Permuting Two Arrays HackerRank Problem Solution

Permuting Two Arrays HackerRank Problem Solution

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In this Permuting Two Arrays HackerRank Problem,

There are two -element arrays of integers,  and . Permute them into some  and  such that the relation  holds for all  where .

There will be  queries consisting of , , and . For each query, return YES if some permutation ,  satisfying the relation exists. Otherwise, return NO.

Example


A valid  is  and :  and . Return YES.

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

Input Format

The first line contains an integer , the number of queries.

The next  sets of  lines are as follows:

  • The first line contains two space-separated integers  and , the size of both arrays  and , and the relation variable.
  • The second line contains  space-separated integers .
  • The third line contains  space-separated integers .

Constraints

Sample Input

STDIN       Function
-----       --------
2           q = 2
3 10        A[] and B[] size n = 3, k = 10
2 1 3       A = [2, 1, 3]
7 8 9       B = [7, 8, 9]
4 5         A[] and B[] size n = 4, k = 5
1 2 2 1     A = [1, 2, 2, 1]
3 3 3 4     B = [3, 3, 3, 4]

Sample Output

YES
NO

Explanation

There are two queries:

  1. Permute these into  and  so that the following statements are true:
  2. , , and . To permute  and  into a valid  and , there must be at least three numbers in  that are greater than .

Permuting Two Arrays Hacker Rank Problem Solutions

I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.

Problem Solution in C#

public static string TwoArrays(int k, List<int> A, List<int> B)
{
    A.Sort();
    B.Sort();
    B.Reverse();


    for (int i = 0; i < A.Count; i++)
    {
        if (A[i] + B[i] < k)
        {
            return "NO";
        }
    }
    return "YES";
}

Problem Solution in JavaScript

function twoArrays(k, A, B) {
    // Write your code here
   
    const a = A.sort((a,b) => a - b)
    const b = B.sort((a,b) => b - a)
   
    for(let i = 0; i < a.length; i++) if(a[i] + b[i] < k) return 'NO';
   
    return 'YES'
   
}

Problem Solution in PHP

function twoArrays($k, $A, $B) {
    // Write your code here
    $c = count($A) - 1;
    $result = true;
    sort($A);
    rsort($B);
    for($i = 0; $i <= $c; $i++){
        $sum = $A[$i] + $B[$i];
        if($sum < $k){
            $result = false;
        }
    }
    return $result == false ? 'NO' : 'YES';


}

Problem Solution in Python

def twoArrays(k, A, B):
    A = sorted(A)
    B = sorted(B, reverse=True)
   
    if any(list(map(lambda x,y: x+y<k, A, B))):
        return 'NO'
    return 'YES'

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