Minimize the sum of product of two arrays with permutations allowed

06 Jul, 2022
Given two arrays, A and B, of equal size n, the task is to find the minimum value of A[0] * B[0] + A[1] * B[1] +...+ A[n-1] * B[n-1]. Shuffling of elements of arrays A and B is allowed.
Examples : 
Input : A[] = {3, 1, 1} and B[] = {6, 5, 4}.
Output : 23
Minimum value of S = 1*6 + 1*5 + 3*4 = 23.

Input : A[] = { 6, 1, 9, 5, 4 } and B[] = { 3, 4, 8, 2, 4 }
Output : 80.
Minimum value of S = 1*8 + 4*4 + 5*4 + 6*3 + 9*2 = 80.
The idea is to multiply minimum element of one array to maximum element of another array. Algorithm to solve this problem:  
Sort both the arrays A and B.Traverse the array and for each element, multiply A[i] and B[n – i – 1] and add to the total.
Sort both the arrays A and B.
Traverse the array and for each element, multiply A[i] and B[n – i – 1] and add to the total.
Note: We are adding multiplication of elements which can lead to overflow conditions.
Below image is an illustration of the above approach: 
Below is the implementation of the above approach: 
C++
Java
Python3
C#
PHP
Javascript
// C++ program to calculate minimum sum of product// of two arrays.#include <bits/stdc++.h>using namespace std;   // Returns minimum sum of product of two arrays// with permutations allowedlong long int minValue(int A[], int B[], int n){    // Sort A and B so that minimum and maximum    // value can easily be fetched.    sort(A, A + n);    sort(B, B + n);       // Multiplying minimum value of A and maximum    // value of B    long long int result = 0;    for (int i = 0; i < n; i++)        result += (A[i] * B[n - i - 1]);       return result;}   // Driven Codeint main(){    int A[] = { 3, 1, 1 };    int B[] = { 6, 5, 4 };    int n = sizeof(A) / sizeof(A[0]);    cout << minValue(A, B, n) << endl;    return 0;}
// Java program to calculate minimum// sum of product of two arrays.import java.io.*;import java.util.*; class GFG {     // Returns minimum sum of product of two arrays    // with permutations allowed    static long minValue(int A[], int B[], int n)    {        // Sort A and B so that minimum and maximum        // value can easily be fetched.        Arrays.sort(A);        Arrays.sort(B);         // Multiplying minimum value of A        // and maximum value of B        long result = 0;        for (int i = 0; i < n; i++)            result += (A[i] * B[n - i - 1]);         return result;    }     // Driven Code    public static void main(String[] args)    {        int A[] = { 3, 1, 1 };        int B[] = { 6, 5, 4 };        int n = A.length;        ;        System.out.println(minValue(A, B, n));    }} // This code is contributed by vt_m
# Python program to calculate minimum sum of product# of two arrays. # Returns minimum sum of product of two arrays# with permutations allowed  def minValue(A, B, n):     # Sort A and B so that minimum and maximum    # value can easily be fetched.    A.sort()    B.sort()     # Multiplying minimum value of A and maximum    # value of B    result = 0    for i in range(n):        result += (A[i] * B[n - i - 1])     return result  # Driven ProgramA = [3, 1, 1]B = [6, 5, 4]n = len(A)print (minValue(A, B, n)) # Contributed by: Afzal Ansari
// C# program to calculate minimum// sum of product of two arrays.using System; class GFG {     // Returns minimum sum of product    // of two arrays with permutations    // allowed    static long minValue(int[] a, int[] b,                                   int n)    {                 // Sort A and B so that minimum        // and maximum value can easily        // be fetched.        Array.Sort(a);        Array.Sort(b);         // Multiplying minimum value of        // A and maximum value of B        long result = 0;                 for (int i = 0; i < n; i++)            result += (a[i] * b[n - i - 1]);         return result;    }     // Driven Code    public static void Main()    {                 int[] a = { 3, 1, 1 };        int[] b = { 6, 5, 4 };        int n = a.Length;                 Console.Write(minValue(a, b, n));    }} // This code is contributed by nitin mittal.
<?php// PHP program to calculate minimum// sum of product of two arrays. // Returns minimum sum of// product of two arrays// with permutations allowedfunction minValue($A, $B, $n){    // Sort A and B so that minimum    // and maximum value can easily    // be fetched.    sort($A); sort($A , $n);    sort($B); sort($B , $n);     // Multiplying minimum value of    // A and maximum value of B    $result = 0;    for ($i = 0; $i < $n; $i++)        $result += ($A[$i] *                    $B[$n - $i - 1]);     return $result;} // Driver Code$A = array( 3, 1, 1 );$B = array( 6, 5, 4 );$n = sizeof($A) / sizeof($A[0]);echo minValue($A, $B, $n) ; // This code is contributed by nitin mittal.?>
<script> // JavaScript program to calculate minimum// sum of product of two arrays. // Returns minimum sum of product of two arrays// with permutations allowedfunction minValue(A, B, n){         // Sort A and B so that minimum and maximum    // value can easily be fetched.    A.sort(function(a,b){        return a - b;    });    B.sort(function(a,b){        return a - b;    });     // Multiplying minimum value of A    // and maximum value of B    let result = 0;    for(let i = 0; i < n; i++)        result += (A[i] * B[n - i - 1]);     return result;} // Driver Codelet A = [ 3, 1, 1 ];let B = [ 6, 5, 4 ];let n = A.length; document.write(minValue(A, B, n));     // This code is contributed by souravghosh0416                          </script>
23

Time Complexity : O(n log n).
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avsadityavardhan
sachuverma
souravghosh0416
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