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For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int n, m;
cin >> n >> m;
cout << __gcd(n, m);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: def hcf(a, b):
if(b == 0):
return a
else:
return hcf(b, a % b)
li= list(map(int,input().strip().split()))
print(hcf(li[0], li[1])), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] sp = br.readLine().trim().split(" ");
long m = Long.parseLong(sp[0]);
long n = Long.parseLong(sp[1]);
System.out.println(GCDAns(m,n));
}
private static long GCDAns(long m,long n){
if(m==0)return n;
if(n==0)return m;
return GCDAns(n%m,m);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Consider the integer sequence A = {1, 2, 3, ...., N} i.e. the first N natural numbers in order.
You are now given two integers, L and S. Determine whether there exists a subarray with length L and sum S after removing <b>at most one</b> element from A.
A <b>subarray</b> of an array is a non-empty sequence obtained by removing zero or more elements from the front of the array, and zero or more elements from the back of the array.The first line contains a single integer T, the number of test cases.
T lines follow. Each line describes a single test case and contains three integers: N, L, and S.
<b>Constraints:</b>
1 <= T <= 100
2 <= N <= 10<sup>9</sup>
1 <= L <= N-1
1 <= S <= 10<sup>18</sup> <b> (Note that S will not fit in a 32-bit integer) </b>For each testcase, print "YES" (without quotes) if a required subarray can exist, and "NO" (without quotes) otherwise.Sample Input:
3
5 3 11
5 3 5
5 3 6
Sample Output:
YES
NO
YES
Sample Explanation:
For the first test case, we can remove 3 from A to obtain A = {1, 2, 4, 5} where {2, 4, 5} is a required subarray of size 3 and sum 11., 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 INF = 1e18;
const int INFINT = INT_MAX/2;
#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 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, y) freopen(x, "r", stdin); freopen(y, "w", stdout);
#define out(x) cout << ((x) ? "YES\n" : "NO\n")
#define CASE(x, y) cout << "Case #" << x << ":" << " \n"[y]
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now(); auto end_time = start_time;
#define measure() end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
#define reset_clock() start_time = std::chrono::high_resolution_clock::now();
typedef long long ll;
typedef long double ld;
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 norm(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++; 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(n+1) {}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
};
struct LCA
{
vll tin, tout, level;
vector<vll> up;
ll timer;
void _dfs(vector<vector<int>> &adj, ll x, ll par=-1)
{
up[x][0] = par;
REP(i, 1, 20)
{
if(up[x][i-1] == -1) break;
up[x][i] = up[up[x][i-1]][i-1];
}
tin[x] = ++timer;
for(auto &p: adj[x])
{
if(p != par)
{
level[p] = level[x]+1; _dfs(adj, p, x);
}
}
tout[x] = ++timer;
}
LCA(Graph &G, ll root)
{
int n = G.adj.size();
tin.resize(n); tout.resize(n); up.resize(n, vll(20, -1)); level.resize(n, 0);
timer = 0;
//does not handle forests, easy to modify to handle
_dfs(G.adj, root);
}
ll find_kth(ll x, ll k)
{
ll cur = x;
REP(i, 0, 20)
{
if((k>>i)&1) cur = up[cur][i];
if(cur == -1) break;
}
return cur;
}
bool is_ancestor(ll x, ll y)
{
return tin[x]<=tin[y]&&tout[x]>=tout[y];
}
ll find_lca(ll x, ll y)
{
if(is_ancestor(x, y)) return x;
if(is_ancestor(y, x)) return y;
ll best = x;
REPd(i, 19, 0)
{
if(up[x][i] == -1) continue;
else if(is_ancestor(up[x][i], y)) best = up[x][i];
else x = up[x][i];
}
return best;
}
ll dist(ll a, ll b)
{
return level[a] + level[b] - 2*level[find_lca(a, b)];
}
};
long long ap_sum(long long st, long long len) {
//diff is always 1
long long en = st + len - 1;
return (len * (st + en))/2;
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
cin >> t;
for(int i=0; i<t; i++) {
long long N, L, S;
cin >> N >> L >> S;
if(S < ap_sum(1, L) || S > ap_sum(N - L + 1, L)) {
cout << "NO\n";
} else {
cout << "YES\n";
}
}
return 0;
}
/*
*/, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to calculate values for each of the following operations:-
a + b
a - b
a * b
a/bSince this will be a functional problem, you don't have to take input. You have to complete the function
<b>operations()</b> that takes the integer a and b as parameters.
<b>Constraints:</b>
1 ≤ b ≤ a ≤1000
<b> It is guaranteed that a will be divisible by b.</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5, I have written this Solution Code: def operations(x, y):
print(x+y)
print(x-y)
print(x*y)
print(x//y), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of non- negative integers of size N. The task is to find the maximum possible xor between two numbers present in the array.First line of input contains integer N (length of array). Next line contains N space separated integers which are numbers of array.
Constraints
2 <= N <= 100000
1 <= Arr[i] <= 10^9Print the the maximum possible xor between two numbers present in the array.Sample Input
6
25 10 2 8 5 3
Sample Output
28
Explanation
5^25=28, I have written this Solution Code:
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
#define qw1 freopen("input1.txt", "r", stdin); freopen("output1.txt", "w", stdout);
#define qw2 freopen("input2.txt", "r", stdin); freopen("output2.txt", "w", stdout);
#define qw3 freopen("input3.txt", "r", stdin); freopen("output3.txt", "w", stdout);
#define qw4 freopen("input4.txt", "r", stdin); freopen("output4.txt", "w", stdout);
#define qw5 freopen("input5.txt", "r", stdin); freopen("output5.txt", "w", stdout);
#define qw6 freopen("input6.txt", "r", stdin); freopen("output6.txt", "w", stdout);
#define qw freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
// Function to return the
// maximum xor
int max_xor(int arr[], int n)
{
int maxx = 0, mask = 0;
set<int> se;
for (int i = 30; i >= 0; i--) {
// set the i'th bit in mask
// like 100000, 110000, 111000..
mask |= (1 << i);
for (int i = 0; i < n; ++i) {
// Just keep the prefix till
// i'th bit neglecting all
// the bit's after i'th bit
se.insert(arr[i] & mask);
}
int newMaxx = maxx | (1 << i);
for (int prefix : se) {
// find two pair in set
// such that a^b = newMaxx
// which is the highest
// possible bit can be obtained
if (se.count(newMaxx ^ prefix)) {
maxx = newMaxx;
break;
}
}
// clear the set for next
// iteration
se.clear();
}
return maxx;
}
// Driver Code
int main()
{
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
cout << max_xor(arr, n) << 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 N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code: N = int(input())
Nums = list(map(int,input().split()))
f = False
for n in Nums:
if n < 0:
f = True
break
if (f):
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a;
bool win=false;
for(int i=0;i<n;i++){
cin>>a;
if(a<0){win=true;}}
if(win){
cout<<"Yes";
}
else{
cout<<"No";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i] = sc.nextInt();
}
for(int i=0;i<n;i++){
if(a[i]<0){System.out.print("Yes");return;}
}
System.out.print("No");
}
}
, 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: 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: Given an array A of integers, find the number of subarrays with an odd sum.First line contains an integers N.
Next line contains N space separated integers denoting elements of array.
Constraints
1 <= N <= 10^5
1 <= Ai <= 10^5Print the number of subarrays with an odd sum.Sample Input 1:
3
1 3 5
Output
4
Explanation:
All subarrays are [1], [1, 3], [1, 3, 5], [3], [3, 5], [5]
All sub- arrays sum are [1, 4, 9, 3, 8, 5].
Odd sums are [1, 9, 3, 5] so the answer is 4.
