Minimum sum of absolute difference of pairs of two arrays

Last Updated : 08 Jul, 2022

Given two arrays a[] and b[] of equal length n. The task is to pair each element of array a to an element in array b, such that sum S of absolute differences of all the pairs is minimum.
Suppose, two elements a[i] and a[j] (i != j) of a are paired with elements b[p] and b[q] of b respectively,
then p should not be equal to q.

Examples:

Input :  a[] = {3, 2, 1}
         b[] = {2, 1, 3}
Output : 0
Explanation :
 1st pairing: |3 - 2| + |2 - 1| + |1 - 3|
         = 1 + 1 + 2 = 4
 2nd pairing: |3 - 2| + |1 - 1| + |2 - 3|
         = 1 + 0 + 1 = 2
 3rd pairing: |2 - 2| + |3 - 1| + |1 - 3|
         = 0 + 2 + 2 = 4
 4th pairing: |1 - 2| + |2 - 1| + |3 - 3|
         = 1 + 1 + 0 = 2
 5th pairing: |2 - 2| + |1 - 1| + |3 - 3|
         = 0 + 0 + 0 = 0
 6th pairing: |1 - 2| + |3 - 1| + |2 - 3|
         = 1 + 2 + 1 = 4
 Therefore, 5th pairing has minimum sum of
 absolute difference.

Input :  n = 4
         a[] = {4, 1, 8, 7}
         b[] = {2, 3, 6, 5}
Output : 6

The solution to the problem is a simple greedy approach. It consists of two steps.

Step 1 : Sort both the arrays in O (n log n) time.
Step 2 : Find absolute difference of each pair of corresponding elements (elements at same index) of both arrays and add the result to the sum S. The time complexity of this step is O(n).

Hence, the overall time complexity of the program is O(n log n).

C++

// C++ program to find minimum sum of absolute
// differences of two arrays.
#include <bits/stdc++.h>
using namespace std;

// Returns minimum possible pairwise absolute
// difference of two arrays.
long long int findMinSum(long long int a[],long long int b[], int n)
{
    // Sort both arrays
    sort(a, a+n);
    sort(b, b+n);

    // Find sum of absolute differences
    long long int sum= 0 ;
    for (int i=0; i<n; i++)
        sum = sum + abs(a[i]-b[i]);

    return sum;
}

// Driver code
int main()
{
    // Both a[] and b[] must be of same size.
    long long int a[] = {4, 1, 8, 7};
    long long int b[] = {2, 3, 6, 5};
    int n = sizeof(a)/sizeof(a[0]);
    printf("%lld\n", findMinSum(a, b, n));
    return 0;
}

// C program to find minimum sum of absolute
// differences of two arrays.
#include <stdio.h>
#include<stdlib.h>

void merge(int arr[], int l, int m, int r)
{
  int i, j, k;
  int n1 = m - l + 1;
  int n2 = r - m;

  /* create temp arrays */
  int L[n1], R[n2];

  /* Copy data to temp arrays L[] and R[] */
  for (i = 0; i < n1; i++)
    L[i] = arr[l + i];
  for (j = 0; j < n2; j++)
    R[j] = arr[m + 1 + j];

  /* Merge the temp arrays back into arr[l..r]*/
  i = 0; // Initial index of first subarray
  j = 0; // Initial index of second subarray
  k = l; // Initial index of merged subarray
  while (i < n1 && j < n2) {
    if (L[i] <= R[j]) {
      arr[k] = L[i];
      i++;
    }
    else {
      arr[k] = R[j];
      j++;
    }
    k++;
  }

  /* Copy the remaining elements of L[], if there
    are any */
  while (i < n1) {
    arr[k] = L[i];
    i++;
    k++;
  }

  /* Copy the remaining elements of R[], if there
    are any */
  while (j < n2) {
    arr[k] = R[j];
    j++;
    k++;
  }
}

/* l is for left index and r is right index of the
sub-array of arr to be sorted */
void mergeSort(int arr[], int l, int r)
{
  if (l < r) {
    // Same as (l+r)/2, but avoids overflow for
    // large l and h
    int m = l + (r - l) / 2;

    // Sort first and second halves
    mergeSort(arr, l, m);
    mergeSort(arr, m + 1, r);

    merge(arr, l, m, r);
  }
}

// Returns minimum possible pairwise absolute
// difference of two arrays.
int findMinSum(int a[],int b[], int n)
{
  // Sort both arrays
  mergeSort(a,0,n-1);
  mergeSort(b,0,n-1);

  // Find sum of absolute differences
  int sum= 0 ;
  for (int i=0; i<n; i++)
    sum = sum + abs(a[i]-b[i]);

  return sum;
}

// Driver code
int main()
{

  // Both a[] and b[] must be of same size.
  int a[] = {4, 1, 8, 7};
  int b[] = {2, 3, 6, 5};
  int n = sizeof(a)/sizeof(a[0]);
  printf("%d\n", findMinSum(a, b, n));
  return 0;
}

// This code is contributed by rexomkar.

