HackerRank Permuting Two Arrays problem solution

In this HackerRank Permuting Two Arrays problem solution there are two n element arrays of integers A and B. we need to permute them into some A’ and B’ such that the relation A'[i] + B'[i] >= k holds for all I where 0 <= I < n. and there are q queries consisting of A, B, and k. for each query return YES if some permutation A’.B’ satisfying the relation exists. otherwise, return NO.

Problem solution in Python.

#!/usr/bin/env python3

def rl(T=str):
    return list(map(T,input().split()))

T, = rl(int)
for _ in range((T)):
    N,K = rl(int)
    A = rl(int)
    B = rl(int)
    A.sort()
    B.sort(reverse=True)
    bad = len([a+b for a,b in zip(A,B) if a+b<K ])>0
    print("NO" if bad else "YES")

Problem solution in Java.

import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while (t-- > 0) {
            int n = in.nextInt();
            int k = in.nextInt();
            int[] a = new int[n];
            int[] b = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = in.nextInt();
            }
            for (int i = 0; i < n; i++) {
                b[i] = in.nextInt();
            }
            Arrays.sort(a);
            Arrays.sort(b);
            solve(a, b, k);
        }
    }

    private static void solve(int[] a, int[] b, int k) {
        for(int i=0, j=a.length-1; i<a.length; i++, j--){
            if(a[i]+b[j] < k) {
                System.out.println("NO");
                return;
            }
        }
        System.out.println("YES");
    }
}

Problem solution in C++.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <stdlib.h>
using namespace std;

int compare1(const void * a, const void * b)
{
  return ( *(int*)b - *(int*)a );
}

int compare2(const void * a, const void * b)
{
  return ( *(int*)a - *(int*)b );
}

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    
    long t;
    long n, k;
    long a[1000], b[1000];
    
    cin>>t;
    
    for(int i=0; i < t; i++)
    {
        cin>>n>>k;
        
        for(int j=0; j<n; j++)
        {
            cin>>a[j];
        }
        
        qsort (a, n, sizeof(long), compare1);
        
        for(int j=0; j<n; j++)
            cin>>b[j];
        
        qsort (b, n, sizeof(long), compare2);
        
        bool flag=true;
        
        for(int i=0; i<n; i++)
            if(a[i]+b[i]<k)
            {
                flag = false;
                break;
            }
        
        if(flag)
            cout<<"YESn";
        else
             cout<<"NOn";
    }
    
    return 0;
}

Problem solution in C.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int compare (const void * a, const void * b)
{
  return ( *(int*)a - *(int*)b );
}

int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    int T;
    int dummy;
    dummy=scanf("%d",&T);
    
    int N;
    int K;
    
    int i;
    for(i=0; i<T; i++)
    {
        dummy=scanf("%d",&N);
        int A[N];
        int B[N];
        
        dummy=scanf("%d",&K);
        
        int j;
        for(j=0; j<N; j++)
        {
            dummy=scanf("%d",A+j);
        }
        for(j=0; j<N; j++)
        {
            dummy=scanf("%d",B+j);
        }
        
        /* input fetched */
        
        //sorting arrays...
        qsort (A, N, sizeof(int), compare);
        qsort (B, N, sizeof(int), compare);
        
        int k;
        int isPossible = 1;
               
        for (k=0; k<N; k++)
        {
            if ((isPossible == 1) && (A[k]+B[N-1-k] < K))
                isPossible = 0;
        }
        
        if (isPossible == 1)
            printf("YESn");
        else
            printf("NOn");
        
    }
    return 0;
}