum product.<b>Sample Input 1</b>
1 2 1 1
<b>Sample Output 1</b>
2
<b>Sample Input 2</b>
3 5 -4 -2
<b>Sample Output 2</b>
-6
<b>Sample Input 3</b>
-1000000000 0 -1000000000 0
<b>Sample Output 3</b>
1000000000000000000, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
long long a,b,c,d;
cin>>a>>b>>c>>d;
cout<<max(max(a*c,a*d),max(b*c,b*d))<<"\n";
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => b - a)
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: def bubbleSort(arr):
arr.sort(reverse = True)
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int t;
for(int i=1;i<n;i++){
if(a[i]>a[i-1]){
for(int j=i;j>0;j--){
if(a[j]>a[j-1]){
t=a[j];
a[j]=a[j-1];
a[j-1]=t;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code:
// author-Shivam gupta
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
const int N = 3e5+5;
#define read(type) readInt<type>()
#define max1 100001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
using vl = vector<ll>;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
long long phi[max1], result[max1],F[max1];
// Precomputation of phi[] numbers. Refer below link
// for details : https://goo.gl/LUqdtY
void computeTotient()
{
// Refer https://goo.gl/LUqdtY
phi[1] = 1;
for (int i=2; i<max1; i++)
{
if (!phi[i])
{
phi[i] = i-1;
for (int j = (i<<1); j<max1; j+=i)
{
if (!phi[j])
phi[j] = j;
phi[j] = (phi[j]/i)*(i-1);
}
}
}
for(int i=1;i<=100000;i++)
{
for(int j=i;j<=100000;j+=i)
{ int p=j/i;
F[j]+=(i*phi[p])%mod;
F[j]%=mod;
}
}
}
int gcd(int a, int b, int& x, int& y) {
if (b == 0) {
x = 1;
y = 0;
return a;
}
int x1, y1;
int d = gcd(b, a % b, x1, y1);
x = y1;
y = x1 - y1 * (a / b);
return d;
}
bool find_any_solution(int a, int b, int c, int &x0, int &y0, int &g) {
g = gcd(abs(a), abs(b), x0, y0);
if (c % g) {
return false;
}
x0 *= c / g;
y0 *= c / g;
if (a < 0) x0 = -x0;
if (b < 0) y0 = -y0;
return true;
}
int main() {
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n,greater<int>());
FOR(i,n){
out1(a[i]);}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After the decimation, the world went into chaos. People had to rebuild the planet so Shield came up with a donation strategy. They feel all the rich guys need to donate more than the poor guys. So, they make a rule. They would make a donation list in which the donation of each person would be shown. But the rule is that a person can’t pay less than what has already been paid before them. Find the extra amount each person will pay, and also, tell shield the total amount of donation.The first line contains n, the total number of people donating. The next line contains n space-separated integers denoting the amount of money paid by the people. The amounts are mentioned in the order in which the people paid.
<b>Constraints:-</b>
1 <= n <= 100000
0 <= money <= 100000The first line contains the extra money that each student has to pay after their teacher applied the rule. The second line contains the total amount collected by the teacher at the end.Sample Input 1:-
10
1 2 3 2 4 3 6 6 7 6
Sample Output 1:-
0 0 0 1 0 1 0 0 0 1
43
Sample Input 2:-
7
10 20 30 40 30 20 10
Sample Output 2:-
0 0 0 0 10 20 30
220, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
InputStreamReader read = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(read);
int n = Integer.parseInt(in.readLine());
int []donationList = new int[n];
int[] defaulterList = new int[n];
long totalDonations = 0L;
StringTokenizer st = new StringTokenizer(in.readLine());
for(int i=0 ; i<n ; i++){
donationList[i] = Integer.parseInt(st.nextToken());
}
int max = Integer.MIN_VALUE;
for(int i=0; i<n; i++){
totalDonations += donationList[i];
max = Math.max(max,donationList[i]);
if(i>0){
if(donationList[i] >= max)
defaulterList[i] = 0;
else
defaulterList[i] = max - donationList[i];
}
totalDonations += defaulterList[i];
}
for(int i=0; i<n ;i++){
System.out.print(defaulterList[i]+" ");
}
System.out.println();
System.out.print(totalDonations);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After the decimation, the world went into chaos. People had to rebuild the planet so Shield came up with a donation strategy. They feel all the rich guys need to donate more than the poor guys. So, they make a rule. They would make a donation list in which the donation of each person would be shown. But the rule is that a person can’t pay less than what has already been paid before them. Find the extra amount each person will pay, and also, tell shield the total amount of donation.The first line contains n, the total number of people donating. The next line contains n space-separated integers denoting the amount of money paid by the people. The amounts are mentioned in the order in which the people paid.
