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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 "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: def fizzbuzz(n):
for i in range (1,n+1):
if (i%3==0 and i%5==0):
print("FizzBuzz",end=' ')
elif i%3==0:
print("Fizz",end=' ')
elif i%5==0:
print("Buzz",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 "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){printf("FizzBuzz ");}
else if(i%5==0){printf("Buzz ");}
else if(i%3==0){printf("Fizz ");}
else{printf("%d ",i);}
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: #include <bits/stdc++.h>
// #define ll long long
using namespace std;
#define ma 10000001
bool a[ma];
int main()
{
int n;
cin>>n;
for(int i=0;i<=n;i++){
a[i]=false;
}
for(int i=2;i<=n;i++){
if(a[i]==false){
for(int j=i+i;j<=n;j+=i){
a[j]=true;
}
}
}
int cnt=0;
for(int i=2;i<=n;i++){
if(a[i]==false){cnt++;}
}
cout<<cnt;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, 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 IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
long n = Integer.parseInt(br.readLine());
long i=2,j,count,noOfPrime=0;
if(n<=1)
System.out.println("0");
else{
while(i<=n)
{
count=0;
for(j=2; j<=Math.sqrt(i); j++)
{
if( i%j == 0 ){
count++;
break;
}
}
if(count==0){
noOfPrime++;
}
i++;
}
System.out.println(noOfPrime);
}
}
}, 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 number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: function numberOfPrimes(N)
{
let arr = new Array(N+1);
for(let i = 0; i <= N; i++)
arr[i] = 0;
for(let i=2; i<= N/2; i++)
{
if(arr[i] === -1)
{
continue;
}
let p = i;
for(let j=2; p*j<= N; j++)
{
arr[p*j] = -1;
}
}
//console.log(arr);
let count = 0;
for(let i=2; i<= N; i++)
{
if(arr[i] === 0)
{
count++;
}
}
//console.log(arr);
return count;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import math
n = int(input())
n=n+1
if n<3:
print(0)
else:
primes=[1]*(n//2)
for i in range(3,int(math.sqrt(n))+1,2):
if primes[i//2]:primes[i*i//2::i]=[0]*((n-i*i-1)//(2*i)+1)
print(sum(primes)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of N nodes and an integer K, your task is to delete the Kth node from the end of the linked list<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>deleteElement()</b> that takes head node and K as parameter.
Constraints:
1 <=K<=N<= 1000
1 <=Node.data<= 1000Return the head of the modified linked listInput 1:
5 3
1 2 3 4 5
Output 1:
1 2 4 5
Explanation:
After deleting 3rd node from the end of the linked list, 3 will be deleted and the list will be as 1, 2, 4, 5.
Input 2:-
5 5
8 1 8 3 6
Output 2:-
1 8 3 6 , I have written this Solution Code: public static Node deleteElement(Node head,int k) {
int cnt=0;
Node temp=head;
while(temp!=null){
cnt++;
temp=temp.next;
}
temp=head;
k=cnt-k;
if(k==0){head=head.next; return head;}
temp=head;
cnt=0;
while(cnt!=k-1){
temp=temp.next;
cnt++;
}
temp.next=temp.next.next;
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print the pattern of "*" in the form of the Right Angle Triangle.
See the below example for clarity.Since this is a functional problem, you don't have to worry about the input. It will be handled by driver code. You just have to complete <b>printTriangle()</b>.
In the custom input area, you can provide any positive integer and check whether your code is working or not.Print the right angle triangle of height 5 as shown.Sample Input:
No Input
Sample Output:-
*
* *
* * *
* * * *
* * * * *, I have written this Solution Code: class Solution {
public static void printTriangle(){
System.out.println("*");
System.out.println("* *");
System.out.println("* * *");
System.out.println("* * * *");
System.out.println("* * * * *");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print the pattern of "*" in the form of the Right Angle Triangle.
See the below example for clarity.Since this is a functional problem, you don't have to worry about the input. It will be handled by driver code. You just have to complete <b>printTriangle()</b>.
In the custom input area, you can provide any positive integer and check whether your code is working or not.Print the right angle triangle of height 5 as shown.Sample Input:
No Input
Sample Output:-
*
* *
* * *
* * * *
* * * * *, I have written this Solution Code: j=1
for i in range(0,5):
for k in range(0,j):
print("*",end=" ")
if(j<=4):
print()
j=j+1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a side of a square, your task is to calculate and print its area.The first line of the input contains the side of the square.
<b>Constraints:</b>
1 <= side <=100You just have to print the area of a squareSample Input:-
3
Sample Output:-
9
Sample Input:-
6
Sample Output:-
36, I have written this Solution Code: def area(side_of_square):
print(side_of_square*side_of_square)
def main():
N = int(input())
area(N)
if __name__ == '__main__':
main(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a side of a square, your task is to calculate and print its area.The first line of the input contains the side of the square.
<b>Constraints:</b>
1 <= side <=100You just have to print the area of a squareSample Input:-
3
Sample Output:-
9
Sample Input:-
6
Sample Output:-
36, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int side = Integer.parseInt(br.readLine());
System.out.print(side*side);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine().trim();
char n[] = s.toCharArray();
int posMin = 0;
for(int i=0;i<n.length;i++){
if(n[posMin]>n[i])
posMin = i;
}
for(int i=0;i<n.length;i++)
n[i] = n[posMin];
System.out.print(String.valueOf(n));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
string s;
cin>>s;
char c='z';
for(auto r:s)
c=min(c,r);
for(int i=0;i<s.length();++i)
cout<<c;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, I have written this Solution Code: s=input()
l=len(s)
k=min(s)
m=k*l
print(m), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are fighting against a monster.
The monster has health M while you have initial health Y.
To defeat him, you need health strictly greater than the monster's health.
You have access to two types of pills, red and green. Eating a red pill adds R to your health while eating a green pill multiplies your health by G.