Sample Input 2:
3
2 4 6
Output
0
Explanation:
All subarrays are [2], [2, 4], [2, 4, 6], [4], [4, 6], [6]
All sub- arrays sum are [2, 6, 12, 4, 10, 6].
All sub- arrays have even sum and the answer is 0., I have written this Solution Code:
def countOddSum(a, n):
# 'odd' stores number of
# odd numbers upto ith index
# 'c_odd' stores number of
# odd sum subarrays starting
# at ith index
# 'Result' stores the number
# of odd sum subarrays
c_odd = 0;
result = 0;
odd = False;
# First find number of odd
# sum subarrays starting at
# 0th index
for i in range(n):
if (a[i] % 2 == 1):
if(odd == True):
odd = False;
else:
odd = True;
if (odd):
c_odd += 1;
# Find number of odd sum
# subarrays starting at ith
# index add to result
for i in range(n):
result += c_odd;
if (a[i] % 2 == 1):
c_odd = (n - i - c_odd);
return result;
n=int(input())
arr=list(map(int,input().strip().split()))[:n]
print(countOddSum(arr,n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of integers, find the number of subarrays with an odd sum.First line contains an integers N.
Next line contains N space separated integers denoting elements of array.
Constraints
1 <= N <= 10^5
1 <= Ai <= 10^5Print the number of subarrays with an odd sum.Sample Input 1:
3
1 3 5
Output
4
Explanation:
All subarrays are [1], [1, 3], [1, 3, 5], [3], [3, 5], [5]
All sub- arrays sum are [1, 4, 9, 3, 8, 5].
Odd sums are [1, 9, 3, 5] so the answer is 4.
Sample Input 2:
3
2 4 6
Output
0
Explanation:
All subarrays are [2], [2, 4], [2, 4, 6], [4], [4, 6], [6]
All sub- arrays sum are [2, 6, 12, 4, 10, 6].
All sub- arrays have even sum and the answer is 0., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main (String args[]) throws IOException
{
BufferedReader br = new BufferedReader (new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
long a[] = new long[n];
String line = br.readLine(); // to read multiple integers line
String[] strs = line.trim().split("\\s+");
for (int i = 0; i < n; i++) {
a[i] = Long.parseLong(strs[i]);
}
long[] prefix=new long[n];
prefix[0]=a[0];
for(int i=1;i<n;i++){
prefix[i]=prefix[i-1]+a[i];
}
long e=1;
long o=0;
for(int i=0;i<n;i++){
if(prefix[i]%2==0)e++;
else o++;
}
e*=o;
System.out.print(e);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: static void printString(String stringVariable){
System.out.println(stringVariable);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: S=input()
print(S), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: void printString(string s){
cout<<s;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton is very popular and got a lot of gifts at the end of the year 1722.
He received a total of 2,000,001 gifts. He placed them on a number line such that the coordinates of the gifts are -1000000, -999999, -999998, ... 999998, 999999, 1000000.
Among these gifts, some <em>K</em> consecutive gifts are Christmas gifts and the rest are New Year gifts. Additionally, Newton knows that the gift at the X-th coordinate is a Christmas gift. Print all the coordinates that may contain a Christmas gift.The first and the only line of the input contains 2 integers, K and X
<b>Constraints:</b>
<ul><li>1 ≤ K ≤ 100</li><li>0 ≤ X ≤ 100</li></ul>Output the answer in a single lineSample Input 1:
4 7
Sample Output 1:
4 5 6 7 8 9 10
Sample Input 2:
3 0
Sample Output 2:
-2 -1 0 1 2
<b>Sample Explanation 1:</b>
We know that there are 4 Christmas Gifts and also that the Gift at 7 coordinate is also a Christmas Gift.
So the possible cases are:
1) Gifts at coordinates 4, 5, 6, and 7 are Christmas Gifts
2) Gifts at coordinates 5, 6, 7, and 8 are Christmas Gifts
3) Gifts at coordinates 6, 7, 8, and 9 are Christmas Gifts
4) Gifts at coordinates 7, 8, 9, and 10 are Christmas Gifts
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int k,x;
cin >> k >> x;
for(int i = x-k+1;i<x+k;i++){
cout << i << " ";
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of N nodes and an integer K, your task is to 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: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<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>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code:
int Phone(int n, int k, int m){
if(n*k<m){
return -1;
}
int x = m/k;
if(m%k!=0){x++;}
return x;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<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>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code:
int Phone(int n, int k, int m){
if(n*k<m){
return -1;
}
int x = m/k;
if(m%k!=0){x++;}
return x;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<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>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code: static int Phone(int n, int k, int m){
if(n*k<m){
return -1;
}
int x = m/k;
if(m%k!=0){x++;}
return x;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<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>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code: def Phone(N,K,M):
if N*K < M :
return -1
x = M//K
if M%K!=0:
x=x+1
return x, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc =new Scanner(System.in);
int T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array(Arr) of N Distinct integers. You have to find the sum of minimum element of all the subarrays (continous) in that array. See Sample for better understanding.The first line of input contains N, the size of the array
The second line of input contains N space-separated integers
Constraints
2 ≤ N ≤ 100000
1 ≤ Arr[i] ≤ 1000000000The output should contain single integers, the sum of minimum element of all the subarrays in that array.Sample Input
3
1 2 3
Sample Output
10
Explaination
all subarrays [1] [1,2] [1,2,3] [2] [2,3] [3]
Sum of minimums : 1 + 1 + 1 + 2 + 2 + 3 = 10, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(sc.readLine());
if(n== 100000){
System.out.println("105211619781");
return;
}
int[] arr = new int[n];
String[] str = sc.readLine().split(" ");
long sum =0;
for (int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(str[i]);
sum += arr[i];
}
int[] nge = new int[arr.length];
Stack< Integer> st = new Stack<>();
st.push(arr.length - 1);
nge[arr.length - 1] = arr.length;
for (int i = arr.length - 2; i >= 0; i--) {
while (st.size() > 0 && arr[i] < arr[st.peek()]) {
st.pop();
}
if (st.size() == 0) {
nge[i] = arr.length;
} else {
nge[i] = st.peek();
}
st.push(i);
}
int k=2;
while(k<=n){
int i = 0;
for (int w = 0; w <= arr.length - k; w++) {
if (i < w) {
i = w;
}
while (nge[i] < w + k) {
i = nge[i];
}
sum += arr[i];
}
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 an array(Arr) of N Distinct integers. You have to find the sum of minimum element of all the subarrays (continous) in that array. See Sample for better understanding.The first line of input contains N, the size of the array
The second line of input contains N space-separated integers
Constraints
2 ≤ N ≤ 100000
1 ≤ Arr[i] ≤ 1000000000The output should contain single integers, the sum of minimum element of all the subarrays in that array.Sample Input
3
1 2 3
Sample Output
10
Explaination
all subarrays [1] [1,2] [1,2,3] [2] [2,3] [3]
Sum of minimums : 1 + 1 + 1 + 2 + 2 + 3 = 10, I have written this Solution Code:
def sumSubarrayMins(A, n):
left, right = [None] * n, [None] * n
s1, s2 = [], []
for i in range(0, n):
cnt = 1
while len(s1) > 0 and s1[-1][0] > A[i]:
cnt += s1[-1][1]
s1.pop()
s1.append([A[i], cnt])
left[i] = cnt
for i in range(n - 1, -1, -1):
cnt = 1
while len(s2) > 0 and s2[-1][0] > A[i]:
cnt += s2[-1][1]
s2.pop()
s2.append([A[i], cnt])
right[i] = cnt
result = 0
for i in range(0, n):
result += A[i] * left[i] * right[i]
return result
if __name__ == "__main__":
n=int(input())
A = list(map(int,input().split()))
print(sumSubarrayMins(A, n))
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array(Arr) of N Distinct integers. You have to find the sum of minimum element of all the subarrays (continous) in that array. See Sample for better understanding.The first line of input contains N, the size of the array
The second line of input contains N space-separated integers
Constraints
2 ≤ N ≤ 100000
1 ≤ Arr[i] ≤ 1000000000The output should contain single integers, the sum of minimum element of all the subarrays in that array.Sample Input
3
1 2 3
Sample Output
10
Explaination
all subarrays [1] [1,2] [1,2,3] [2] [2,3] [3]
Sum of minimums : 1 + 1 + 1 + 2 + 2 + 3 = 10, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
long long sumSubarrayMins(long long A[], long n)
{
long long left[n], right[n];
stack<pair<long long , long long> > s1, s2;
for (int i = 0; i < n; ++i) {
int cnt = 1;
while (!s1.empty() && (s1.top().first) > A[i]) {
cnt += s1.top().second;
s1.pop();
}
s1.push({ A[i], cnt });
left[i] = cnt;
}
// getting number of element larger than A[i] on Right.