<b>Constraints:-</b>
1 <= n <= 100000
0 <= money <= 100000The first line contains the extra money that each student has to pay after their teacher applied the rule. The second line contains the total amount collected by the teacher at the end.Sample Input 1:-
10
1 2 3 2 4 3 6 6 7 6
Sample Output 1:-
0 0 0 1 0 1 0 0 0 1
43
Sample Input 2:-
7
10 20 30 40 30 20 10
Sample Output 2:-
0 0 0 0 10 20 30
220, I have written this Solution Code: n = int(input())
a = input().split()
b = int(a[0])
sum = 0
for i in a:
if int(i)<b:
print(b-int(i),end=' ')
else:
b = int(i)
print(0,end=' ')
sum = sum+b
print()
print(sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After the decimation, the world went into chaos. People had to rebuild the planet so Shield came up with a donation strategy. They feel all the rich guys need to donate more than the poor guys. So, they make a rule. They would make a donation list in which the donation of each person would be shown. But the rule is that a person can’t pay less than what has already been paid before them. Find the extra amount each person will pay, and also, tell shield the total amount of donation.The first line contains n, the total number of people donating. The next line contains n space-separated integers denoting the amount of money paid by the people. The amounts are mentioned in the order in which the people paid.
<b>Constraints:-</b>
1 <= n <= 100000
0 <= money <= 100000The first line contains the extra money that each student has to pay after their teacher applied the rule. The second line contains the total amount collected by the teacher at the end.Sample Input 1:-
10
1 2 3 2 4 3 6 6 7 6
Sample Output 1:-
0 0 0 1 0 1 0 0 0 1
43
Sample Input 2:-
7
10 20 30 40 30 20 10
Sample Output 2:-
0 0 0 0 10 20 30
220, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
void solve(){
int n;
cin>>n;
int a[n];
int ma=0;
int cnt=0;
//map<int,int> m;
for(int i=0;i<n;i++){
cin>>a[i];
ma=max(ma,a[i]);
cout<<ma-a[i]<<" ";
cnt+=ma-a[i];
cnt+=a[i];
//m[a[i]]++;
}
cout<<endl;
cout<<cnt<<endl;
}
signed main(){
int t;
t=1;
while(t--){
solve();}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public 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 {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton wants to take revenge from two apples fallen on his head. So, he applies force F<sub>1</sub> on first apple (mass M<sub>1</sub>) resulting in acceleration of A<sub>1</sub> and F<sub>2</sub> on second apple (mass M<sub>2</sub>) resulting in acceleration of A<sub>2</sub>. Given M<sub>1</sub>, A<sub>1</sub>, M<sub>2</sub>, A<sub>2</sub>. Calculate total force applied by him on two apples.
<b>Note:</b> F = M*A is the equation of relation between force, mass and acceleration.First line contains four integers M<sub>1</sub>, A<sub>1</sub>, M<sub>2</sub>, A<sub>2</sub>.