Determine if it is possible to defeat the monster by eating at most one pill of any kind.Input contains four integers M (0 <= M <= 10<sup>4</sup>), Y (0 <= Y <= 10<sup>4</sup>), R (0 <= R <= 10<sup>4</sup>) and G (1 <= G <= 10<sup>4</sup>).Print 1 if it is possible to defeat the monster and 0 if it is impossible to defeat it.Sample Input 1:
10 2 2 2
Sample Output 1:
0
Sample Input 2:
8 7 2 1
Sample Output 2:
1
Explanation for Sample 2:
You can eat a red pill to make your health : 7 + 2 = 9 > 8., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException
{
InputStreamReader pk = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(pk);
String[] strNums;
int num[] = new int[4];
strNums = in.readLine().split(" ");
for (int i = 0; i < strNums.length; i++) {
num[i] = Integer.parseInt(strNums[i]);
}
if((num[0]<(num[1]+num[2])) || (num[0]<(num[1]*num[3])))
System.out.println("1");
else
System.out.println("0");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are fighting against a monster.
The monster has health M while you have initial health Y.
To defeat him, you need health strictly greater than the monster's health.
You have access to two types of pills, red and green. Eating a red pill adds R to your health while eating a green pill multiplies your health by G.
Determine if it is possible to defeat the monster by eating at most one pill of any kind.Input contains four integers M (0 <= M <= 10<sup>4</sup>), Y (0 <= Y <= 10<sup>4</sup>), R (0 <= R <= 10<sup>4</sup>) and G (1 <= G <= 10<sup>4</sup>).Print 1 if it is possible to defeat the monster and 0 if it is impossible to defeat it.Sample Input 1:
10 2 2 2
Sample Output 1:
0
Sample Input 2:
8 7 2 1
Sample Output 2:
1
Explanation for Sample 2:
You can eat a red pill to make your health : 7 + 2 = 9 > 8., I have written this Solution Code: a = []
a = list(map(int, input().split()))
if a[1]+a[2] > a[0] or a[1]*a[3] > a[0]:
defeat = 1
else:
defeat = 0
print(defeat), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are fighting against a monster.
The monster has health M while you have initial health Y.
To defeat him, you need health strictly greater than the monster's health.
You have access to two types of pills, red and green. Eating a red pill adds R to your health while eating a green pill multiplies your health by G.
Determine if it is possible to defeat the monster by eating at most one pill of any kind.Input contains four integers M (0 <= M <= 10<sup>4</sup>), Y (0 <= Y <= 10<sup>4</sup>), R (0 <= R <= 10<sup>4</sup>) and G (1 <= G <= 10<sup>4</sup>).Print 1 if it is possible to defeat the monster and 0 if it is impossible to defeat it.Sample Input 1:
10 2 2 2
Sample Output 1:
0
Sample Input 2:
8 7 2 1
Sample Output 2:
1
Explanation for Sample 2:
You can eat a red pill to make your health : 7 + 2 = 9 > 8., I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include <sys/resource.h>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define f first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now();
#define measure() auto end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
typedef long long ll;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifdef LOCALY
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
// DEBUG FUNCTIONS END
// CUSTOM HASH TO SPEED UP UNORDERED MAP AND TO AVOID FORCED CLASHES
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()); // FOR RANDOM NUMBER GENERATION
ll mod_exp(ll a, ll b, ll c)
{
ll res=1; a=a%c;
while(b>0)
{
if(b%2==1)
res=(res*a)%c;
b/=2;
a=(a*a)%c;
}
return res;
}
ll mymod(ll a,ll b)
{
return (((a = a%b) < 0) ? a + b : a);
}
ll gcdExtended(ll,ll,ll *,ll *);
ll modInverse(ll a, ll m)
{
ll x, y;
ll g = gcdExtended(a, m, &x, &y);
g++; //this line was added just to remove compiler warning
ll res = (x%m + m) % m;
return res;
}
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
if (a == 0)
{
*x = 0, *y = 1;
return b;
}
ll x1, y1;
ll gcd = gcdExtended(b%a, a, &x1, &y1);
*x = y1 - (b/a) * x1;
*y = x1;
return gcd;
}
struct Graph
{
vector<vector<int>> adj;
Graph(int n)
{
adj.resize(n+1);
}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
};
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll M, Y, R, G;
cin >> M >> Y >> R >> G;
ll maxm = Y;
maxm = max(maxm, Y+R);
maxm = max(maxm, Y*G);
cout << (M < maxm) << "\n";
return 0;
}
/*
1. Check borderline constraints. Can a variable you are dividing by be 0?
2. Use ll while using bitshifts
3. Do not erase from set while iterating it
4. Initialise everything
5. Read the task carefully, is something unique, sorted, adjacent, guaranteed??
6. DO NOT use if(!mp[x]) if you want to iterate the map later
7. Are you using i in all loops? Are the i's conflicting?
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a positive integer N. Print the square of N.The input consists of a single integer N.
<b> Constraints: </b>
1 ≤ N ≤ 50Print a single integer – the value of N<sup>2</sup>.Sample Input 1:
5
Sample Output 1:
25, I have written this Solution Code:
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
int n = in.nextInt();
out.println(n*n);
out.flush();
}
static FastScanner in = new FastScanner();
static PrintWriter out = new PrintWriter(System.out);
static int oo = Integer.MAX_VALUE;
static long ooo = Long.MAX_VALUE;
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
void ini() throws FileNotFoundException {
br = new BufferedReader(new FileReader("input.txt"));
}
String next() {
while(!st.hasMoreTokens())
try { st = new StringTokenizer(br.readLine()); }
catch(IOException e) {}
return st.nextToken();
}
String nextLine(){
try{ return br.readLine(); }
catch(IOException e) { } return "";
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int a[] = new int[n];
for(int i=0;i<n;i++) a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
long[] readArrayL(int n) {
long a[] = new long[n];
for(int i=0;i<n;i++) a[i] = nextLong();
return a;
}
double nextDouble() {
return Double.parseDouble(next());
}
double[] readArrayD(int n) {
double a[] = new double[n];
for(int i=0;i<n;i++) a[i] = nextDouble();
return a;
}
}
static final Random random = new Random();
static void ruffleSort(int[] a){
int n = a.length;
for(int i=0;i<n;i++){
int j = random.nextInt(n), temp = a[j];
a[j] = a[i]; a[i] = temp;
}
Arrays.sort(a);
}
static void ruffleSortL(long[] a){
int n = a.length;
for(int i=0;i<n;i++){
int j = random.nextInt(n);
long temp = a[j];
a[j] = a[i]; a[i] = temp;
}
Arrays.sort(a);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a positive integer N. Print the square of N.The input consists of a single integer N.