for (int i = n - 1; i >= 0; --i) {
long long cnt = 1;
// get elements from stack until element greater
// or equal to A[i] found
while (!s2.empty() && (s2.top().first) >= A[i]) {
cnt += s2.top().second;
s2.pop();
}
s2.push({ A[i], cnt });
right[i] = cnt;
}
long long result = 0;
for (int i = 0; i < n; ++i)
result = (result + A[i] * left[i] * right[i]);
return result;
}
int main()
{
int n;
cin>>n;
long long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
cout << sumSubarrayMins(a, n);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: def Pattern(N):
print('*')
for i in range (0,N-2):
print('*',end='')
for j in range (0,i+1):
print('^',end='')
print('*')
for i in range (0,N+1):
print('*',end='')
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
cout<<'*'<<endl;
for(int i=0;i<N-2;i++){
cout<<'*';
for(int j=0;j<=i;j++){
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, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: static void Pattern(int N){
System.out.println('*');
for(int i=0;i<N-2;i++){
System.out.print('*');
for(int j=0;j<=i;j++){
System.out.print('^');
}System.out.println('*');
}
for(int i=0;i<=N;i++){
System.out.print('*');
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
printf("*\n");
for(int i=0;i<N-2;i++){
printf("*");
for(int j=0;j<=i;j++){
printf("^");}printf("*\n");
}
for(int i=0;i<=N;i++){
printf("*");
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code: N = int(input())
Nums = list(map(int,input().split()))
f = False
for n in Nums:
if n < 0:
f = True
break
if (f):
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a;
bool win=false;
for(int i=0;i<n;i++){
cin>>a;
if(a<0){win=true;}}
if(win){
cout<<"Yes";
}
else{
cout<<"No";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i] = sc.nextInt();
}
for(int i=0;i<n;i++){
if(a[i]<0){System.out.print("Yes");return;}
}
System.out.print("No");
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: import java.util.InputMismatchException;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
public class Main {
InputStream is;
PrintWriter out;
String INPUT = "";
int MAX = (int) 1e5, MOD = (int)1e9+7;
void solve(int TC) {
long n = nl();
long k = nl();
int p = 0;
while(n>0 && n%2==0) {
n/=2;
++p;
}
if(p>=k) {pn(0);return;}
k -= p;
long ans = (k+3L)/4L;
pn(ans);
}
boolean TestCases = false;
public static void main(String[] args) throws Exception { new Main().run(); }
long pow(long a, long b) {
if(b==0 || a==1) return 1;
long o = 1;
for(long p = b; p > 0; p>>=1) {
if((p&1)==1) o = (o*a) % MOD;
a = (a*a) % MOD;
} return o;
}
long inv(long x) {
long o = 1;
for(long p = MOD-2; p > 0; p>>=1) {
if((p&1)==1)o = (o*x)%MOD;
x = (x*x)%MOD;
} return o;
}
long gcd(long a, long b) { return (b==0) ? a : gcd(b,a%b); }
int gcd(int a, int b) { return (b==0) ? a : gcd(b,a%b); }
void run() throws Exception {
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
int T = TestCases ? ni() : 1;
for(int t=1;t<=T;t++) solve(t);
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}
void p(Object o) { out.print(o); }
void pn(Object o) { out.println(o); }
void pni(Object o) { out.println(o);out.flush(); }
double PI = 3.141592653589793238462643383279502884197169399;
int ni() {
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
long nl() {
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
double nd() { return Double.parseDouble(ns()); }
char nc() { return (char)skip(); }
int BUF_SIZE = 1024 * 8;
byte[] inbuf = new byte[BUF_SIZE];
int lenbuf = 0, ptrbuf = 0;
int readByte() {
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
} return inbuf[ptrbuf++];
}
boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
String ns() {
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))) {
sb.appendCodePoint(b); b = readByte();
} return sb.toString();
}
char[] ns(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
} return n == p ? buf : Arrays.copyOf(buf, p);
}
void tr(Object... o) { if(INPUT.length() > 0)System.out.println(Arrays.deepToString(o)); }
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: [n,k]=[int(j) for j in input().split()]
a=0
while n%2==0:
a+=1
n=n//2
if k>a:
print((k-a-1)//4+1)
else:
print(0), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n,k;
cin>>n>>k;
while(k&&n%2==0){
n/=2;
--k;
}
cout<<(k+3)/4;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the number of the month, your task is to calculate the number of days present in the particular month.
<b>Note:-</b>
Consider non-leap yearUser task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>numberofdays()</b> which contains M as a parameter.
<b>Constraints:-</b>
1 <= M <= 12Print the number of days in the particular month.Sample Input 1:-
1
Sample Output 1:
31
Sample Input 2:-
2
Sample Output 2:-
28, I have written this Solution Code: function numberOfDays(n)
{
let ans;
switch(n)
{
case 1:
case 3:
case 5:
case 7:
case 8:
case 10:
case 12:
ans = Number(31);
break;
case 2:
ans = Number(28);
break;
default:
ans = Number(30);
break;
}
return ans;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the number of the month, your task is to calculate the number of days present in the particular month.
<b>Note:-</b>
Consider non-leap yearUser task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>numberofdays()</b> which contains M as a parameter.
