Maximize array sum after K negations

Maximize array sum after K negations | Set 1

Difficulty Level : Easy
Last Updated : 21 Jan, 2021

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++

// 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 code
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 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 code
    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 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 code
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 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 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)

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.

C++

// C++ program to find maximum array sum
// after at most k negations.
#include <bits/stdc++.h>
 
using namespace std;
 
int sol(int arr[], int n, int k)
{
    int sum = 0;
    int i = 0;
   
    // Sorting given array using in-built
    // sort function
    sort(arr, arr + n);
    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 < n; j++) 
    {
        sum += arr[j];
    }
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { -2, 0, 5, -1, 2 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
     
    cout << sol(arr, n, 4) << endl;
 
    return 0;
}
 
// This code is contributed by pratham76

Java

// 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;
    }
   
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { -2, 0, 5, -1, 2 };
        System.out.println(sol(arr, 4));
    }
}

Python3

# Python3 program to find maximum array
# sum after at most k negations
 
 
def sol(arr, k):
 
    # Sorting given array using
    # in-built java sort function
    arr.sort()
 
    Sum = 0
    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 += 1
 
    # Calculating sum
    for j in range(len(arr)):
        Sum += arr[j]
 
    return Sum
 
 
# Driver code
arr = [-2, 0, 5, -1, 2]
 
print(sol(arr, 4))
 
# This code is contributed by avanitrachhadiya2155

C#

// C# program to find maximum array sum 
// after at most k negations.
using System;
  
class GFG{
  
static int sol(int []arr, int k)
{
     
    // Sorting given array using 
    // in-built java sort function
    Array.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;
}
 
// Driver code
public static void Main(string[] args)
{
    int []arr = { -2, 0, 5, -1, 2 };
     
    Console.Write(sol(arr, 4));
}
}
 
// This code is contributed by rutvik_56

Output :

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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