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.

Below is C++ implementation of this approach:

## C/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 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 int result = 0; for (int i = 0; i < n; i++) result += (A[i] * B[n - i - 1]); return result; } // Driven Program 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; }

## 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. sorted(A) sorted(B) # 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

Output:

23

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

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