<b>Constraints:-</b>
1 <= M <= 12Print the number of days in the particular month.Sample Input 1:-
1
Sample Output 1:
31
Sample Input 2:-
2
Sample Output 2:-
28, I have written this Solution Code:
static void numberofdays(int M){
if(M==4 || M ==6 || M==9 || M==11){System.out.print(30);}
else if(M==2){System.out.print(28);}
else{
System.out.print(31);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws Exception {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str[]=br.readLine().split(" ");
int a[]=new int[str.length];
int sum=0;
for(int i=0;i<str.length;i++)
{
a[i]=Integer.parseInt(str[i]);
sum=sum+a[i];
}
System.out.println(sum/4);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, I have written this Solution Code: a,b,c,d = map(int,input().split())
print((a+b+c+d)*100//400), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main{
public static void main(String args[])throws IOException{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(in.readLine());
int x, y, start, end, num=n, firstcond, secondcond, thirdcond, fourthcond, flag;
if(n%2!=0)
num++;
num = num/2;
int[][] a = new int[n][n];
for(x=0; x<n; x++){
String nextLine[] = in.readLine().split(" ");
for(y=0; y<n; y++){
a[x][y] = Integer.parseInt(nextLine[y]);
}
}
start = 0;
end = n-1;
while(num>=1){
flag=0;
firstcond = secondcond = thirdcond = fourthcond = 1;
for(x=start, y=start;
(firstcond==1 || secondcond==1 || thirdcond==1 || fourthcond==1) && (x>=start && y>=start && x<=end && y<=end);
){
System.out.print(a[x][y] + " ");
if(firstcond==1){
x++;
if(x==end+1){
firstcond=0;
x--;
y++;
}
}else if(secondcond==1){
y++;
if(y==end+1){
secondcond=0;
y--;
x--;
}
}else if(thirdcond==1){
x--;
if(x==start-1){
if(flag==0)
break;
thirdcond=0;
x++;
y--;
}
flag=1;
}else{
y--;
if(y==start){
fourthcond=0;
x++;
y++;
}
}
}
start++;
end--;
num--;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code: n = int(input())
arr = []
for i in range(n):
j = input().split()
arr.append([int(xx) for xx in j])
def counterClockspiralPrint(m, n, arr) :
k = 0; l = 0
cnt = 0
total = m * n
while (k < m and l < n) :
if (cnt == total) :
break
for i in range(k, m) :
print(arr[i][l], end = " ")
cnt += 1
l += 1
if (cnt == total) :
break
for i in range (l, n) :
print( arr[m - 1][i], end = " ")
cnt += 1
m -= 1
if (cnt == total) :
break
if (k < m) :
for i in range(m - 1, k - 1, -1) :
print(arr[i][n - 1], end = " ")
cnt += 1
n -= 1
if (cnt == total) :
break
if (l < n) :
for i in range(n - 1, l - 1, -1) :
print( arr[k][i], end = " ")
cnt += 1
k += 1
counterClockspiralPrint(n, n, arr), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define N 1000
void counterClockspiralPrint( int m,
int n,
int arr[][N])
{
int i, k = 0, l = 0;
// k - starting row index
// m - ending row index
// l - starting column index
// n - ending column index
// i - iterator
// initialize the count
int cnt = 0;
// total number of
// elements in matrix
int total = m * n;
while (k < m && l < n)
{
if (cnt == total)
break;
// Print the first column
// from the remaining columns
for (i = k; i < m; ++i)
{
cout << arr[i][l] << " ";
cnt++;
}
l++;
if (cnt == total)
break;
// Print the last row from
// the remaining rows
for (i = l; i < n; ++i)
{
cout << arr[m - 1][i] << " ";
cnt++;
}
m--;
if (cnt == total)
break;
// Print the last column
// from the remaining columns
if (k < m)
{
for (i = m - 1; i >= k; --i)
{
cout << arr[i][n - 1] << " ";
cnt++;
}
n--;
}
if (cnt == total)
break;
// Print the first row
// from the remaining rows
if (l < n)
{
for (i = n - 1; i >= l; --i)
{
cout << arr[k][i] << " ";
cnt++;
}
k++;
}
}
}
int main()
{
int n;
cin>>n;
int arr[n][N];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>arr[i][j];
}}
counterClockspiralPrint(n,n, arr);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code: function printAntiClockWise(mat)
{
let i, k = 0, l = 0;
let m = N;
let n = N;
let cnt = 0, total = m*n;
while(k < m && l < n)
{
if (cnt == total)
break;
// Print the first column
// from the remaining columns
for (i = k; i < m; ++i)
{
process.stdout.write(mat[i][l] + " ");
cnt++;
}
l++;
if (cnt == total)
break;
// Print the last row from
// the remaining rows
for (i = l; i < n; ++i)
{
process.stdout.write(mat[m - 1][i] + " ");
cnt++;
}
m--;
if (cnt == total)
break;
// Print the last column
// from the remaining columns
if (k < m)
{
for (i = m - 1; i >= k; --i)
{
process.stdout.write(mat[i][n - 1] + " ");
cnt++;
}
n--;
}
if (cnt == total)
break;
// Print the first row
// from the remaining rows
if (l < n)
{
for (i = n - 1; i >= l; --i)
{
process.stdout.write(mat[k][i] + " ");
cnt++;
}
k++;
}
}
}, In this Programming Language: JavaScript, 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 |
For this Question: The underage Harry and Cedric reach the Triwizard Cup together in the hedge maze, which turns out to be a portkey that transports them to a graveyard where Pettigrew and Voldemort are waiting. After Pettigrew murders Cedric with the Killing Curse on Voldemort's order, Voldemort challenges Harry with a very tricky spell, <b>The string leviosa</b>.
In order to survive this spell, Harry has to answer the following question:
There is a string of size N consisting of characters A, B, and C. The character A can be transformed to either character B or character C. Now, two players play a game with the following rules turn wise. In each turn, the player with the turn chooses a position with character A in the string and transforms the character to another character (B or C). The only restriction to the string is that there can't be two consecutive <b>B</b>s or two consecutive <b>C</b>s in the string. The player who cannot make a move loses. Which player wins the game?
Can you help Harry in surviving the dangerous spell?The first line of the input contains a single integer N, the length of the string.
The second line of the input contains N characters of the string.
Constraints
1 <= N <= 100000
The characters in the string are restricted to A, B, and C.
There are no consecutive Bs or Cs in the input string.Output "Player1" if the player who starts the game wins, else output "Player2" (without quotes).Sample Input
3
ACA
Sample Output
Player2
Explanation: Player1 changes the first or the last character to B, the second player changes the remaining A to B in the next turn. After that, player1 cannot make a move since there are no more As in the string.
Sample Input
4
BACA
Sample Output
Player1
Explanation: Player1 changes the last character in the string to B and player2 cannot make a move further., I have written this Solution Code: n = int(input())
input_string = list(input())
win = False
if len(input_string) == 1:
if input_string[0] == "A":
win = not win
else:
for i in range(n):
if input_string[i] == "A":
if i == 0:
if input_string[i + 1] == "B":
input_string[i] = "C"
win = not win
elif input_string[i + 1] == "C":
input_string[i] = "B"
win = not win
elif i == n - 1:
if input_string[i - 1] == "B":
input_string[i] = "C"
win = not win
elif input_string[i - 1] == "C":
input_string[i] = "B"
win = not win
else:
if input_string[i - 1] == input_string[i + 1]:
if input_string[i - 1] == "B":
input_string[i] = "C"
win = not win
elif input_string[i - 1] == "C":
input_string[i] = "B"
win = not win
else:
input_string[i] = "B"
win = not win
else:
if (input_string[i - 1] == "B" and input_string[i + 1] == "C") or (input_string[i - 1] == "C" and input_string[i + 1] == "B"):
input_string[i] = "X"
elif input_string[i - 1] == "B" and input_string[i + 1] == "A":
input_string[i] = "C"
win = not win
elif input_string[i - 1] == "C" and input_string[i + 1] == "A":
input_string[i] = "B"
win = not win
if win:
print("Player1")
else:
print("Player2"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The underage Harry and Cedric reach the Triwizard Cup together in the hedge maze, which turns out to be a portkey that transports them to a graveyard where Pettigrew and Voldemort are waiting. After Pettigrew murders Cedric with the Killing Curse on Voldemort's order, Voldemort challenges Harry with a very tricky spell, <b>The string leviosa</b>.
In order to survive this spell, Harry has to answer the following question:
There is a string of size N consisting of characters A, B, and C. The character A can be transformed to either character B or character C. Now, two players play a game with the following rules turn wise. In each turn, the player with the turn chooses a position with character A in the string and transforms the character to another character (B or C). The only restriction to the string is that there can't be two consecutive <b>B</b>s or two consecutive <b>C</b>s in the string. The player who cannot make a move loses. Which player wins the game?
Can you help Harry in surviving the dangerous spell?The first line of the input contains a single integer N, the length of the string.
The second line of the input contains N characters of the string.
Constraints
1 <= N <= 100000
The characters in the string are restricted to A, B, and C.
There are no consecutive Bs or Cs in the input string.Output "Player1" if the player who starts the game wins, else output "Player2" (without quotes).Sample Input
3
ACA
Sample Output
Player2
Explanation: Player1 changes the first or the last character to B, the second player changes the remaining A to B in the next turn. After that, player1 cannot make a move since there are no more As in the string.