<b> Constraints: </b>
1 ≤ N ≤ 50Print a single integer – the value of N<sup>2</sup>.Sample Input 1:
5
Sample Output 1:
25, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
cout<<(n*n)<<'\n';
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a positive integer N. Print the square of N.The input consists of a single integer N.
<b> Constraints: </b>
1 ≤ N ≤ 50Print a single integer – the value of N<sup>2</sup>.Sample Input 1:
5
Sample Output 1:
25, I have written this Solution Code: n=int(input())
print(n*n), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a large integer N. Find the sum of its digits. Eg:- if the integer is 1234, the answer is 1+2+3+4=10.The first and only line of input contains the integer N.
Constraints
The number of digits in N won't exceed 100000.Output a single integer, the sum of digits in N.Sample Input
1234
Sample Output
10
Sample Input
11111111111111111111
Sample Output
20, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str=br.readLine();
int sum=0;
for(int i=0;i<str.length();i++){
char c=str.charAt(i);
int k=c-'0';
sum+=k;}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a large integer N. Find the sum of its digits. Eg:- if the integer is 1234, the answer is 1+2+3+4=10.The first and only line of input contains the integer N.
Constraints
The number of digits in N won't exceed 100000.Output a single integer, the sum of digits in N.Sample Input
1234
Sample Output
10
Sample Input
11111111111111111111
Sample Output
20, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
string s; cin>>s;
int ans = 0;
For(i, 0, sz(s)){
ans += (s[i]-'0');
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a large integer N. Find the sum of its digits. Eg:- if the integer is 1234, the answer is 1+2+3+4=10.The first and only line of input contains the integer N.
Constraints
The number of digits in N won't exceed 100000.Output a single integer, the sum of digits in N.Sample Input
1234
Sample Output
10
Sample Input
11111111111111111111
Sample Output
20, I have written this Solution Code: s = input()
count = 0
for x in s:count+=int(x)
print(count) ;, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Give integer values M and N. Using the character ‘*’ draw a rectangle of height M and breadth N.The input contains two integers M and N.
Constraints:-
3 <= M, N <= 20Print the rectangle as shown in the exampleSample Input:-
3 3
Sample Output:-
***
* *
***
Sample Input:-
5 3
Sample Output:-
***
* *
* *
* *
***, 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 M=sc.nextInt();
int N=sc.nextInt();
for(int col=0;col<N;col++){
System.out.print("*");
}
System.out.println();
for(int row=0;row<M-2;row++){
System.out.print("*");
for(int col=0;col<N-2;col++){
System.out.print(" ");
}
System.out.print("*");
System.out.println();
}
for(int col=0;col<N;col++){
System.out.print("*");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Give integer values M and N. Using the character ‘*’ draw a rectangle of height M and breadth N.The input contains two integers M and N.
Constraints:-
3 <= M, N <= 20Print the rectangle as shown in the exampleSample Input:-
3 3
Sample Output:-
***
* *
***
Sample Input:-
5 3
Sample Output:-
***
* *
* *
* *
***, I have written this Solution Code: li = list(map(int,input().strip().split()))
m=li[0]
n=li[1]
for i in range(0,n):
if i==n-1:
print("*",end="\n")
else:
print("*",end="")
for i in range(1,m-1):
print("*",end="")
for j in range(0,n-2):
print(" ",end="")
print("*",end="\n")
for i in range(0,n):
print("*",end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Give integer values M and N. Using the character ‘*’ draw a rectangle of height M and breadth N.The input contains two integers M and N.
Constraints:-
3 <= M, N <= 20Print the rectangle as shown in the exampleSample Input:-
3 3
Sample Output:-
***
* *
***
Sample Input:-
5 3
Sample Output:-
***
* *
* *
* *
***, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int m,n;
cin>>m>>n;
for(int i=0;i<n;i++){
cout<<'*';
}
cout<<endl;
m-=2;
while(m--){
cout<<'*';
for(int i=0;i<n-2;i++){
cout<<' ';
}
cout<<'*'<<endl;
}
for(int i=0;i<n;i++){
cout<<'*';
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: public static void For_Loop(int n){
for(int i=1;i<=n;i++){
if(i%2==1){System.out.print("odd ");}
else{
System.out.print("even ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: n = int(input())
for i in range(1, n+1):
if(i%2)==0:
print("even ",end="")
else:
print("odd ",end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are N integers A[1], A[2],. , A[N]. You need to find whether their product is divisible by 4.The first line of input contains a single integer N.
The next line contains N singly spaced integers, A[1], A[2],. , A[N].
Constraints
1 <= N <= 10
1 <= A[i] <= 1000000000Output "YES" if the product is divisible by 4, else output "NO". (without quotes)Sample Input
5
3 5 2 5 1
Sample Output
NO
Explanation: Product = 150 and 150 is not divisible by 4.
Sample Input
4
1 3 8 2
Sample Output
YES
Explanation: Product = 48 and 48 is divisible by 4., I have written this Solution Code: n = int(input())
arr = []
x = 4
prod = 1
a = input().split()
for i in range(0,n):
prod = prod*int(a[i])
if prod % 4 == 0:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are N integers A[1], A[2],. , A[N]. You need to find whether their product is divisible by 4.The first line of input contains a single integer N.
The next line contains N singly spaced integers, A[1], A[2],. , A[N].
