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.

Function Description

Complete the twoArrays function in the editor below. It should return a string, either YES or NO.

twoArrays has the following parameter(s):

• int k: an integer
• int A[n]: an array of integers
• int B[n]: an array of integers

Returns
– string: either YES or NO

HackerRank Permuting Two Arrays 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")

Permuting Two Arrays 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;
}