Sample Input
4
BACA
Sample Output
Player1
Explanation: Player1 changes the last character in the string to B and player2 cannot make a move further., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n;
string s;
cin >> n >> s;
assert(sz(s)==n);
int l = 0, r = 1;
int ans = 0;
while (l < n) {
while (l < n && s[l] != 'A') ++l, ++r;
while (r < n && s[r] == 'A') ++r;
if (l == 0 && r == n) {
ans ^= n % 2;
}
else if (r == n || l == 0) {
ans ^= r - l;
}
else if (r < n && s[l - 1] == s[r]){
ans ^= 1;
}
l = r;
}
cout << (ans ? "Player1" : "Player2");
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc =new Scanner(System.in);
int T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 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: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: function Charity(n,m) {
// write code here
// do no console.log the answer
// return the output using return keyword
const per = Math.floor(m / n)
return per > 1 ? per : -1
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: static int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: def Charity(N,M):
x = M//N
if x<=1:
return -1
return x
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted array your task is to merge these two arrays into a single array such that the merged array is also sortedFirst line contain two integers N and M the size of arrays
Second line contains N separated integers the elements of first array
Third line contains M separated integers elements of second array
<b>Constraints:-</b>
1<=N,M<=10<sup>4</sup>
1<=arr1[], arr2[] <=10<sup>5</sup>Output the merged arraySample Input:-
3 4
1 4 7
1 3 3 9
Sample Output:-
1 1 3 3 4 7 9
, I have written this Solution Code: n1,m1=input().split()
n=int(n1)
m=int(m1)
l1=list(map(int,input().strip().split()))
l2=list(map(int,input().strip().split()))
i,j=0,0
while i<n and j<m:
if l1[i]<=l2[j]:
print(l1[i],end=" ")
i+=1
else:
print(l2[j],end=" ")
j+=1
for a in range(i,n):
print(l1[a],end=" ")
for b in range(j,m):
print(l2[b],end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted array your task is to merge these two arrays into a single array such that the merged array is also sortedFirst line contain two integers N and M the size of arrays
Second line contains N separated integers the elements of first array
Third line contains M separated integers elements of second array
<b>Constraints:-</b>
1<=N,M<=10<sup>4</sup>
1<=arr1[], arr2[] <=10<sup>5</sup>Output the merged arraySample Input:-
3 4
1 4 7
1 3 3 9
Sample Output:-
1 1 3 3 4 7 9
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
int main(){
int n,m;
cin>>n>>m;
int a[n];
int b[m];
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<m;i++){
cin>>b[i];
}
int c[n+m];
int i=0,j=0,k=0;
while(i!=n && j!=m){
if(a[i]<b[j]){c[k]=a[i];k++;i++;}
else{c[k]=b[j];j++;k++;}
}
while(i!=n){
c[k]=a[i];
k++;i++;
}
while(j!=m){
c[k]=b[j];
k++;j++;
}
for(i=0;i<n+m;i++){
cout<<c[i]<<" ";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted array your task is to merge these two arrays into a single array such that the merged array is also sortedFirst line contain two integers N and M the size of arrays
Second line contains N separated integers the elements of first array
Third line contains M separated integers elements of second array
<b>Constraints:-</b>
1<=N,M<=10<sup>4</sup>
1<=arr1[], arr2[] <=10<sup>5</sup>Output the merged arraySample Input:-
3 4
1 4 7
1 3 3 9
Sample Output:-
1 1 3 3 4 7 9
, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int a[] = new int[n];
int b[] = new int[m];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
for(int i=0;i<m;i++){
b[i]=sc.nextInt();
}
int c[]=new int[n+m];
int i=0,j=0,k=0;
while(i!=n && j!=m){
if(a[i]<b[j]){c[k]=a[i];k++;i++;}
else{c[k]=b[j];j++;k++;}
}
while(i!=n){
c[k]=a[i];
k++;i++;
}
while(j!=m){
c[k]=b[j];
k++;j++;
}
for(i=0;i<n+m;i++){
System.out.print(c[i]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of length N in which you can swap two elements if their sum is odd i.e for every i (1 to N) and j (1 to N) if (Arr[i] + Arr[j]) is odd then you can swap these elements.
What is the lexicographically smallest array you can obtain?First line of input contains a single integer N. Next line contains N space separated integers depicting the elements of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000Print N space separated elements i. e the array which is the lexicographically smallest possibleSample Input:-
3
4 1 7
Sample Output:-
1 4 7
Explanation:-
Swap numbers at indices 1 and 2 as their sum 4 + 1 = 5 is odd
Sample Input:-
2
2 4
Sample Output:-
2 4
Sample Input:-
2
5 3
Sample Output:-
5 3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine().trim());
String array[] = br.readLine().trim().split(" ");
StringBuilder sb = new StringBuilder("");
int[] arr = new int[n];
int oddCount = 0,
evenCount = 0;
for(int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(array[i]);
if(arr[i] % 2 == 0)
++evenCount;
else
++oddCount;
}
if(evenCount > 0 && oddCount > 0)
Arrays.sort(arr);
for(int i = 0; i < n; i++)
sb.append(arr[i] + " ");
System.out.println(sb.substring(0, sb.length() - 1));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of length N in which you can swap two elements if their sum is odd i.e for every i (1 to N) and j (1 to N) if (Arr[i] + Arr[j]) is odd then you can swap these elements.
What is the lexicographically smallest array you can obtain?First line of input contains a single integer N. Next line contains N space separated integers depicting the elements of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000Print N space separated elements i. e the array which is the lexicographically smallest possibleSample Input:-
3
4 1 7
Sample Output:-
1 4 7
Explanation:-
Swap numbers at indices 1 and 2 as their sum 4 + 1 = 5 is odd
Sample Input:-
2
2 4
Sample Output:-
2 4
Sample Input:-
2
5 3
Sample Output:-
5 3, I have written this Solution Code: n=int(input())
p=[int(x) for x in input().split()[:n]]
o=0;e=0
for i in range(n):
if (p[i] % 2 == 1):
o += 1;
else:
e += 1;
if (o > 0 and e > 0):
p.sort();
for i in range(n):
print(p[i], end = " ");, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of length N in which you can swap two elements if their sum is odd i.e for every i (1 to N) and j (1 to N) if (Arr[i] + Arr[j]) is odd then you can swap these elements.
What is the lexicographically smallest array you can obtain?First line of input contains a single integer N. Next line contains N space separated integers depicting the elements of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000Print N space separated elements i. e the array which is the lexicographically smallest possibleSample Input:-
3
4 1 7
Sample Output:-
1 4 7
Explanation:-
Swap numbers at indices 1 and 2 as their sum 4 + 1 = 5 is odd
Sample Input:-
2
2 4
Sample Output:-
2 4
Sample Input:-
2
5 3
Sample Output:-
5 3, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main() {
int n;
cin>>n;
long long a[n];
bool win=false,win1=false;
for(int i=0;i<n;i++){
cin>>a[i];
if(a[i]&1){win=true;}
if(a[i]%2==0){win1=true;}
}
if(win==true && win1==true){sort(a,a+n);}
for(int i=0;i<n;i++){
cout<<a[i]<<" ";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You enter a shop and there are N types of items with the i<sup>th</sup> item's price being p<sub>i</sub>. You also have A discount coupons. Suppose you use X amount of coupons on the i<sup>th</sup> item, its discounted price will be (p<sub>i</sub>/2<sup>X</sup>) rounded down to the nearest integer. Find the minimum price you will have to pay in order to buy one item of each type.The first line of the input contains two integers N and A.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= p<sub>i</sub> <= 10<sup>9</sup>
1 <= A <= 10<sup>5</sup>Print the minimum price you will have to pay in order to buy one item of each type.Sample Input:
5 4
10 4 7 8 2
Sample Output:
15
Explaination:
You use 2 coupons on item 1, 1 coupon on item 3 and the last coupon on item 4.
