Maximize array sum after K negations

Given an array of size n and a number k. We must modify array K number of times. Here modify array means in each operation we can replace any array element arr[i] by -arr[i]. We need to perform this operation in such a way that after K operations, sum of array must be maximum?

Examples :

Input : arr[] = {-2, 0, 5, -1, 2} 
        K = 4
Output: 10
// Replace (-2) by -(-2), array becomes {2, 0, 5, -1, 2}
// Replace (-1) by -(-1), array becomes {2, 0, 5, 1, 2}
// Replace (0) by -(0), array becomes {2, 0, 5, 1, 2}
// Replace (0) by -(0), array becomes {2, 0, 5, 1, 2}

Input : arr[] = {9, 8, 8, 5} 
        K = 3
Output: 20

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

This problem has very simple solution, we just have to replace the minimum element arr[i] in array by -arr[i] for current operation. In this way we can make sum of array maximum after K operations. Once interesting case is, once minimum element becomes 0, we don’t need to make any more changes.

C++

// C++ program to to maximize array sum after
// k operations.
#include<bits/stdc++.h>
using namespace std;

// This function does k operations on array
// in a way that maximize the array sum.
// index --> stores the index of current minimum
//           element for j'th operation
int maximumSum(int arr[], int n, int k)
{
    // Modify array K number of times
    for (int i=1; i<=k; i++)
    {
        int min = INT_MAX;
        int index = -1;

        // Find minimum element in array for
        // current operation and modify it
        // i.e; arr[j] --> -arr[j]
        for (int j=0; j<n; j++)
        {
            if (arr[j] < min)
            {
                min = arr[j];
                index = j;
            }
        }

        // this the condition if we find 0 as
        // minimum element, so it will useless to
        // replace 0 by -(0) for remaining operations
        if (min == 0)
            break;

        // Modify element of array
        arr[index] = -arr[index];
    }

    // Calculate sum of array
    int sum = 0;
    for (int i=0; i<n; i++)
        sum += arr[i];
    return sum;
}

// Driver program to test the case
int main()
{
    int arr[] = {-2, 0, 5, -1, 2};
    int k = 4;
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << maximumSum(arr, n, k);
    return 0;
}

Java

// Java program to to maximize array 
// sum after k operations.

class GFG
{
    // This function does k operations 
    // on array in a way that maximize 
    // the array sum. index --> stores 
    // the index of current minimum
    // element for j'th operation
    static int maximumSum(int arr[], int n, int k)
    {
        // Modify array K number of times
        for (int i = 1; i <= k; i++)
        {
            int min = +2147483647;
            int index = -1;
    
            // Find minimum element in array for
            // current operation and modify it
            // i.e; arr[j] --> -arr[j]
            for (int j=0; j<n; j++)
            {
                if (arr[j] < min)
                {
                    min = arr[j];
                    index = j;
                }
            }
    
            // this the condition if we find 0 as
            // minimum element, so it will useless to
            // replace 0 by -(0) for remaining operations
            if (min == 0)
                break;
    
            // Modify element of array
            arr[index] = -arr[index];
        }
    
        // Calculate sum of array
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += arr[i];
        return sum;
    }
    
    
    // Driver program
    public static void main(String arg[])
    {
    int arr[] = {-2, 0, 5, -1, 2};
        int k = 4;
        int n = arr.length;
        System.out.print(maximumSum(arr, n, k));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python3 program to to maximize 
# array sum after k operations.

# This function does k operations on array
# in a way that maximize the array sum.
# index --> stores the index of current 
# minimum element for j'th operation
def maximumSum(arr, n, k):

    # Modify array K number of times
    for i in range(1, k + 1):
    
        min = +2147483647
        index = -1

        # Find minimum element in array for
        # current operation and modify it
        # i.e; arr[j] --> -arr[j]
        for j in range(n):
        
            if (arr[j] < min):
            
                min = arr[j]
                index = j

        # this the condition if we find 0 as
        # minimum element, so it will useless to
        # replace 0 by -(0) for remaining operations
        if (min == 0):
            break

        # Modify element of array
        arr[index] = -arr[index]
    

    # Calculate sum of array
    sum = 0
    for i in range(n):
        sum += arr[i]
    return sum

# Driver program
arr = [-2, 0, 5, -1, 2]
k = 4
n = len(arr)
print(maximumSum(arr, n, k))

# This code is contributed by Anant Agarwal.

C#

// C# program to to maximize array 
// sum after k operations.
using System;

class GFG
{
    
    // This function does k operations 
    // on array in a way that maximize 
    // the array sum. index --> stores 
    // the index of current minimum
    // element for j'th operation
    static int maximumSum(int []arr, int n, 
                          int k)
    {
        
        // Modify array K number of times
        for (int i = 1; i <= k; i++)
        {
            int min = +2147483647;
            int index = -1;
    
            // Find minimum element in array for
            // current operation and modify it
            // i.e; arr[j] --> -arr[j]
            for (int j = 0; j < n; j++)
            {
                if (arr[j] < min)
                {
                    min = arr[j];
                    index = j;
                }
            }
    
            // this the condition if we find
            // 0 as minimum element, so it
            // will useless to replace 0 by -(0)
            // for remaining operations
            if (min == 0)
                break;
    
            // Modify element of array
            arr[index] = -arr[index];
        }
    
        // Calculate sum of array
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += arr[i];
        return sum;
    }
    
    // Driver code
    public static void Main()
    {
        int []arr = {-2, 0, 5, -1, 2};
        int k = 4;
        int n = arr.Length;
        Console.Write(maximumSum(arr, n, k));
    }
}

// This code is contributed by Nitin Mittal.

PHP

<?php
// PHP program to to maximize 
// array sum after k operations.

// This function does k operations 
// on array in a way that maximize 
// the array sum. index --> stores
// the index of current minimum
// element for j'th operation
function maximumSum($arr, $n, $k)
{
    $INT_MAX = 0;
    // Modify array K
    // number of times
    for ($i = 1; $i <= $k; $i++)
    {
        $min = $INT_MAX;
        $index = -1;

        // Find minimum element in 
        // array for current operation
        // and modify it i.e; 
        // arr[j] --> -arr[j]
        for ($j = 0; $j < $n; $j++)
        {
            if ($arr[$j] < $min)
            {
                $min = $arr[$j];
                $index = $j;
            }
        }

        // this the condition if we
        // find 0 as minimum element,so 
        // it will useless to replace 0 
        // by -(0) for remaining operations
        if ($min == 0)
            break;

        // Modify element of array
        $arr[$index] = -$arr[$index];
    }

    // Calculate sum of array
    $sum = 0;
    for ($i = 0; $i < $n; $i++)
        $sum += $arr[$i];
    return $sum;
}

// Driver Code
$arr = array(-2, 0, 5, -1, 2);
$k = 4;
$n = sizeof($arr) / sizeof($arr[0]);
echo maximumSum($arr, $n, $k);
    
// This code is contributed
// by nitin mittal. 
?>

Output :

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

Maximize array sum after K negations | Set 2

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Practice Tags :

Arrays

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