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 image is an illustration of the above approach:
Below is the implementation of the above 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; } |
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 int 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 int result = 0; for (int i = 0; i < n; i++) result += (A[i] * B[n - i - 1]); return result; } // Driven Program 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 int 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 int result = 0; for (int i = 0; i < n; i++) result += (a[i] * b[n - i - 1]); return result; } // Driven Program 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).
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