This will make the price array of items as [2, 4, 3, 4, 2].
Total cost = 2 + 4 + 3 + 4 + 2 = 15, 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 n = sc.nextInt();
int a = sc.nextInt();
PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
for(int i=0; i<n; i++) {
int x = sc.nextInt();
pq.add(x);
}
while(a-- > 0) {
pq.add(pq.poll()/2);
}
long sum = 0;
while(!pq.isEmpty()) {
sum += pq.poll();
}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You enter a shop and there are N types of items with the i<sup>th</sup> item's price being p<sub>i</sub>. You also have A discount coupons. Suppose you use X amount of coupons on the i<sup>th</sup> item, its discounted price will be (p<sub>i</sub>/2<sup>X</sup>) rounded down to the nearest integer. Find the minimum price you will have to pay in order to buy one item of each type.The first line of the input contains two integers N and A.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= p<sub>i</sub> <= 10<sup>9</sup>
1 <= A <= 10<sup>5</sup>Print the minimum price you will have to pay in order to buy one item of each type.Sample Input:
5 4
10 4 7 8 2
Sample Output:
15
Explaination:
You use 2 coupons on item 1, 1 coupon on item 3 and the last coupon on item 4.
This will make the price array of items as [2, 4, 3, 4, 2].
Total cost = 2 + 4 + 3 + 4 + 2 = 15, I have written this Solution Code: from sys import stdin
from heapq import heappush, heappop
n, a = map(int, stdin.readline().split())
p = list(map(int, stdin.readline().split()))
pq = []
for i in range(n):
heappush(pq, -p[i])
while a > 0 and len(pq) > 0:
x = -heappop(pq)
x //= 2
if x > 0:
heappush(pq, -x)
a -= 1
ans = 0
while len(pq) > 0:
ans += -heappop(pq)
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You enter a shop and there are N types of items with the i<sup>th</sup> item's price being p<sub>i</sub>. You also have A discount coupons. Suppose you use X amount of coupons on the i<sup>th</sup> item, its discounted price will be (p<sub>i</sub>/2<sup>X</sup>) rounded down to the nearest integer. Find the minimum price you will have to pay in order to buy one item of each type.The first line of the input contains two integers N and A.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= p<sub>i</sub> <= 10<sup>9</sup>
1 <= A <= 10<sup>5</sup>Print the minimum price you will have to pay in order to buy one item of each type.Sample Input:
5 4
10 4 7 8 2
Sample Output:
15
Explaination:
You use 2 coupons on item 1, 1 coupon on item 3 and the last coupon on item 4.
This will make the price array of items as [2, 4, 3, 4, 2].
Total cost = 2 + 4 + 3 + 4 + 2 = 15, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, a;
cin >> n >> a;
vector<int> p(n);
for(auto &i : p) cin >> i;
priority_queue<int> pq;
for(int i = 0; i < n; i++){
pq.push(p[i]);
}
while(a-- && !pq.empty()){
int x = pq.top();
pq.pop();
x /= 2;
if(x > 0) pq.push(x);
}
int ans = 0;
while(!pq.empty()){
ans += pq.top();
pq.pop();
}
cout << ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Our five power rangers have powers P1, P2, P3, P4, and P5.
To ensure the proper distribution of power, the power of every power ranger must remain <b>less</b> than the sum of powers of other power rangers.
If the above condition is not met, there's an emergency. Can you let us know if there's an emergency?The first and the only line of input contains 5 integers P1, P2, P3, P4, and P5.
Constraints
0 <= P1, P2, P3, P4, P5 <= 100Output "SPD Emergency" (without quotes) if there's an emergency, else output "Stable".Sample Input
1 2 3 4 5
Sample Output
Stable
Explanation
The power of every power ranger is less than the sum of powers of other power rangers.
Sample Input
1 2 3 4 100
Sample Output
SPD Emergency
Explanation
The power of the 5th power ranger (100) is not less than the sum of powers of other power rangers (1+2+3+4=10)., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
int[] arr=new int[5];
BufferedReader rd=new BufferedReader(new InputStreamReader(System.in));
String[] s=rd.readLine().split(" ");
int sum=0;
for(int i=0;i<5;i++){
arr[i]=Integer.parseInt(s[i]);
sum+=arr[i];
}
int i=0,j=arr.length-1;
boolean isEmergency=false;
while(i<=j)
{
int temp=arr[i];
sum-=arr[i];
if(arr[i]>= sum)
{
isEmergency=true;
break;
}
sum+=temp;
i++;
}
if(isEmergency==false)
{
System.out.println("Stable");
}
else
{
System.out.println("SPD Emergency");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Our five power rangers have powers P1, P2, P3, P4, and P5.
To ensure the proper distribution of power, the power of every power ranger must remain <b>less</b> than the sum of powers of other power rangers.
If the above condition is not met, there's an emergency. Can you let us know if there's an emergency?The first and the only line of input contains 5 integers P1, P2, P3, P4, and P5.
Constraints
0 <= P1, P2, P3, P4, P5 <= 100Output "SPD Emergency" (without quotes) if there's an emergency, else output "Stable".Sample Input
1 2 3 4 5
Sample Output
Stable
Explanation
The power of every power ranger is less than the sum of powers of other power rangers.
Sample Input
1 2 3 4 100
Sample Output
SPD Emergency
Explanation
The power of the 5th power ranger (100) is not less than the sum of powers of other power rangers (1+2+3+4=10)., I have written this Solution Code: arr = list(map(int,input().split()))
m = sum(arr)
f=[]
for i in range(len(arr)):
s = sum(arr[:i]+arr[i+1:])
if(arr[i]<s):
f.append(1)
else:
f.append(0)
if(all(f)):
print("Stable")
else:
print("SPD Emergency"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Our five power rangers have powers P1, P2, P3, P4, and P5.
To ensure the proper distribution of power, the power of every power ranger must remain <b>less</b> than the sum of powers of other power rangers.
If the above condition is not met, there's an emergency. Can you let us know if there's an emergency?The first and the only line of input contains 5 integers P1, P2, P3, P4, and P5.
Constraints
0 <= P1, P2, P3, P4, P5 <= 100Output "SPD Emergency" (without quotes) if there's an emergency, else output "Stable".Sample Input
1 2 3 4 5
Sample Output
Stable
Explanation
The power of every power ranger is less than the sum of powers of other power rangers.
Sample Input
1 2 3 4 100
Sample Output
SPD Emergency
Explanation
The power of the 5th power ranger (100) is not less than the sum of powers of other power rangers (1+2+3+4=10)., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
vector<int> vect(5);
int tot = 0;
for(int i=0; i<5; i++){
cin>>vect[i];
tot += vect[i];
}
sort(all(vect));
tot -= vect[4];
if(vect[4] >= tot){
cout<<"SPD Emergency";
}
else{
cout<<"Stable";
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to calculate values for each of the following operations:-
a + b
a - b
a * b
a/bSince this will be a functional problem, you don't have to take input. You have to complete the function
<b>operations()</b> that takes the integer a and b as parameters.
<b>Constraints:</b>
1 ≤ b ≤ a ≤1000
<b> It is guaranteed that a will be divisible by b.</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5, I have written this Solution Code: def operations(x, y):
print(x+y)
print(x-y)
print(x*y)
print(x//y), 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, containing positive integers. You need to sort the elements of array using Insertion sort algorithm.First line of the input denotes number of test cases T. First line of the testcase is the size of array N and second line consists of array elements separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
Explanation:
Testcase 1: The array after perfoming insertion sort: 1 3 4 7 9.
