# Minimum sum of product of two arrays

• Difficulty Level : Medium
• Last Updated : 13 Jun, 2022

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 = 5
Output : -31
Explanation:
Here n = 3 and k = 5.
So, we modified a, 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 = 3
Output : 25
Explanation:
Here, total numbers are 5 and total
modifications allowed are 3. So, modify
a, 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 ``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 ``#include` `// 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

 ` 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

 ``

Output :

`25`

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

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