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.

## C++

// 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; }

## Java

// Java program to find minimum sum // of product of two arrays with k // operations allowed on first array. import java.math.*; class GFG { // Function to find the minimum product static int minproduct(int a[], int b[], int n, int k) { int diff = 0, res = 0; int temp = 0; 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 = Math.abs(pro - temp); if (d > diff) diff = d; } return res - diff; } // Driver function public static void main(String[] args) { int a[] = { 2, 3, 4, 5, 4 }; int b[] = { 3, 4, 2, 3, 2 }; int n = 5, k = 3; System.out.println(minproduct(a, b, n, k)); } } // This code is contributed by Prerna Saini

Output :

25

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