Constraints
1 <= N <= 10
1 <= A[i] <= 1000000000Output "YES" if the product is divisible by 4, else output "NO". (without quotes)Sample Input
5
3 5 2 5 1
Sample Output
NO
Explanation: Product = 150 and 150 is not divisible by 4.
Sample Input
4
1 3 8 2
Sample Output
YES
Explanation: Product = 48 and 48 is divisible by 4., 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 two=0,four=0;
for(int i=0;i<n;i++){
if(a[i]%2==0){
two++;
}
if(a[i]%4==0){
four++;
}
}
if(four>=1 || two>=2){
System.out.print("YES");
}
else{
System.out.print("NO");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are N integers A[1], A[2],. , A[N]. You need to find whether their product is divisible by 4.The first line of input contains a single integer N.
The next line contains N singly spaced integers, A[1], A[2],. , A[N].
Constraints
1 <= N <= 10
1 <= A[i] <= 1000000000Output "YES" if the product is divisible by 4, else output "NO". (without quotes)Sample Input
5
3 5 2 5 1
Sample Output
NO
Explanation: Product = 150 and 150 is not divisible by 4.
Sample Input
4
1 3 8 2
Sample Output
YES
Explanation: Product = 48 and 48 is divisible by 4., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define int long long
#define ld long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
#define MOD 1000000007
#define INF 1000000000000000007LL
const int N = 200005;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int n; cin>>n;
int ct = 0;
For(i, 0, n){
int a; cin>>a;
while(a%2==0){
ct++;
a/=2;
}
}
if(ct>=2){
cout<<"YES";
}
else{
cout<<"NO";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){cout<<"FizzBuzz"<<" ";}
else if(i%5==0){cout<<"Buzz ";}
else if(i%3==0){cout<<"Fizz ";}
else{cout<<i<<" ";}
}
}, In this Programming Language: C++, 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 "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, 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 x= sc.nextInt();
fizzbuzz(x);
}
static void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("FizzBuzz ");}
else if(i%5==0){System.out.print("Buzz ");}
else if(i%3==0){System.out.print("Fizz ");}
else{System.out.print(i+" ");}
}
}
}, 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 "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: def fizzbuzz(n):
for i in range (1,n+1):
if (i%3==0 and i%5==0):
print("FizzBuzz",end=' ')
elif i%3==0:
print("Fizz",end=' ')
elif i%5==0:
print("Buzz",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 "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){printf("FizzBuzz ");}
else if(i%5==0){printf("Buzz ");}
else if(i%3==0){printf("Fizz ");}
else{printf("%d ",i);}
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a decimal number n. You need to find the gray code of the number n and convert it into decimal.
Binary to Gray conversion :
1. The Most Significant Bit (MSB) of the gray code is always equal to the MSB of the given binary code.
2. Other bits of the output gray code can be obtained by XORing binary code bit at that index and previous index.
Eg: Gray code of 01001 is 01101The first line contains an integer T, the number of test cases. For each test case, there is an integer N denoting the number.
<b>Constraints</b>
1 <= T <= 100
1 <= N <= 10^9For each test case, print gray code equivalent of N in a separate line.Sample Input:
4
10
13
5
101
Sample Output:
15
11
7
87
Explanation:
5: 101 in binary
Gray: 111, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(t-- > 0) {
long n = Long.parseLong(br.readLine());
System.out.println(greyCode(n));
}
}
static long greyCode(long n) {
return n ^ (n>>1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a decimal number n. You need to find the gray code of the number n and convert it into decimal.
Binary to Gray conversion :
1. The Most Significant Bit (MSB) of the gray code is always equal to the MSB of the given binary code.
2. Other bits of the output gray code can be obtained by XORing binary code bit at that index and previous index.
Eg: Gray code of 01001 is 01101The first line contains an integer T, the number of test cases. For each test case, there is an integer N denoting the number.
<b>Constraints</b>
1 <= T <= 100
1 <= N <= 10^9For each test case, print gray code equivalent of N in a separate line.Sample Input:
4
10
13
5
101
Sample Output:
15
11
7
87
Explanation:
5: 101 in binary
Gray: 111, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 500 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
void solve(){
int x; cin >> x;
cout << (x ^ (x >> 1)) << endl;
}
void testcases(){
int tt = 1;
cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a decimal number n. You need to find the gray code of the number n and convert it into decimal.
Binary to Gray conversion :
1. The Most Significant Bit (MSB) of the gray code is always equal to the MSB of the given binary code.
2. Other bits of the output gray code can be obtained by XORing binary code bit at that index and previous index.
Eg: Gray code of 01001 is 01101The first line contains an integer T, the number of test cases. For each test case, there is an integer N denoting the number.
<b>Constraints</b>
1 <= T <= 100
1 <= N <= 10^9For each test case, print gray code equivalent of N in a separate line.Sample Input:
4
10
13
5
101
Sample Output:
15
11
7
87
Explanation:
5: 101 in binary
Gray: 111, I have written this Solution Code: t=int(input())
for _ in range(t):
a=int(input())
x=list(map(int,(list(bin(a)[2:]))))
y=[x[0]]
for i in range(1,len(x)):
y.append(x[i-1]^x[i])
c=''
p=0
s=0
#print(y)
for i in range(len(y)-1,-1,-1):
s+=y[i]*2**p
p+=1
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print the pattern of "*" in the form of the Right Angle Triangle.
See the below example for clarity.Since this is a functional problem, you don't have to worry about the input. It will be handled by driver code. You just have to complete <b>printTriangle()</b>.
In the custom input area, you can provide any positive integer and check whether your code is working or not.Print the right angle triangle of height 5 as shown.Sample Input:
No Input
Sample Output:-
*
* *
* * *
* * * *
* * * * *, I have written this Solution Code: class Solution {
public static void printTriangle(){
System.out.println("*");
System.out.println("* *");
System.out.println("* * *");
System.out.println("* * * *");
System.out.println("* * * * *");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print the pattern of "*" in the form of the Right Angle Triangle.
See the below example for clarity.Since this is a functional problem, you don't have to worry about the input. It will be handled by driver code. You just have to complete <b>printTriangle()</b>.
