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Rearrange array such that sum of same indexed elements is atmost K
  • Last Updated : 09 Nov, 2020

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 Ai + Bi 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 ith 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: 



Below is the implementation of the above approach:

C++

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// 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 conditon 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;
}

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Java

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// 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 conditon 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

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Python3

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# 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 conditon 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

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C#

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// 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 conditon 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

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Output: 

3 2 1 1 1










 

Time Complexity: O(N) 
Auxiliary Space Complexity: O(1)

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