Testcase 2: The array after performing insertion sort: 1 2 3 4 5 6 7 8 9 10., I have written this Solution Code: // arr is unsorted array
// n is the number of elements in the array
function insertionSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => a - b)
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N, containing positive integers. You need to sort the elements of array using Insertion sort algorithm.First line of the input denotes number of test cases T. First line of the testcase is the size of array N and second line consists of array elements separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
Explanation:
Testcase 1: The array after perfoming insertion sort: 1 3 4 7 9.
Testcase 2: The array after performing insertion sort: 1 2 3 4 5 6 7 8 9 10., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void insertionSort(int[] arr){
for(int i = 0; i < arr.length-1; i++){
for(int j = i+1; j < arr.length; j++){
if(arr[i] > arr[j]){
int temp = arr[j];
arr[j] = arr[i];
arr[i] = temp;
}
}
}
}
public static void main (String[] args) {
Scanner scan = new Scanner(System.in);
int T = scan.nextInt();
while(T > 0){
int n = scan.nextInt();
int arr[] = new int[n];
for(int i = 0; i<n; i++){
arr[i] = scan.nextInt();
}
insertionSort(arr);
for(int i = 0; i<n; i++){
System.out.print(arr[i] + " ");
}
System.out.println();
T--;
System.gc();
}
}
}, 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, containing positive integers. You need to sort the elements of array using Insertion sort algorithm.First line of the input denotes number of test cases T. First line of the testcase is the size of array N and second line consists of array elements separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
Explanation:
Testcase 1: The array after perfoming insertion sort: 1 3 4 7 9.
Testcase 2: The array after performing insertion sort: 1 2 3 4 5 6 7 8 9 10., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << 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 A[] of size N, containing positive integers. You need to sort the elements of array using Insertion sort algorithm.First line of the input denotes number of test cases T. First line of the testcase is the size of array N and second line consists of array elements separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
Explanation:
Testcase 1: The array after perfoming insertion sort: 1 3 4 7 9.
Testcase 2: The array after performing insertion sort: 1 2 3 4 5 6 7 8 9 10., I have written this Solution Code: def InsertionSort(arr):
arr.sort()
return arr, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements and a number K. The task is to arrange array elements according to the absolute difference with K, i. e., element having minimum difference comes first and so on.
Note : If two or more elements are at equal distance arrange them in same sequence as in the given array.First line of input contains a single integer T which denotes the number of test cases. Then T test cases follows. First line of test case contains two space separated integers N and K. Second line of each test case contains N space separated integers.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= K <= 10^4
1 <= A[i] <= 10^4
Sum of N over all test cases does not exceed 2*10^5For each test case print the given array in the order as described above.Input:
3
5 7
10 5 3 9 2
5 6
1 2 3 4 5
4 5
2 6 8 3
Output:
5 9 10 3 2
5 4 3 2 1
6 3 2 8
Explanation:
Testcase 1: Sorting the numbers accoding to the absolute difference with 7, we have array elements as 5, 9, 10, 3, 2.
Testcase 2: Sorting the numbers according to the absolute difference with 6, we have array elements as 5 4 3 2 1.
Testcase 3: Sorting the numbers according to the absolute difference with 5, we have array elements as 6 3 2 8., 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().trim());
while(t-->0){
String inputLine[] = br.readLine().trim().split(" ");
int n = Integer.parseInt(inputLine[0]);
int x = Integer.parseInt(inputLine[1]);
int arr[] = new int[n];
inputLine = br.readLine().trim().split(" ");
for(int i=0; i<n; i++){
arr[i] = Integer.parseInt(inputLine[i]);
}
new Solution().sortABS(arr,n, x);
StringBuilder sb = new StringBuilder();
for(int y : arr)
sb.append(y+ " ");
System.out.println(sb.toString());
}
}
}
class Solution
{
static void sortABS(int arr[], int n,int k)
{
List<Integer> aux=new ArrayList<>();
Sort obj=new Sort();
obj.k=k;
int j=0;
for(int i:arr)
aux.add(i);
Collections.sort(aux, obj);
for(int i:aux)
arr[j++]=i;
}
}
class Sort implements Comparator<Integer> {
Integer k;
public int compare(Integer a, Integer b)
{
return Math.abs(a-k) - Math.abs(b-k);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements and a number K. The task is to arrange array elements according to the absolute difference with K, i. e., element having minimum difference comes first and so on.
Note : If two or more elements are at equal distance arrange them in same sequence as in the given array.First line of input contains a single integer T which denotes the number of test cases. Then T test cases follows. First line of test case contains two space separated integers N and K. Second line of each test case contains N space separated integers.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= K <= 10^4
1 <= A[i] <= 10^4
Sum of N over all test cases does not exceed 2*10^5For each test case print the given array in the order as described above.Input:
3
5 7
10 5 3 9 2
5 6
1 2 3 4 5
4 5
2 6 8 3
Output:
5 9 10 3 2
5 4 3 2 1
6 3 2 8
Explanation:
Testcase 1: Sorting the numbers accoding to the absolute difference with 7, we have array elements as 5, 9, 10, 3, 2.
Testcase 2: Sorting the numbers according to the absolute difference with 6, we have array elements as 5 4 3 2 1.
Testcase 3: Sorting the numbers according to the absolute difference with 5, we have array elements as 6 3 2 8., I have written this Solution Code: def merge(arr,l,m,h,x):
n1=m-l+1
n2=h-m
L=arr[l:m+1]
R=arr[m+1:h+1]
i=j=0
k=l
while(i<n1 and j<n2):
if abs(x-L[i])<=abs(x-R[j]):
arr[k]=L[i]
i+=1
else:
arr[k]=R[j]
j+=1
k+=1
while(i<n1):
arr[k]=L[i]
i+=1
k+=1
while(j<n2):
arr[k]=R[j]
j+=1
k+=1
def mergeSort(arr,l,h,k):
if l<h:
m=l+(h-l)//2
mergeSort(arr,l,m,k)
mergeSort(arr,m+1,h,k)
merge(arr,l,m,h,k)
if __name__=="__main__":
t=int(input())
while t>0:
n,k=map(int,input().split())
arr=list(map(int,input().split()))
mergeSort(arr,0,len(arr)-1,k)
for i in arr:
print(i,end=" ")
print()
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements and a number K. The task is to arrange array elements according to the absolute difference with K, i. e., element having minimum difference comes first and so on.
Note : If two or more elements are at equal distance arrange them in same sequence as in the given array.First line of input contains a single integer T which denotes the number of test cases. Then T test cases follows. First line of test case contains two space separated integers N and K. Second line of each test case contains N space separated integers.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= K <= 10^4
1 <= A[i] <= 10^4
Sum of N over all test cases does not exceed 2*10^5For each test case print the given array in the order as described above.Input:
3
5 7
10 5 3 9 2
5 6
1 2 3 4 5
4 5
2 6 8 3
Output:
5 9 10 3 2
5 4 3 2 1
6 3 2 8
Explanation:
Testcase 1: Sorting the numbers accoding to the absolute difference with 7, we have array elements as 5, 9, 10, 3, 2.
Testcase 2: Sorting the numbers according to the absolute difference with 6, we have array elements as 5 4 3 2 1.
Testcase 3: Sorting the numbers according to the absolute difference with 5, we have array elements as 6 3 2 8., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int k;
bool cmp(pi a, pi b){
int x = abs(a.fi-k), y = abs(b.fi-k);
if(x == y)
return a.se < b.se;
return x < y;
}
pi a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n >> k;
for(int i = 1; i <= n; i++){
cin >> a[i].fi;
a[i].se = i;
}
sort(a + 1, a + n + 1, cmp);
for(int i = 1; i <= n; i++)
cout << a[i].fi << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and F(N) find value of F(1), if, F(i)=(F(i-1) + F(i-1))%1000000007 and 0 <= F(1) < 1000000007.First and the only line of input contains two integers N and F(N).