In the custom input area, you can provide any positive integer and check whether your code is working or not.Print the right angle triangle of height 5 as shown.Sample Input:
No Input
Sample Output:-
*
* *
* * *
* * * *
* * * * *, I have written this Solution Code: j=1
for i in range(0,5):
for k in range(0,j):
print("*",end=" ")
if(j<=4):
print()
j=j+1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: def get_opposite_face(n):
return 7-n
t = int(input())
for n in range(t):
print(get_opposite_face(int(input()))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
cout << 7-n << endl;
}
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 cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t-->0){
int n=Integer.parseInt(br.readLine());
System.out.println(6-n+1);
}
}
}, 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 a circular linked list consisting of N nodes, your task is to exchange the first and last node of the list.
<b>Note:
Examples in Sample Input and Output just show how a linked list will look depending on the questions. Do not copy-paste as it is in custom input</b><b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>exchangeNodes()</b> that takes head node as parameter.
Constraints:
3 <= N <= 1000
1 <= Node. data<= 1000Return the head of the modified linked list.Sample Input 1:-
3
1- >2- >3
Sample Output 1:-
3- >2- >1
Sample Input 2:-
4
1- >2- >3- >4
Sample Output 2:-
4- >2- >3- >1, I have written this Solution Code: public static Node exchangeNodes(Node head) {
Node p = head;
while (p.next.next != head)
p = p.next;
p.next.next = head.next;
head.next = p.next;
p.next = head;
head = head.next;
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: static void simpleSum(int a, int b, int c){
System.out.println(a+b+c);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: void simpleSum(int a, int b, int c){
cout<<a+b+c;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: x = input()
a, b, c = x.split()
a = int(a)
b = int(b)
c = int(c)
print(a+b+c), 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: 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: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, I have written this Solution Code: function supermarket(prices, n, k) {
// write code here
// do not console.log the answer
// return sorted array
const newPrices = prices.sort((a, b) => a - b).slice(2)
let kk = k
let price = 0;
let i = 0
while (kk-- && i < newPrices.length) {
price += newPrices[i]
i += 1
}
return price
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, 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().trim().split(" ");
int n=Integer.parseInt(str[0]);
int k=Integer.parseInt(str[1]);
str= br.readLine().trim().split(" ");
int arr[]=new int[n];
for(int i=0;i<n;i++)
arr[i]=Integer.parseInt(str[i]);
Arrays.sort(arr);
long sum=0;
for(int i=2;i<k+2;i++)
sum+=arr[i];
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, I have written this Solution Code: a=input().split()
for i in range(len(a)):
a[i]=int(a[i])
b=input().split()
for i in range(len(b)):
b[i]=int(b[i])
b.sort()
b.reverse()
b.pop()
b.pop()
s=0
while a[1]>0:
s+=b.pop()
a[1]-=1
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
#define ll long long
#define pb push_back
#define endl '\n'
#define pii pair<int,int>
#define vi vector<int>
#define all(a) (a).begin(),(a).end()
#define F first
#define S second
#define sz(x) (int)x.size()
#define hell 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define lbnd lower_bound
#define ubnd upper_bound
#define bs binary_search
#define mp make_pair
using namespace std;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
void solve(){
int n, k;
cin >> n >> k;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
int ans = 0;
for(int i = 3; i <= 2 + k; i++)
ans += a[i];
cout << ans << endl;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms: ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you have to perform only two types of operations on the number:-
1) If N is odd increase it by 1
2) If N is even divide it by 2
Your task is to find the number of operations it takes to make N equal to 1.<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>MakeOne()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations it takes to make the number N equal to 1.Sample Input:-
7
Sample Output:-
4
Explanation:-
7 - > 8 - > 4 - > 2 - > 1
Sample Input:-
3
Sample Output:-
3, I have written this Solution Code: int MakeOne(int N){
int cnt=0;
int a =N;
while(a!=1){
if(a&1){a++;}
else{a/=2;}
cnt++;
}
return cnt;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you have to perform only two types of operations on the number:-
1) If N is odd increase it by 1
2) If N is even divide it by 2
Your task is to find the number of operations it takes to make N equal to 1.<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>MakeOne()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations it takes to make the number N equal to 1.Sample Input:-
7
Sample Output:-
4
Explanation:-
7 - > 8 - > 4 - > 2 - > 1
Sample Input:-
3
Sample Output:-
3, I have written this Solution Code: static int MakeOne(int N){
int cnt=0;
int a =N;
while(a!=1){
if(a%2==1){a++;}
else{a/=2;}
cnt++;
}
return cnt;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you have to perform only two types of operations on the number:-
1) If N is odd increase it by 1
2) If N is even divide it by 2
Your task is to find the number of operations it takes to make N equal to 1.<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>MakeOne()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations it takes to make the number N equal to 1.Sample Input:-
7
Sample Output:-
4
Explanation:-
7 - > 8 - > 4 - > 2 - > 1
Sample Input:-
3
Sample Output:-
3, I have written this Solution Code: int MakeOne(int N){
int cnt=0;
int a =N;
while(a!=1){
if(a&1){a++;}
else{a/=2;}
cnt++;
}
return cnt;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you have to perform only two types of operations on the number:-
1) If N is odd increase it by 1
2) If N is even divide it by 2
Your task is to find the number of operations it takes to make N equal to 1.<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>MakeOne()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations it takes to make the number N equal to 1.Sample Input:-
7
Sample Output:-
4
Explanation:-
7 - > 8 - > 4 - > 2 - > 1
Sample Input:-
3
Sample Output:-
3, I have written this Solution Code: def MakeOne(N):
cnt=0
while N!=1:
if N%2==1:
N=N+1
else:
N=N//2
cnt=cnt+1
return cnt
, 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, and an integer k. Find the maximum force by involving only k elements. The Force of an element is the square of its value.
<b>Note:</b>
Elements are not needed to be continuous.The first line of the input contains two integers denoting n and k.
