Minimum sum of product of two arrays

3.6

Find the minimum sum of Products of two arrays of the same size, given that k modifications are allowed on the first array. In each modification, one array element of the first array can either be increased or decreased by 2.

Examples:

Input : ar[] = {1, 2, -3}
        b[]  = {-2, 3, -5}
           k = 5
Output : -31
Explanation:
Here n = 3 and k = 5. 
So, we modified a[2], which is -3 and 
increased it by 10 (as 5 modifications 
are allowed).
Final sum will be :
(1 * -2) + (2 * 3) + (7 * -5)
   -2    +    6    -    35
             -31
(which is the minimum sum of the array 
with given conditions)

Input : a[] = {2, 3, 4, 5, 4}
        b[] = {3, 4, 2, 3, 2}
Output : 25
Explanation: 
Here, total numbers are 5 and total 
modifications allowed are 3. So, modify 
a[1], which is 3 and decreased it by 6 
(as 3 modifications are allowed).
Final sum will be :
(2 * 3) + (-3 * 4) + (4 * 2) + (5 * 3) + (4 * 2)
   6    –    12    +    8    +    15   +    8
                        25
(which is the minimum sum of the array with 
given conditions)

Since we need to minimize the product sum, we find the maximum product and reduce it. By taking some examples, we observe that making 2*k changes to only one element is enough to get the minimum sum. Based on this observation, we consider every element as the element on which we apply all k operations and keep track of the element that reduces result to minimum.

// CPP program to find minimum sum of product 
// of two arrays with k operations allowed on
// first array.
#include <bits/stdc++.h>
using namespace std;

// Function to find the minimum product
int minproduct(int a[], int b[], int n, int k)
{
    int diff = 0, res = 0;
    int temp;
    for (int i = 0; i < n; i++) {

        // Find product of current elements and update
        // result.
        int pro = a[i] * b[i];
        res = res + pro;

        // If both product and b[i] are negative,
        // we must increase value of a[i] to minimize
        // result.
        if (pro < 0 && b[i] < 0)
            temp = (a[i] + 2 * k) * b[i];

        // If both product and a[i] are negative,
        // we must decrease value of a[i] to minimize
        // result.
        else if (pro < 0 && a[i] < 0)
            temp = (a[i] - 2 * k) * b[i];

        // Similar to above two cases for positive
        // product.
        else if (pro > 0 && a[i] < 0)
            temp = (a[i] + 2 * k) * b[i];
        else if (pro > 0 && a[i] > 0)
            temp = (a[i] - 2 * k) * b[i];

        // Check if current difference becomes higher
        // than the maximum difference so far.
        int d = abs(pro - temp);
        if (d > diff)
            diff = d;        
    }

    return res - diff;
}

// Driver function
int main()
{
    int a[] = { 2, 3, 4, 5, 4 };
    int b[] = { 3, 4, 2, 3, 2 };
    int n = 5, k = 3;
    cout << minproduct(a, b, n, k) 
         << endl;
    return 0;
}

Output :

25

This article is contributed by Abhishek Sharma. 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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