1 <= M<sub>1</sub>, A<sub>1</sub>, M<sub>2</sub>, A<sub>2</sub> <= 100Output total force applied by Newton.INPUT:
1 2 3 4
OUTPUT:
14
Explanation:
Total force is equal to 1*2 + 3*4 = 14., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
int m1,a1,m2,a2;
cin >> m1 >> a1 >> m2 >> a2;
cout << (m1*a1)+(m2*a2) << endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers your task is to calculate the maximum integer among the given integers.The input contains three integers a, b, and c
<b>Constraint:</b>
1<=integers<=10000Print the maximum integer among the given integers.Sample Input:-
2 6 3
Sample Output:-
6
Sample Input:-
48 100 100
Sample Output:
100, I have written this Solution Code: a,b,c=[int(a) for a in input().split()]
print(max(a,b,c)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers your task is to calculate the maximum integer among the given integers.The input contains three integers a, b, and c
<b>Constraint:</b>
1<=integers<=10000Print the maximum integer among the given integers.Sample Input:-
2 6 3
Sample Output:-
6
Sample Input:-
48 100 100
Sample Output:
100, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int p = scanner.nextInt();
int tm = scanner.nextInt();
int r = scanner.nextInt();
int intrst = MaxInteger(p,tm,r);
System.out.println(intrst);
}
static int MaxInteger(int a ,int b, int c){
if(a>=b && a>=c){return a;}
if(b>=a && b>=c){return b;}
return c;}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n, find the highest power of 2 that is smaller than or equal to n.The first line of the input contains the value n.
Constraints
1 <= n <= 1e18Print the highest power of 2 that is smaller than or equal to n.Sample Input
10
Sample Output
8
Explanation
8 is the highest power of 2 which is less than equal to 10, 16 is also there but it's higher than 10., I have written this Solution Code: a=int(input())
x=1
while(x<a):
x=x*2
print(x//2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n, find the highest power of 2 that is smaller than or equal to n.The first line of the input contains the value n.
Constraints
1 <= n <= 1e18Print the highest power of 2 that is smaller than or equal to n.Sample Input
10
Sample Output
8
Explanation
8 is the highest power of 2 which is less than equal to 10, 16 is also there but it's higher than 10., I have written this Solution Code: /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static long log2(long N){
long result = (long)(Math.log(N) / Math.log(2));
return result;
}
public static void main (String[] args) throws java.lang.Exception
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
long n;
String s = br.readLine();
String ip[] = s.split(" ");
n = Long.parseLong(ip[0]);
double p = (double)log2(n);
double x = 2;
double res = Math.pow(x,p);
System.out.println((long)res);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n, find the highest power of 2 that is smaller than or equal to n.The first line of the input contains the value n.
Constraints
1 <= n <= 1e18Print the highest power of 2 that is smaller than or equal to n.Sample Input
10
Sample Output
8
Explanation
8 is the highest power of 2 which is less than equal to 10, 16 is also there but it's higher than 10., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
int32_t main() {
int n;
cin >> n;
int p = (int)log2(n);
cout << (int)pow(2, p) << "\n";
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 find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: def compound_interest(principle, rate, time):
Amount = principle * (pow((1 + rate / 100), time))
CI = Amount - principle
print( '%.2f'%CI)
principle,rate,time=map(int, input().split())
compound_interest(principle,rate,time), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: function calculateCI(P, R, T)
{
let interest = P * (Math.pow(1.0 + R/100.0, T) - 1);
return interest.toFixed(2);
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int p,r,t;
cin>>p>>r>>t;
double rate= (float)r/100;
double amt = (float)p*(pow(1+rate,t));
cout << fixed << setprecision(2) << (amt - p);
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 find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] s= br.readLine().split(" ");
double[] darr = new double[s.length];
for(int i=0;i<s.length;i++){
darr[i] = Double.parseDouble(s[i]);
}
double ans = darr[0]*Math.pow(1+darr[1]/100,darr[2])-darr[0];
System.out.printf("%.2f",ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the Armstrong numbers which are present between 1 to N.
<b>A number is said to Armstrong if it is equal to sum of cube of its digits. </b>The input contains a single integer N.
