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.

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++

`// 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 allowed ` `long` `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 Code ` `int` `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

`// 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

`# 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 Program` `A ` `=` `[` `3` `, ` `1` `, ` `1` `]` `B ` `=` `[` `6` `, ` `5` `, ` `4` `]` `n ` `=` `len` `(A)` `print` `minValue(A, B, n)` `# Contributed by: Afzal Ansari` |

## C#

`// 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` `// PHP program to calculate minimum ` `// sum of product of two arrays.` `// Returns minimum sum of ` `// product of two arrays ` `// with permutations allowed` `function` `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. ` `?>` |

**Output :**

23

**Time Complexity : **O(n log n).

This article is contributed by **Anuj Chauhan(anuj0503)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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