The next line contains n integers denoting elements of the array.
<b>Constraints:</b>
1 < = k < = n < = 1000
-10^7 <= A[i] <= 10^7Output the maximum force.Sample Input 1:
4 4
1 2 3 4
Sample Output 1:
30
Sample Input 2:
2 1
1 10
Sample Output 2:
100
<b>Explanation:</b>
Force = 1*1 + 2*2 + 3*3 + 4*4 = 30, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int m = 100001;
int main(){
int n,k;
cin>>n>>k;
long long a[n],sum=0;
for(int i=0;i<n;i++){
cin>>a[i];
if(a[i]<0){
a[i]=-a[i];
}
}
sort(a,a+n);
for(int i=0;i<k;i++){
sum+=a[n-i-1]*a[n-i-1];
}
cout<<sum;
}
, 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, and an integer k. Find the maximum force by involving only k elements. The Force of an element is the square of its value.
<b>Note:</b>
Elements are not needed to be continuous.The first line of the input contains two integers denoting n and k.
The next line contains n integers denoting elements of the array.
<b>Constraints:</b>
1 < = k < = n < = 1000
-10^7 <= A[i] <= 10^7Output the maximum force.Sample Input 1:
4 4
1 2 3 4
Sample Output 1:
30
Sample Input 2:
2 1
1 10
Sample Output 2:
100
<b>Explanation:</b>
Force = 1*1 + 2*2 + 3*3 + 4*4 = 30, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
String[] NK = br.readLine().split(" ");
String[] inputs = br.readLine().split(" ");
int N = Integer.parseInt(NK[0]);
int K = Integer.parseInt(NK[1]);
long[] arr = new long[N];
long answer = 0;
for(int i = 0; i < N; i++){
arr[i] = Math.abs(Long.parseLong(inputs[i]));
}
quicksort(arr, 0, N-1);
for(int i = (N-K); i < N;i++){
answer += (arr[i]*arr[i]);
}
System.out.println(answer);
}
static void quicksort(long[] arr, int start, int end){
if(start < end){
int pivot = partition(arr, start, end);
quicksort(arr, start, pivot-1);
quicksort(arr, pivot+1, end);
}
}
static int partition(long[] arr, int start, int end){
long pivot = arr[end];
int i = start - 1;
for(int j = start; j < end; j++){
if(arr[j] < pivot){
i++;
swap(arr, i, j);
}
}
swap(arr, i+1, end);
return (i+1);
}
static void swap(long[] arr, int i, int j){
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}, 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, and an integer k. Find the maximum force by involving only k elements. The Force of an element is the square of its value.
<b>Note:</b>
Elements are not needed to be continuous.The first line of the input contains two integers denoting n and k.
The next line contains n integers denoting elements of the array.
<b>Constraints:</b>
1 < = k < = n < = 1000
-10^7 <= A[i] <= 10^7Output the maximum force.Sample Input 1:
4 4
1 2 3 4
Sample Output 1:
30
Sample Input 2:
2 1
1 10
Sample Output 2:
100
<b>Explanation:</b>
Force = 1*1 + 2*2 + 3*3 + 4*4 = 30, I have written this Solution Code: x,y = map(int,input().split())
arr = list(map(int,input().split()))
s=0
for i in range(x):
if arr[i]<0:
arr[i]=abs(arr[i])
arr=sorted(arr,reverse=True)
for i in range(0,y):
s = s+arr[i]*arr[i]
print(s)
, 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: 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: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: static int focal_length(int R, char Mirror)
{
int f=R/2;
if((R%2==1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: def focal_length(R,Mirror):
f=R/2;
if(Mirror == ')'):
f=-f
if R%2==1:
f=f-1
return int(f)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a Binary Tree, your task is to convert it to a Doubly Linked List. The left and right pointers in nodes are to be used as previous and next pointers respectively in converted Double linked list. The order of nodes in Double linked list must be same as Inorder of the given Binary Tree. The first node of Inorder traversal (left most node in Binary tree) must be head node of the Doubly linked list.<b>User Task:</b>
Since this will be a functional problem. You don't have to take input. You just have to complete the function <b>BToDLL()</b> that takes "root" node of binary tree as parameter.
Constraint:-
1 <= Number of Nodes <= 1000
1 <= Node.data <= 1000
The printing is done by the driver code you just need to complete the function.Sample Input:-
3
1 2 3
Sample Output:-
2 1 3
Sample Input:-
5
6 5 4 3 2
Sample Output:-
3 5 2 6 4 , I have written this Solution Code: static void BToDLL(Node root)
{
// Base cases
if (root == null)
return ;
// Recursively convert right subtree
BToDLL(root.right);
// insert root into DLL
root.right = head;
// Change left pointer of previous head
if (head != null)
(head).left = root;
// Change head of Doubly linked list
head = root;
// Recursively convert left subtree
BToDLL(root.left);
} , In this Programming Language: Java, 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: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.<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>StrangeNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 1000Return the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: def StrangeNumber(N):
return 9*(N-1)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.<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>StrangeNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 1000Return the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: int StrangeNumber(int N){
return 9*(N-1);
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.<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>StrangeNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 1000Return the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: int StrangeNumber(int N){
return 9*(N-1);
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.<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>StrangeNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 1000Return the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: static int StrangeNumber(int N){
return 9*(N-1);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: static void simpleSum(int a, int b, int c){
System.out.println(a+b+c);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: void simpleSum(int a, int b, int c){
cout<<a+b+c;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: x = input()
a, b, c = x.split()
a = int(a)
b = int(b)
c = int(c)
print(a+b+c), 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 8*8 empty chessboard in which a knight is placed at a position (X, Y). Your task is to find the number of positions in the chessboard knight can jump into in a single move .
Note:- Rows and Columns are numbered through 1 to N.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Knight()</b> that takes integers X and Y as arguments.
