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)


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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