# Minimum sum of product of two arrays

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 :a[] = {1, 2, -3} b[] = {-2, 3, -5} k = 5Output :-31Explanation: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} k = 3Output :25Explanation: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;` `}` `// This code is contributed by Sania Kumari Gupta` |

## C

`// C program to find minimum sum of product` `// of two arrays with k operations allowed on` `// first array.` `#include <stdio.h>` `#include<stdlib.h>` `// 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;` ` ` `printf` `(` `"%d "` `,minproduct(a, b, n, k));` ` ` `return` `0;` `}` `// This code is contributed by Sania Kumari Gupta` |

## 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 Sania Kumari Gupta` |

## Python3

`# Python program to find` `# minimum sum of product ` `# of two arrays with k` `# operations allowed on` `# first array.` `# Function to find the minimum product` `def` `minproduct(a,b,n,k):` ` ` `diff ` `=` `0` ` ` `res ` `=` `0` ` ` `for` `i ` `in` `range` `(n): ` ` ` `# Find product of current` ` ` `# elements and update result.` ` ` `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` `and` `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.` ` ` `elif` `(pro < ` `0` `and` `a[i] < ` `0` `):` ` ` `temp ` `=` `(a[i] ` `-` `2` `*` `k) ` `*` `b[i]` ` ` `# Similar to above two cases` ` ` `# for positive product.` ` ` `elif` `(pro > ` `0` `and` `a[i] < ` `0` `):` ` ` `temp ` `=` `(a[i] ` `+` `2` `*` `k) ` `*` `b[i]` ` ` `elif` `(pro > ` `0` `and` `a[i] > ` `0` `):` ` ` `temp ` `=` `(a[i] ` `-` `2` `*` `k) ` `*` `b[i]` ` ` `# Check if current difference` ` ` `# becomes higher` ` ` `# than the maximum difference so far.` ` ` `d ` `=` `abs` `(pro ` `-` `temp)` ` ` `if` `(d > diff):` ` ` `diff ` `=` `d ` ` ` `return` `res ` `-` `diff` `# Driver function` `a ` `=` `[ ` `2` `, ` `3` `, ` `4` `, ` `5` `, ` `4` `]` `b ` `=` `[ ` `3` `, ` `4` `, ` `2` `, ` `3` `, ` `2` `]` `n ` `=` `5` `k ` `=` `3` `print` `(minproduct(a, b, n, k))` `# This code is contributed` `# by Azkia Anam.` |

## C#

`// C# program to find minimum sum ` `// of product of two arrays with k ` `// operations allowed on first array.` `using` `System;` `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()` ` ` `{` ` ` `int` `[]a = { 2, 3, 4, 5, 4 };` ` ` `int` `[]b = { 3, 4, 2, 3, 2 };` ` ` `int` `n = 5, k = 3;` ` ` ` ` `Console.WriteLine(minproduct(a, b, n, k));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## PHP

`<?php` `// PHP program to find minimum sum of product ` `// of two arrays with k operations allowed on` `// first array.` `// Function to find the minimum product` `function` `minproduct( ` `$a` `, ` `$b` `, ` `$n` `, ` `$k` `)` `{` ` ` `$diff` `= 0; ` `$res` `= 0;` ` ` `$temp` `;` ` ` `for` `( ` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) {` ` ` `// Find product of current` ` ` `// elements and update` ` ` `// result.` ` ` `$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 ` `and` `$b` `[` `$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` `(` `$pro` `< 0 ` `and` `$a` `[` `$i` `] < 0)` ` ` `$temp` `= (` `$a` `[` `$i` `] - 2 * ` `$k` `) * ` `$b` `[` `$i` `];` ` ` `// Similar to above two ` ` ` `// cases for positive` ` ` `// product.` ` ` `else` `if` `(` `$pro` `> 0 ` `and` `$a` `[` `$i` `] < 0)` ` ` `$temp` `= (` `$a` `[` `$i` `] + 2 * ` `$k` `) * ` `$b` `[` `$i` `];` ` ` `else` `if` `(` `$pro` `> 0 ` `and` `$a` `[` `$i` `] > 0)` ` ` `$temp` `= (` `$a` `[` `$i` `] - 2 * ` `$k` `) * ` `$b` `[` `$i` `];` ` ` `// Check if current difference becomes higher` ` ` `// than the maximum difference so far.` ` ` `$d` `= ` `abs` `(` `$pro` `- ` `$temp` `);` ` ` `if` `(` `$d` `> ` `$diff` `)` ` ` `$diff` `= ` `$d` `; ` ` ` `}` ` ` `return` `$res` `- ` `$diff` `;` `}` ` ` `// Driver Code` ` ` `$a` `= ` `array` `(2, 3, 4, 5, 4 ,0);` ` ` `$b` `=` `array` `(3, 4, 2, 3, 2);` ` ` `$n` `= 5; ` ` ` `$k` `= 3;` ` ` `echo` `minproduct(` `$a` `, ` `$b` `, ` `$n` `, ` `$k` `);` ` ` `// This code is contributed by anuj_67.` `?>` |

## Javascript

`<script>` `// Javascript program to find minimum sum` `// of product of two arrays with k` `// operations allowed on first array.` `// Function to find the minimum product` `function` `minproduct(a, b, n, k)` `{` ` ` `let diff = 0, res = 0;` ` ` `let temp = 0;` ` ` `for` `(let i = 0; i < n; i++)` ` ` `{` ` ` ` ` `// Find product of current elements` ` ` `// and update result.` ` ` `let 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.` ` ` `let d = Math.abs(pro - temp);` ` ` `if` `(d > diff)` ` ` `diff = d; ` ` ` `}` ` ` ` ` `return` `res - diff;` `}` ` ` `// Driver code` ` ` `let a = [ 2, 3, 4, 5, 4 ];` ` ` `let b = [ 3, 4, 2, 3, 2 ];` ` ` `let n = 5, k = 3;` ` ` `document.write(minproduct(a, b, n, k));` ` ` ` ` `// This code is contributed by sanjoy_62.` `</script>` |

**Output :**

25

**Time Complexity:** O(n)**Auxiliary Space:** O(1)

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