Constraints:-
1 <= X <= 8
1 <= Y <= 8Return the number of positions Knight can jump into in a single move.Sample input:-
4 5
Sample Output:-
8
Explanation:-
Positions:- (3, 3), (5, 3), (3, 7), (5, 7), (6, 6), (6, 4), (2, 6), (2, 4)
Sample input:-
1 1
Sample Output:-
2
Explanation:-
Positions:- (3, 2), (2, 3), I have written this Solution Code: def Knight(X,Y):
cnt=0
if(X>2):
if(Y>1):
cnt=cnt+1
if(Y<8):
cnt=cnt+1
if(Y>2):
if(X>1):
cnt=cnt+1
if(X<8):
cnt=cnt+1
if(X<7):
if(Y>1):
cnt=cnt+1
if(Y<8):
cnt=cnt+1
if(Y<7):
if(X>1):
cnt=cnt+1
if(X<8):
cnt=cnt+1
return cnt;
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an 8*8 empty chessboard in which a knight is placed at a position (X, Y). Your task is to find the number of positions in the chessboard knight can jump into in a single move .
Note:- Rows and Columns are numbered through 1 to N.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Knight()</b> that takes integers X and Y as arguments.
Constraints:-
1 <= X <= 8
1 <= Y <= 8Return the number of positions Knight can jump into in a single move.Sample input:-
4 5
Sample Output:-
8
Explanation:-
Positions:- (3, 3), (5, 3), (3, 7), (5, 7), (6, 6), (6, 4), (2, 6), (2, 4)
Sample input:-
1 1
Sample Output:-
2
Explanation:-
Positions:- (3, 2), (2, 3), I have written this Solution Code: static int Knight(int X, int Y){
int cnt=0;
if(X>2){
if(Y>1){cnt++;}
if(Y<8){cnt++;}
}
if(Y<7){
if(X>1){cnt++;}
if(X<8){cnt++;}
}
if(X<7){
if(Y>1){cnt++;}
if(Y<8){cnt++;}
}
if(Y>2){
if(X>1){cnt++;}
if(X<8){cnt++;}
}
return cnt;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an 8*8 empty chessboard in which a knight is placed at a position (X, Y). Your task is to find the number of positions in the chessboard knight can jump into in a single move .
Note:- Rows and Columns are numbered through 1 to N.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Knight()</b> that takes integers X and Y as arguments.
Constraints:-
1 <= X <= 8
1 <= Y <= 8Return the number of positions Knight can jump into in a single move.Sample input:-
4 5
Sample Output:-
8
Explanation:-
Positions:- (3, 3), (5, 3), (3, 7), (5, 7), (6, 6), (6, 4), (2, 6), (2, 4)
Sample input:-
1 1
Sample Output:-
2
Explanation:-
Positions:- (3, 2), (2, 3), I have written this Solution Code: int Knight(int X, int Y){
int cnt=0;
if(X>2){
if(Y>1){cnt++;}
if(Y<8){cnt++;}
}
if(Y<7){
if(X>1){cnt++;}
if(X<8){cnt++;}
}
if(X<7){
if(Y>1){cnt++;}
if(Y<8){cnt++;}
}
if(Y>2){
if(X>1){cnt++;}
if(X<8){cnt++;}
}
return cnt;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an 8*8 empty chessboard in which a knight is placed at a position (X, Y). Your task is to find the number of positions in the chessboard knight can jump into in a single move .
Note:- Rows and Columns are numbered through 1 to N.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Knight()</b> that takes integers X and Y as arguments.
Constraints:-
1 <= X <= 8
1 <= Y <= 8Return the number of positions Knight can jump into in a single move.Sample input:-
4 5
Sample Output:-
8
Explanation:-
Positions:- (3, 3), (5, 3), (3, 7), (5, 7), (6, 6), (6, 4), (2, 6), (2, 4)
Sample input:-
1 1
Sample Output:-
2
Explanation:-
Positions:- (3, 2), (2, 3), I have written this Solution Code: int Knight(int X, int Y){
int cnt=0;
if(X>2){
if(Y>1){cnt++;}
if(Y<8){cnt++;}
}
if(Y<7){
if(X>1){cnt++;}
if(X<8){cnt++;}
}
if(X<7){
if(Y>1){cnt++;}
if(Y<8){cnt++;}
}
if(Y>2){
if(X>1){cnt++;}
if(X<8){cnt++;}
}
return cnt;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of N nodes, your task is to check if the given list is circular or not.
<b>Note: Sample Input and Output just show how a linked list will look depending on the questions. Do not copy-paste as it is in custom input</b><b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>check()</b> that takes the head node as a parameter.
<b>Note: 0 and 1 in sample input just show given LL is CLL or not. 1 denotes it is CLL, 0 denotes not</b>
Constraints:
1 <=N <= 1000
1 <= Node.data<= 1000
Return 1 if the given linked list is circular else return 0.Sample Input 1:-
3 0
1 2 3
Sample Output 1:-
0
Explanation:-
1->2->3
Sample Input 2:-
3 1
1 2 3
Sample Output 2:-
1
Explanation:-
1->2->3->1.......
, I have written this Solution Code: public static int check(Node head) {
if (head == null)
return 1;
Node node = head.next;
while (node != null && node != head)
node = node.next;
if(node==head){return 1;}
else {return 0;}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to find the maximum value of the sum of its subarray modulo M, i. e., find the sum of each subarray mod M and print the maximum value of this after modulo operation.The first line of input contains two space-separated integers N and M, the next line of input contains N space-separated integers depicting value of the array.
