Minimum sum of absolute difference of pairs of two arrays
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 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( int a[], 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; } |
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#
// 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. ?> |
Output :
6
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