Constraints:-
1 < = N < = 1000Print all the number which are armstrong numbers less than equal to N.Sample Input:-
2
Sample Output:-
1
Sample input:-
4
Sample Output:
1, I have written this Solution Code: import math
n= int(input())
for i in range (1,n+1):
arm=i
summ=0
while(arm!=0):
rem=math.pow(arm%10,3)
summ=summ+rem
arm=math.floor(arm/10)
if summ == i :
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the Armstrong numbers which are present between 1 to N.
<b>A number is said to Armstrong if it is equal to sum of cube of its digits. </b>The input contains a single integer N.
Constraints:-
1 < = N < = 1000Print all the number which are armstrong numbers less than equal to N.Sample Input:-
2
Sample Output:-
1
Sample input:-
4
Sample Output:
1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
bool checkArmstrong(int n)
{
int temp = n, sum = 0;
while(n > 0)
{
int d = n%10;
sum = sum + d*d*d;
n = n/10;
}
if(sum == temp)
return true;
return false;
}
int main()
{
int n;
cin>>n;
for(int i = 1; i <= n; i++)
{
if(checkArmstrong(i) == true)
cout << i << " ";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the Armstrong numbers which are present between 1 to N.
<b>A number is said to Armstrong if it is equal to sum of cube of its digits. </b>The input contains a single integer N.
Constraints:-
1 < = N < = 1000Print all the number which are armstrong numbers less than equal to N.Sample Input:-
2
Sample Output:-
1
Sample input:-
4
Sample Output:
1, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int digitsum,num,digit;
for(int i=1;i<=n;i++){
digitsum=0;
num=i;
while(num>0){
digit=num%10;
digitsum+=digit*digit*digit;
num/=10;
}
if(digitsum==i){System.out.print(i+" ");}
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: n=int(input())
a=map(int,input().split())
b=[]
mx=-200000
cnt=0
for i in a:
if i>mx:
cnt+=1
mx=i
print(cnt), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: function numberOfRoofs(arr)
{
let count=1;
let max = arr[0];
for(let i=1;i<arrSize;i++)
{
if(arr[i] > max)
{
count++;
max = arr[i];
}
}
return count;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
Scanner s=new Scanner(System.in);
int n=s.nextInt();
int []a=new int[n];
for(int i=0;i<n;i++){
a[i]=s.nextInt();
}
int count=1;
int max = a[0];
for(int i=1;i<n;i++)
{
if(a[i] > max)
{
count++;
max = a[i];
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to find the count of repeated elements. Print the repeated elements in ascending order along with their frequency.
Have a look at the example for more understanding.The first line of input contains a single integer N, the next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000For each duplicate element in sorted order in a new line, First, print the duplicate element and then print its number of occurence space- separated.
Note:- It is guaranteed that at least one duplicate element will exist in the given array.Sample Input:-
5
3 2 1 1 2
Sample Output:-
1 2
2 2
Sample Input:-
5
1 1 1 1 5
Sample Output:-
1 4
Explaination:
test 1: Only 1 and 2 are repeated. Both are repeated twice. So, we print:
1 -> frequency of 1
2 -> frequency of 2
1 is printed before 2 as it is smaller than 2, I have written this Solution Code: import numpy as np
from collections import defaultdict
n=int(input())
a=np.array([input().strip().split()],int).flatten()
d=defaultdict(int)
for i in a:
d[i]+=1
d=sorted(d.items())
for i in d:
if(i[1]>1):
print(i[0],end=" ")
print(i[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 elements, your task is to find the count of repeated elements. Print the repeated elements in ascending order along with their frequency.
Have a look at the example for more understanding.The first line of input contains a single integer N, the next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000For each duplicate element in sorted order in a new line, First, print the duplicate element and then print its number of occurence space- separated.
