Rearrange array such that sum of same indexed elements is atmost K

Given two arrays A[] and B[] consisting of N integers each and an integer K, the task is to rearrange the array B[] such that sum of A_i + B_i is atmost K. If no such arrangement is possible, print -1.

Examples:

Input: A[] = {1, 2, 3, 4, 2}, B[] = {1, 2, 3, 1, 1}, K = 5
Output: 1 3 1 1 2

Input: A[] = {1, 2, 3, 4, 5}, B[] = {2, 3, 4, 5, 6}, K = 6
Output: -1

Approach: The most optimal rearrangement of the array B[] to satisfy the given conditions is to sort the array in descending order.

Proof:

1. Since the sum of every pair of i^th indexed elements can be atmost K. Therefore, mn + num ? X and mx + num1 ? X, where mn and mx are the minimum elements in the array A[] and B[] respectively.
2. Therefore, by induction, it can be proved that mn and mx can be paired.
3. Therefore, sorting the array in descending order is the most optimal rearrangement

Follow the steps below to solve the problem: 

• Sort the array B[] in descending order.
• Traverse the array and check if A_i + B_i is less than or equal to K for every i^th index.
• If the condition fails for any pair, print -1.
• Otherwise, print the array B[]

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to rearrange array such
// that sum of similar indexed elements
// does not exceed K
void rearrangeArray(int A[], int B[],
                    int N, int K)
{
    // Sort the array B[]
    // in descending order
    sort(B, B + N, greater<int>());

    bool flag = true;

    for (int i = 0; i < N; i++) {

        // If condition fails
        if (A[i] + B[i] > K) {
            flag = false;
            break;
        }
    }

    if (!flag) {

        cout << "-1" << endl;
    }
    else {

        // Print the array
        for (int i = 0; i < N; i++) {
            cout << B[i] << " ";
        }
    }
}

// Driver Code
int main()
{
    // Given arrays
    int A[] = { 1, 2, 3, 4, 2 };
    int B[] = { 1, 2, 3, 1, 1 };

    int N = sizeof(A) / sizeof(A[0]);

    int K = 5;

    rearrangeArray(A, B, N, K);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Reverse array
static int[] reverse(int a[])
{
    int i, n = a.length, t;
    for(i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
}

// Function to rearrange array such
// that sum of similar indexed elements
// does not exceed K
static void rearrangeArray(int A[], int B[],
                           int N, int K)
{

    // Sort the array B[]
    // in descending order
    Arrays.sort(B);
    B = reverse(B);

    boolean flag = true;

    for(int i = 0; i < N; i++)
    {

        // If condition fails
        if (A[i] + B[i] > K)
        {
            flag = false;
            break;
        }
    }

    if (!flag)
    {
        System.out.print("-1" + "\n");
    }
    else
    {

        // Print the array
        for(int i = 0; i < N; i++)
        {
            System.out.print(B[i] + " ");
        }
    }
}

// Driver Code
public static void main(String[] args)
{

    // Given arrays
    int A[] = { 1, 2, 3, 4, 2 };
    int B[] = { 1, 2, 3, 1, 1 };

    int N = A.length;

    int K = 5;

    rearrangeArray(A, B, N, K);
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach

# Function to rearrange array such
# that sum of similar indexed elements
# does not exceed K
def rearrangeArray(A, B, N, K):

    # Sort the array B[]
    # in descending order
    B.sort(reverse = True)

    flag = True

    for i in range(N):

        # If condition fails
        if (A[i] + B[i] > K):
            flag = False
            break

    if (flag == False):
        print("-1")
    else:

        # Print the array
        for i in range(N):
            print(B[i], end = " ")

# Driver Code
if __name__ == '__main__':

    # Given arrays
    A = [ 1, 2, 3, 4, 2 ]
    B = [ 1, 2, 3, 1, 1 ]

    N = len(A)
    K = 5;

    rearrangeArray(A, B, N, K)

# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the
// above approach
using System;
class GFG{

// Reverse array
static int[] reverse(int []a)
{
  int i, n = a.Length, t;

  for(i = 0; i < n / 2; i++)
  {
    t = a[i];
    a[i] = a[n - i - 1];
    a[n - i - 1] = t;
  }
  return a;
}

// Function to rearrange array such
// that sum of similar indexed elements
// does not exceed K
static void rearrangeArray(int []A, int []B,
                           int N, int K)
{
  // Sort the array []B
  // in descending order
  Array.Sort(B);
  B = reverse(B);

  bool flag = true;

  for(int i = 0; i < N; i++)
  {
    // If condition fails
    if (A[i] + B[i] > K)
    {
      flag = false;
      break;
    }
  }

  if (!flag)
  {
    Console.Write("-1" + "\n");
  }
  else
  {
    // Print the array
    for(int i = 0; i < N; i++)
    {
      Console.Write(B[i] + " ");
    }
  }
}

// Driver Code
public static void Main(String[] args)
{
  // Given arrays
  int []A = {1, 2, 3, 4, 2};
  int []B = {1, 2, 3, 1, 1};

  int N = A.Length;
  int K = 5;
  rearrangeArray(A, B, N, K);
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript program for the above approach

// Reverse array
function reverse(a)
{
    let i, n = a.length, t;
    for(i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
}

// Function to rearrange array such
// that sum of similar indexed elements
// does not exceed K
function rearrangeArray(A, B, N, K)
{

    // Sort the array B[]
    // in descending order
    B.sort();
    B = reverse(B);

    let flag = true;

    for(let i = 0; i < N; i++)
    {

        // If condition fails
        if (A[i] + B[i] > K)
        {
            flag = false;
            break;
        }
    }

    if (!flag)
    {
         document.write("-1" + "<br/>");
    }
    else
    {

        // Print the array
        for(let i = 0; i < N; i++)
        {
             document.write(B[i] + " ");
        }
    }
}

// Driver Code

// Given arrays
let A = [ 1, 2, 3, 4, 2 ];
let B = [ 1, 2, 3, 1, 1 ];

let N = A.length;
let K = 5;

rearrangeArray(A, B, N, K);

// This code is contribute by target_2

</script>

Output:

3 2 1 1 1

Time Complexity: O(N log N), used for sorting the given array B 
Auxiliary Space Complexity: O(1)