Java

// Java program to find minimum sum of
// absolute differences of two arrays.
import java.util.Arrays;

class MinSum
{
    // Returns minimum possible pairwise 
    // absolute difference of two arrays.
    static long findMinSum(long a[], long b[], long n)
    {
        // Sort both arrays
        Arrays.sort(a);
        Arrays.sort(b);
     
        // Find sum of absolute differences
        long sum = 0 ;
        for (int i = 0; i < n; i++)
            sum = sum + Math.abs(a[i] - b[i]);
     
        return sum;
    }
     
    // Driver code
    public static void main(String[] args) 
    {
        // Both a[] and b[] must be of same size.
        long a[] = {4, 1, 8, 7};
        long b[] = {2, 3, 6, 5};
        int n = a.length;
        System.out.println(findMinSum(a, b, n));
    }    
}
 
// This code is contributed by Raghav Sharma

Python3

# Python3 program to find minimum sum 
# of absolute differences of two arrays.
def findMinSum(a, b, n):

    # Sort both arrays
    a.sort()
    b.sort()

    # Find sum of absolute differences
    sum = 0
    
    for i in range(n):
        sum = sum + abs(a[i] - b[i])

    return sum

# Driver program
    
# Both a[] and b[] must be of same size.
a = [4, 1, 8, 7]
b = [2, 3, 6, 5]
n = len(a)

print(findMinSum(a, b, n))

# This code is contributed by Anant Agarwal.

// C# program to find minimum sum of
// absolute differences of two arrays.
using System;

class MinSum {
    
    // Returns minimum possible pairwise 
    // absolute difference of two arrays.
    static long findMinSum(long []a, long []b,
                           long n)
    {
        
        // Sort both arrays
        Array.Sort(a);
        Array.Sort(b);
    
        // Find sum of absolute differences
        long sum = 0 ;
        for (int i = 0; i < n; i++)
            sum = sum + Math.Abs(a[i] - b[i]);
    
        return sum;
    }
    
    // Driver code
    public static void Main(String[] args) 
    {
        // Both a[] and b[] must be of same size.
        long []a = {4, 1, 8, 7};
        long []b = {2, 3, 6, 5};
        int n = a.Length;
        Console.Write(findMinSum(a, b, n));
    } 
}

// This code is contributed by parashar...

PHP

<?php
// PHP program to find minimum sum 
// of absolute differences of two 
// arrays.

// Returns minimum possible pairwise
// absolute difference of two arrays.
function findMinSum($a, $b, $n)
{
    
    // Sort both arrays
    sort($a); 
    sort($a, $n);
    sort($b); 
    sort($b, $n);

    // Find sum of absolute 
    // differences
    $sum= 0 ;
    for ($i = 0; $i < $n; $i++)
        $sum = $sum + abs($a[$i] - 
                          $b[$i]);

    return $sum;
}

    // Driver Code
    // Both a[] and b[] must
    // be of same size.
    $a = array(4, 1, 8, 7);
    $b = array(2, 3, 6, 5);
    $n = sizeof($a);
    echo(findMinSum($a, $b, $n));

// This code is contributed by nitin mittal.
?>

JavaScript

<script>

// JavaScript program for the above approach

// Function to return the minimum number

// Returns minimum possible pairwise 
// absolute difference of two arrays.
    function findMinSum(a, b, n)
    {
        // Sort both arrays
        a.sort();
        b.sort();
     
        // Find sum of absolute differences
        let sum = 0 ;
        for (let i = 0; i < n; i++)
            sum = sum + Math.abs(a[i] - b[i]);
     
        return sum;
    }
    
// Driver Code

    // Both a[] and b[] must be of same size.
    let a = [4, 1, 8, 7];
    let b = [2, 3, 6, 5];
    let n = a.length;
    document.write(findMinSum(a, b, n));

// This code is contributed by code_hunt.
</script>

Output

Time Complexity: O(n * logn)
Auxiliary Space: O(1)

Minimum increment/decrement to make array non-Increasing

SAHIL BAJAJ 1

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Article Tags :

Greedy
DSA

Practice Tags :

Greedy

Minimum sum of absolute difference of pairs of two arrays

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