Note:- It is guaranteed that at least one duplicate element will exist in the given array.Sample Input:-
5
3 2 1 1 2
Sample Output:-
1 2
2 2
Sample Input:-
5
1 1 1 1 5
Sample Output:-
1 4
Explaination:
test 1: Only 1 and 2 are repeated. Both are repeated twice. So, we print:
1 -> frequency of 1
2 -> frequency of 2
1 is printed before 2 as it is smaller than 2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
InputStreamReader read = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(read);
int noOfElements = Integer.parseInt(in.readLine());
String [] elem = in.readLine().trim().split(" ");
int [] elements = new int [noOfElements];
for(int i=0 ; i< noOfElements; i++){
elements[i] = Integer.parseInt(elem[i]);
}
Arrays.sort(elements);
int count =0;
for(int i = 0 ; i < noOfElements-1 ; i++){
if(elements[i] == elements[i+1]){
count ++;
}
else if(count != 0){
System.out.print(elements[i]+" "+(count+1));
count =0;
System.out.println();
}
}
if(count != 0)
System.out.print(elements[noOfElements-1]+" "+(count+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 elements, your task is to find the count of repeated elements. Print the repeated elements in ascending order along with their frequency.
Have a look at the example for more understanding.The first line of input contains a single integer N, the next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000For each duplicate element in sorted order in a new line, First, print the duplicate element and then print its number of occurence space- separated.
Note:- It is guaranteed that at least one duplicate element will exist in the given array.Sample Input:-
5
3 2 1 1 2
Sample Output:-
1 2
2 2
Sample Input:-
5
1 1 1 1 5
Sample Output:-
1 4
Explaination:
test 1: Only 1 and 2 are repeated. Both are repeated twice. So, we print:
1 -> frequency of 1
2 -> frequency of 2
1 is printed before 2 as it is smaller than 2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin>>n;
int a[n];
map<int,int> m;
for(int i=0;i<n;i++){
cin>>a[i];
m[a[i]]++;
}
for(auto it = m.begin();it!=m.end();it++){
if(it->second>1){
cout<<it->first<<" "<<it->second<<'\n';
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of N integers, find the pair of integers in the array which have minimum XOR value. Report the minimum XOR value.First line denotes N, the size of the array.
Next line denotes N space-separated array elements.
Constraints:
2 <= N <= 100000
0 <= A[i] <= 10^7Print a single integer denoting minimum xor valueSample Input
4
0 2 5 7
Sample Output
2
Explanation:
0 xor 2 = 2
Sample Input
4
0 4 7 9
Sample Output
3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
int ans = 0;
int mini = Integer.MAX_VALUE;
Scanner sc = new Scanner(System.in);
int array_size = sc.nextInt();
int N[] = new int[array_size];
for (int i = 0; i < array_size; i++) {
if(mini == 0){
break;
}
N[i] = sc.nextInt();
for (int j = i - 1; j >= 0; j--) {
ans = N[i] ^ N[j];
if (mini > ans) {
mini = ans;
}
}
}
System.out.println(mini);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of N integers, find the pair of integers in the array which have minimum XOR value. Report the minimum XOR value.First line denotes N, the size of the array.
Next line denotes N space-separated array elements.
Constraints:
2 <= N <= 100000
0 <= A[i] <= 10^7Print a single integer denoting minimum xor valueSample Input
4
0 2 5 7
Sample Output
2
Explanation:
0 xor 2 = 2
Sample Input
4
0 4 7 9
Sample Output
3, I have written this Solution Code: a=int(input())
lis = list(map(int,input().split()))
val=666666789
for i in range(a+1):
for j in range(i+1,a):
temp = lis[i]^lis[j]
if(temp<val):
val = temp
if(val==0):
break
print(val), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of N integers, find the pair of integers in the array which have minimum XOR value. Report the minimum XOR value.First line denotes N, the size of the array.
Next line denotes N space-separated array elements.