<b>Constraints:-</b>
1 < = N < = 100000
1 < = M < = 10000000000
1 < = Arr[i] < = 10000000000Print the maximum value of sum modulo m.Sample Input:-
5 13
6 6 11 15 2
Sample Output:-
12
Explanation:
[6, 6] is subarray is maximum sum modulo 13
Sample Input:-
3 15
1 2 3
Sample Output:-
6
Explanation:
Max sum occurs when we take the whole array, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] s = br.readLine().split(" ");
int N = Integer.parseInt(s[0]);
int M = Integer.parseInt(s[1]);
s = br.readLine().split(" ");
int[] prefix = new int[N];
int preSum = 0;
for(int i=0; i<N; i++){
int curr = Integer.parseInt(s[i]);
preSum += curr;
prefix[i] = preSum;
}
if(M>=prefix[N-1]){
System.out.println(prefix[N-1]);
return;
}
int maxSum = 0;
for(int i=0; i<N; i++){
for(int j=0; j<i; j++){
int curr = prefix[i]%M;
maxSum = Math.max(maxSum, curr);
curr = (prefix[i]-prefix[j])%M;
maxSum = Math.max(maxSum, curr);
if(maxSum==M-1){
break;
}
}
}
System.out.println(maxSum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to find the maximum value of the sum of its subarray modulo M, i. e., find the sum of each subarray mod M and print the maximum value of this after modulo operation.The first line of input contains two space-separated integers N and M, the next line of input contains N space-separated integers depicting value of the array.
<b>Constraints:-</b>
1 < = N < = 100000
1 < = M < = 10000000000
1 < = Arr[i] < = 10000000000Print the maximum value of sum modulo m.Sample Input:-
5 13
6 6 11 15 2
Sample Output:-
12
Explanation:
[6, 6] is subarray is maximum sum modulo 13
Sample Input:-
3 15
1 2 3
Sample Output:-
6
Explanation:
Max sum occurs when we take the whole array, I have written this Solution Code: import bisect
def maximumSum(coll, m):
n = len(coll)
maxSum, prefixSum = 0, 0
sortedPrefixes = []
for endIndex in range(n):
prefixSum = (prefixSum + coll[endIndex]) % m
maxSum = max(maxSum, prefixSum)
startIndex = bisect.bisect_right(sortedPrefixes, prefixSum)
if startIndex < len(sortedPrefixes):
maxSum = max(maxSum, prefixSum - sortedPrefixes[startIndex] + m)
bisect.insort(sortedPrefixes, prefixSum)
return maxSum
a,b=map(int,input().split())
c=list(map(int,input().split()))
print(maximumSum(c,b)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to find the maximum value of the sum of its subarray modulo M, i. e., find the sum of each subarray mod M and print the maximum value of this after modulo operation.The first line of input contains two space-separated integers N and M, the next line of input contains N space-separated integers depicting value of the array.
<b>Constraints:-</b>
1 < = N < = 100000
1 < = M < = 10000000000
1 < = Arr[i] < = 10000000000Print the maximum value of sum modulo m.Sample Input:-
5 13
6 6 11 15 2
Sample Output:-
12
Explanation:
[6, 6] is subarray is maximum sum modulo 13
Sample Input:-
3 15
1 2 3
Sample Output:-
6
Explanation:
Max sum occurs when we take the whole array, I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
using namespace std;
// Return the maximum sum subarray mod m.
int maxSubarray(int arr[], int n, int m)
{
int x, prefix = 0, maxim = 0;
set<int> S;
S.insert(0);
// Traversing the array.
for (int i = 0; i < n; i++)
{
// Finding prefix sum.
prefix = (prefix + arr[i])%m;
// Finding maximum of prefix sum.
maxim = max(maxim, prefix);
// Finding iterator pointing to the first
// element that is not less than value
// "prefix + 1", i.e., greater than or
// equal to this value.
auto it = S.lower_bound(prefix+1);
if (it != S.end())
maxim = max(maxim, prefix - (*it) + m );
// Inserting prefix in the set.
S.insert(prefix);
}
return maxim;
}
// Driver Program
signed main()
{
int n,m;
cin>>n>>m;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
cout << maxSubarray(a, n, m) << 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 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: Given a string S consisting of digits between 1 and 9. Rearrange the string such that the number of even substrings it contains is maximised. A substring is said to be even if the number it represents is even.<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>solve()</b> that takes the string S as parameter.
<b>Constraints</b>
1<=|S|<=1000
S only contains digits between 1 and 9 (both inclusive).Return the number of even substrings present in the string after rearranging it such that the number is maximised.Sample Input:
1232
Sample Output:
7
We will rearrange the string to "1322". Now, the seven even substrings are "132", "32", "2", "2", "1232", "232", "32". Two substrings are considered different if at least one of the starting or ending index differs from the other., I have written this Solution Code:
class Solution {
public long solve(String S) {
int e = 0;
for (int i = 0; i < S.length(); i++) {
int y = S.charAt(i) - '0';
if (y % 2 == 0) {
e++;
}
}
long ans = 0;
for (int i = S.length(); i >= 0; i--) {
if (e > 0) {
ans += i;
e--;
}
}
return 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 the lowercase English word corresponding to the number if it is <=5 else print "Greater than 5".
Numbers <=5 and their corresponding words :
1 = one
2 = two
3 = three
4 = four
5 = fiveThe input contains a single integer N.
Constraint:
1 <= n <= 100Print a string consisting of the lowercase English word corresponding to the number if it is <=5 else print the string "Greater than 5"Sample Input:
4
Sample Output
four
Sample Input:
6
Sample Output:
Greater than 5, I have written this Solution Code: N = int(input())
if N > 5:
print("Greater than 5")
elif(N == 1):
print("one")
elif(N == 2):
print("two")
elif(N == 3):
print("three")
elif(N == 4):
print("four")
elif(N == 5):
print("five"), 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 the lowercase English word corresponding to the number if it is <=5 else print "Greater than 5".
Numbers <=5 and their corresponding words :
1 = one
2 = two
3 = three
4 = four
5 = fiveThe input contains a single integer N.
Constraint:
1 <= n <= 100Print a string consisting of the lowercase English word corresponding to the number if it is <=5 else print the string "Greater than 5"Sample Input:
4
Sample Output
four
Sample Input:
6
Sample Output:
Greater than 5, 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 side = scanner.nextInt();
String area = conditional(side);
System.out.println(area);
}static String conditional(int n){
if(n==1){return "one";}
else if(n==2){return "two";}
else if(n==3){return "three";}
else if(n==4){return "four";}
else if(n==5){return "five";}
else{
return "Greater than 5";}
}}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
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