Maximize array sum after K negations | Set 1

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

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++

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// C++ program 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;
}

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Java

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// Java program 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.

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Python3

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# Python3 program 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.

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C#

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// C# program 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.

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PHP

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<?php
// PHP program 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. 
?>

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Output :

10

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



Approach 2 (Using Sort):
This approach is somewhat better than the above discussed method. In this method we will first sort the given array using java in-built sort function which has worst running time complexity of O(nlogn). Then for a given value of k we will continue to iterate through the array till k remains greater than 0, If value of array at any index is less than 0 we will change its sign and decrement k by 1. If we find a 0 in the array we will immediately set k equal to 0 to maximize our result. In some cases if we have all the values in array greater than 0 we will change sign of positive values, as our array is already sorted we will be changing signs of lower values present in the array which will eventually maximize our sum.

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// Java program to find maximum array sum 
// after at most k negations.
import java.util.Arrays;
  
public class GFG {
  
    static int sol(int arr[], int k)
    {
        // sorting given array using in-built java sort function
        Arrays.sort(arr);
        int sum = 0;
  
        int i = 0;
        while (k > 0) {
  
            // If we find a 0 in our
            // sorted array, we stop
            if (arr[i] == 0)
                k = 0;
  
            else {
                arr[i] = (-1) * arr[i];
                k = k - 1;
            }
  
            i++;
        }
  
        // calculating sum
        for (int j = 0; j < arr.length; j++) {
            sum += arr[j];
        }
        return sum;
    }
    public static void main(String[] args)
    {
        int arr[] = { -2, 0, 5, -1, 2 };
        System.out.println(sol(arr, 4));
    }
}

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Time Complexity : O(n*logn)
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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