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 = 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}
        k = 3
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;
}

// 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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