Constraints:
2 <= N <= 100000
0 <= A[i] <= 10^7Print a single integer denoting minimum xor valueSample Input
4
0 2 5 7
Sample Output
2
Explanation:
0 xor 2 = 2
Sample Input
4
0 4 7 9
Sample Output
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
// Function to find minimum XOR pair
int minXOR(int arr[], int n)
{
// Sort given array
sort(arr, arr + n);
int minXor = INT_MAX;
int val = 0;
// calculate min xor of consecutive pairs
for (int i = 0; i < n - 1; i++) {
val = arr[i] ^ arr[i + 1];
minXor = min(minXor, val);
}
return minXor;
}
// Driver program
int main()
{
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
cout << minXOR(arr, n) << endl;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Calculate inversion count of array of integers.
Inversion count of an array is quantisation of how much unsorted an array is. A sorted array has inversion count 0, while an unsorted array has maximum inversion count.
Formally speaking inversion count = number of pairs i, j such that i < j and a[i] > a[j].The first line contain integers N.
The second line of the input contains N singly spaces integers.
1 <= N <= 100000
1 <= A[i] <= 1000000000Output one integer the inversion count.Sample Input
5
1 1 3 2 2
Sample Output
2
Sample Input
5
5 4 3 2 1
Sample Output
10, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static long count(int[] arr, int l, int h, int[] aux) {
if (l >= h) return 0;
int mid = (l +h) / 2;
long count = 0;
count += count(aux, l, mid, arr);
count += count(aux, mid + 1, h, arr);
count += merge(arr, l, mid, h, aux);
return count;
}
static long merge(int[] arr, int l, int mid, int h, int[] aux) {
long count = 0;
int i = l, j = mid + 1, k = l;
while (i <= mid || j <= h) {
if (i > mid) {
arr[k++] = aux[j++];
} else if (j > h) {
arr[k++] = aux[i++];
} else if (aux[i] <= aux[j]) {
arr[k++] = aux[i++];
} else {
arr[k++] = aux[j++];
count += mid + 1 - i;
}
}
return count;
}
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
String str[];
str = br.readLine().split(" ");
int n = Integer.parseInt(str[0]);
str = br.readLine().split(" ");
int arr[] =new int[n];
for (int j = 0; j < n; j++) {
arr[j] = Integer.parseInt(str[j]);
}
int[] aux = arr.clone();
System.out.print(count(arr, 0, n - 1, aux));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Calculate inversion count of array of integers.
Inversion count of an array is quantisation of how much unsorted an array is. A sorted array has inversion count 0, while an unsorted array has maximum inversion count.
Formally speaking inversion count = number of pairs i, j such that i < j and a[i] > a[j].The first line contain integers N.
The second line of the input contains N singly spaces integers.
1 <= N <= 100000
1 <= A[i] <= 1000000000Output one integer the inversion count.Sample Input
5
1 1 3 2 2
Sample Output
2
Sample Input
5
5 4 3 2 1
Sample Output
10, I have written this Solution Code: count=0
def implementMergeSort(arr,s,e):
global count
if e-s==1:
return
mid=(s+e)//2
implementMergeSort(arr,s,mid)
implementMergeSort(arr,mid,e)
count+=merge_sort_place(arr,s,mid,e)
return count
def merge_sort_place(arr,s,mid,e):
arr3=[]
i=s
j=mid
count=0
while i<mid and j<e:
if arr[i]>arr[j]:
arr3.append(arr[j])
j+=1
count+=(mid-i)
else:
arr3.append(arr[i])
i+=1
while (i<mid):
arr3.append(arr[i])
i+=1
while (j<e):
arr3.append(arr[j])
j+=1
for x in range(len(arr3)):
arr[s+x]=arr3[x]
return count
n=int(input())
arr=list(map(int,input().split()[:n]))
c=implementMergeSort(arr,0,len(arr))
print(c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Calculate inversion count of array of integers.
Inversion count of an array is quantisation of how much unsorted an array is. A sorted array has inversion count 0, while an unsorted array has maximum inversion count.
Formally speaking inversion count = number of pairs i, j such that i < j and a[i] > a[j].The first line contain integers N.
The second line of the input contains N singly spaces integers.