Constraints:
1 <= N <= 1000000000
0 <= F(N) < 1000000007Print a single integer, F(1).Sample Input 1
2 6
Sample Output 1
3
Exlpanation: F(1) = 3, F(2)=(3+3)%1000000007 = 6.
Sample Input 2
3 6
Sample Input 2
500000005
Explanation:
F(1) = 500000005
F(2) = (500000005+500000005)%1000000007 = 3
F(3)= (3+3)%1000000007 = 6, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws NumberFormatException, IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer(br.readLine());
int n=Integer.parseInt(st.nextToken());
int fn=Integer.parseInt(st.nextToken());
long ans=fn;
int P=1000000007;
long inv2n=(long)pow(pow(2,n-1,P)%P,P-2,P);
ans=((ans%P)*(inv2n%P))%P;
System.out.println((ans)%P);
}
static long pow(long x, long y,long P)
{
long res = 1l;
while (y > 0)
{
if ((y & 1) == 1)
res = (res * x)%P;
y = y >> 1;
x = (x * x)%P;
}
return res;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and F(N) find value of F(1), if, F(i)=(F(i-1) + F(i-1))%1000000007 and 0 <= F(1) < 1000000007.First and the only line of input contains two integers N and F(N).
Constraints:
1 <= N <= 1000000000
0 <= F(N) < 1000000007Print a single integer, F(1).Sample Input 1
2 6
Sample Output 1
3
Exlpanation: F(1) = 3, F(2)=(3+3)%1000000007 = 6.
Sample Input 2
3 6
Sample Input 2
500000005
Explanation:
F(1) = 500000005
F(2) = (500000005+500000005)%1000000007 = 3
F(3)= (3+3)%1000000007 = 6, I have written this Solution Code: n,f=map(int,input().strip().split())
mod=10**9+7
m1=pow(2,n-1,mod)
m=pow(m1,mod-2,mod)
print(f*m%mod), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and F(N) find value of F(1), if, F(i)=(F(i-1) + F(i-1))%1000000007 and 0 <= F(1) < 1000000007.First and the only line of input contains two integers N and F(N).
Constraints:
1 <= N <= 1000000000
0 <= F(N) < 1000000007Print a single integer, F(1).Sample Input 1
2 6
Sample Output 1
3
Exlpanation: F(1) = 3, F(2)=(3+3)%1000000007 = 6.
Sample Input 2
3 6
Sample Input 2
500000005
Explanation:
F(1) = 500000005
F(2) = (500000005+500000005)%1000000007 = 3
F(3)= (3+3)%1000000007 = 6, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
long long powerm(long long x, unsigned long long y, long long p)
{
long long res = 1;
x = x % p;
while (y > 0)
{
if (y & 1)
res = (res*x) % p;
y = y>>1;
x = (x*x) % p;
}
return res;
}
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n,fn;
cin>>n>>fn;
int mo=1000000007;
cout<<(fn*powerm(powerm(2,n-1,mo),mo-2,mo))%mo;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given first term A and the common ratio R of a GP (Geometric Progression) and an integer N, your task is to calculate its Nth term.<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>NthGP()</b> that takes the integer A, R and N as parameter.
<b>Constraints</b>
1 <= A, R <= 100
1 <= N <= 10
Return the Nth term of the given GP.
<b>Note:</b> It is guaranteed that the Nth term of this GP will always fit in 10^9.Sample Input:
3 2
5
Sample Output:-
48
Sample Input:-
2 2
10
Sample Output:
1024
<b>Explanation:-</b>
For Test Case 1:- 3 6 12 24 48 96., I have written this Solution Code:
static int NthGP(int l, int b, int h){
int ans=l;
h--;
while(h-->0){
ans*=b;
}
return ans;
}
, 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, find the count of pairs of indices <b>i</b> and <b>j</b> such that <b>1 <= i < j <= N</b> and the <b>mean</b> of elements at these indices is at least <b>K</b>.
More formally, you have to find the number of indices <b>i</b> and <b>j</b> such that <b>1 <= i < j <= N</b> and (A<sub>i</sub> + A<sub>j</sub>)/2 >= KFirst line of the input contains two integers N and K.
The second line contains N space seperated integers A<sub>i</sub>
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= K <= 10<sup>9</sup>
1 <= A<sub>i</sub> <= 10<sup>9</sup>Print the number of pairs satisfying the above condition in array A.Sample Input:
5 6
4 7 8 2 5
Sample Output:
4
Explaination:
The following pairs of indices satisfy the condition (1-based indexing)
(1, 3) -> (4 + 8)/2 = 6
(2, 3) -> (7 + 8)/2 = 7.5
(2, 5) -> (7 + 5)/2 = 6
(3, 5) -> (8 + 5)/2 = 6.5
There are no more pairs of indices that satisfy the above condition., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define fast \
ios_base::sync_with_stdio(false); \
cin.tie(NULL);
#define int long long
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using T = pair<int, int>;
typedef long long ll;
const int mod = 1e9 + 7;
const int INF = 1e9;
void solve() {
int n, k;
cin >> n >> k;
assert(1 <= n <= 1e5);
assert(1 <= k <= 1e9);
vector<int> a(n);
for (auto &i : a) cin >> i, assert(1 <= i <= 1e9);
sort(all(a));
int ans = 0;
int req_sum = 2 * k;
for (int i = 0; i < n; i++) {
int x = req_sum - a[i];
x = max(x, (int)0);
int l = i + 1, r = n - 1, ind = n;
while (l <= r) {
int m = (l + r) / 2;
if (a[m] >= x) {
r = m - 1;
ind = m;
} else
l = m + 1;
}
ans += n - ind;
}
cout << ans;
}
signed main() {
fast int t = 1;
// cin >> t;
for (int i = 1; i <= t; i++) {
solve();
if (i != t) cout << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, find the count of pairs of indices <b>i</b> and <b>j</b> such that <b>1 <= i < j <= N</b> and the <b>mean</b> of elements at these indices is at least <b>K</b>.
More formally, you have to find the number of indices <b>i</b> and <b>j</b> such that <b>1 <= i < j <= N</b> and (A<sub>i</sub> + A<sub>j</sub>)/2 >= KFirst line of the input contains two integers N and K.
The second line contains N space seperated integers A<sub>i</sub>
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= K <= 10<sup>9</sup>
1 <= A<sub>i</sub> <= 10<sup>9</sup>Print the number of pairs satisfying the above condition in array A.Sample Input:
5 6
4 7 8 2 5
Sample Output:
4
Explaination:
The following pairs of indices satisfy the condition (1-based indexing)
(1, 3) -> (4 + 8)/2 = 6
(2, 3) -> (7 + 8)/2 = 7.5
(2, 5) -> (7 + 5)/2 = 6
(3, 5) -> (8 + 5)/2 = 6.5
There are no more pairs of indices that satisfy the above condition., 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 nk=br.readLine();
String nkarr[]=nk.split(" ");
int n =Integer.parseInt(nkarr[0]);
int k =Integer.parseInt(nkarr[1]);
String ss = br.readLine();
String sst[]=ss.split(" ");
int arr[]=new int[n];
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(sst[i]);
}
long val = findMean(arr,n,k);
System.out.println(val);
}
public static long findMean(int []arr,int n,int k){
Arrays.sort(arr);
int low =0;
int high=arr.length-1;
long ans=0;
while(low<high){
long mean =(arr[low]+arr[high])/2;
if(mean>=k){
ans+=(high-low);
high--;
}else{
low++;
}
}
return ans;
}
}, 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: 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: 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 |
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