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

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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