1 <= N <= 100000
1 <= A[i] <= 1000000000Output one integer the inversion count.Sample Input
5
1 1 3 2 2
Sample Output
2
Sample Input
5
5 4 3 2 1
Sample Output
10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
long long _mergeSort(long long arr[], int temp[], int left, int right);
long long merge(long long arr[], int temp[], int left, int mid, int right);
/* This function sorts the input array and returns the
number of inversions in the array */
long long mergeSort(long long arr[], int array_size)
{
int temp[array_size];
return _mergeSort(arr, temp, 0, array_size - 1);
}
/* An auxiliary recursive function that sorts the input array and
returns the number of inversions in the array. */
long long _mergeSort(long long arr[], int temp[], int left, int right)
{
long long mid, inv_count = 0;
if (right > left) {
/* Divide the array into two parts and
call _mergeSortAndCountInv()
for each of the parts */
mid = (right + left) / 2;
/* Inversion count will be sum of
inversions in left-part, right-part
and number of inversions in merging */
inv_count += _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid + 1, right);
/*Merge the two parts*/
inv_count += merge(arr, temp, left, mid + 1, right);
}
return inv_count;
}
/* This funt merges two sorted arrays
and returns inversion count in the arrays.*/
long long merge(long long arr[], int temp[], int left,
int mid, int right)
{
int i, j, k;
long long inv_count = 0;
i = left; /* i is index for left subarray*/
j = mid; /* j is index for right subarray*/
k = left; /* k is index for resultant merged subarray*/
while ((i <= mid - 1) && (j <= right)) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
}
else {
temp[k++] = arr[j++];
/* this is tricky -- see above
explanation/diagram for merge()*/
inv_count = inv_count + (mid - i);
}
}
/* Copy the remaining elements of left subarray
(if there are any) to temp*/
while (i <= mid - 1)
temp[k++] = arr[i++];
/* Copy the remaining elements of right subarray
(if there are any) to temp*/
while (j <= right)
temp[k++] = arr[j++];
/*Copy back the merged elements to original array*/
for (i = left; i <= right; i++)
arr[i] = temp[i];
return inv_count;}
int main(){
int n;
cin>>n;
long long a[n];
for(int i=0;i<n;i++){
cin>>a[i];}
long long ans = mergeSort(a, n);
cout << ans; }
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers <b>a</b> and <b>b</b>, your task is to calculate and print the following four values:-
a+b
a-b
a*b
a/bThe input contains two integers a and b separated by spaces.
<b>Constraints:</b>
1 ≤ b ≤ a ≤ 1000
<b> It is guaranteed that a will be divisible by b</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5
<b>Explanation:-</b>
First operation is a+b so 15+3 = 18
The second Operation is a-b so 15-3 = 12
Third Operation is a*b so 15*3 = 45
Fourth Operation is a/b so 15/3 = 5, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args){
FastReader read = new FastReader();
int a = read.nextInt();
int b = read.nextInt();
System.out.println(a+b);
System.out.println(a-b);
System.out.println(a*b);
System.out.println(a/b);
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader(){
InputStreamReader inr = new InputStreamReader(System.in);
br = new BufferedReader(inr);
}
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());
}
double nextDouble(){
return Double.parseDouble(next());
}
long nextLong(){
return Long.parseLong(next());
}
String nextLine(){
String str = "";
try{
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: Given two integers <b>a</b> and <b>b</b>, your task is to calculate and print the following four values:-
a+b
a-b
a*b
a/bThe input contains two integers a and b separated by spaces.
<b>Constraints:</b>
1 ≤ b ≤ a ≤ 1000
<b> It is guaranteed that a will be divisible by b</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5
<b>Explanation:-</b>
First operation is a+b so 15+3 = 18
The second Operation is a-b so 15-3 = 12
Third Operation is a*b so 15*3 = 45
Fourth Operation is a/b so 15/3 = 5, I have written this Solution Code: a,b=map(int,input().split())
print(a+b)
print(a-b)
print(a*b)
print(a//b), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
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