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Maximum array sum that can be obtained after exactly k changes

  • Difficulty Level : Easy
  • Last Updated : 11 May, 2021

Given an array arr[] of n integers and an integer k. The task is to maximize the sum of the array after performing the given operation exactly k times. In a single operation, any element of the array can be multiplied by -1 i.e. the sign of the element can be changed.

Examples: 

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Input: arr[] = {-5, 4, 1, 3, 2}, k = 4 
Output: 13 
Change the sign of -5 once and then change the sign of 1 three times. 
Thus, it changes to -1 and the sum will be 5 + 4 + (-1) + 3 + 2 = 13.
Input: arr[] = {-1, -1, 1}, k = 1 
Output: 1  



Approach: If the count of negative elements in the array is count,  

  1. If count ≥ k then the sign of exactly k negative numbers will be changed starting from the smallest.
  2. If count < k then change the sign of all the negative elements to positive and for the remaining operations i.e. rem = k – count
    • If rem % 2 = 0 then no changes will be done to the current array as changing the sign of an element twice gives the original number.
    • If rem % 2 = 1 then change the sign of the smallest element from the updated array.
  3. Finally, print the sum of the elements of the updated array.

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Utility function to return the sum
// of the array elements
int sumArr(int arr[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
 
    return sum;
}
 
// Function to return the maximized sum
// of the array after performing
// the given operation exactly k times
int maxSum(int arr[], int n, int k)
{
    // Sort the array elements
    sort(arr, arr + n);
 
    int i = 0;
    // Change signs of the negative elements
    // starting from the smallest
    while (i < n && k > 0 && arr[i] < 0) {
        arr[i] *= -1;
        k--;
        i++;
    }
 
    // If a single operation has to be
    // performed then it must be performed
    // on the smallest positive element
    if (k % 2 == 1) {
 
        // To store the index of the
        // minimum element
        int min = 0;
        for (i = 1; i < n; i++)
 
            // Update the minimum index
            if (arr[min] > arr[i])
                min = i;
 
        // Perform remaining operation
        // on the smallest element
        arr[min] *= -1;
    }
 
    // Return the sum of the updated array
    return sumArr(arr, n);
}
 
// Driver code
int main()
{
    int arr[] = { -5, 4, 1, 3, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 4;
 
    cout << maxSum(arr, n, k) << endl;
 
    return 0;
}

Java




// Java implementation of the above approach
import java.util.Arrays;
 
class GFG
{
 
// Utility function to return the sum
// of the array elements
static int sumArr(int arr[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
 
    return sum;
}
 
// Function to return the maximized sum
// of the array after performing
// the given operation exactly k times
static int maxSum(int arr[], int n, int k)
{
    // Sort the array elements
    Arrays.sort(arr);
 
    int i = 0;
     
    // Change signs of the negative elements
    // starting from the smallest
    while (i < n && k > 0 && arr[i] < 0)
    {
        arr[i] *= -1;
        k--;
        i++;
    }
 
    // If a single operation has to be
    // performed then it must be performed
    // on the smallest positive element
    if (k % 2 == 1)
    {
 
        // To store the index of the
        // minimum element
        int min = 0;
        for (i = 1; i < n; i++)
 
            // Update the minimum index
            if (arr[min] > arr[i])
                min = i;
 
        // Perform remaining operation
        // on the smallest element
        arr[min] *= -1;
    }
 
    // Return the sum of the updated array
    return sumArr(arr, n);
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { -5, 4, 1, 3, 2 };
    int n = arr.length;
    int k = 4;
 
    System.out.println(maxSum(arr, n, k));
}
}
 
/* This code contributed by PrinciRaj1992 */

Python3




# Python 3 implementation of the approach
 
# Utility function to return the sum
# of the array elements
def sumArr(arr, n):
    sum = 0
    for i in range(n):
        sum += arr[i]
 
    return sum
 
# Function to return the maximized sum
# of the array after performing
# the given operation exactly k times
def maxSum(arr, n, k):
     
    # Sort the array elements
    arr.sort(reverse = False)
 
    i = 0
     
    # Change signs of the negative elements
    # starting from the smallest
    while (i < n and k > 0 and arr[i] < 0):
        arr[i] *= -1
        k -= 1
        i += 1
 
    # If a single operation has to be
    # performed then it must be performed
    # on the smallest positive element
    if (k % 2 == 1):
         
        # To store the index of the
        # minimum element
        min = 0
        for i in range(1, n):
             
            # Update the minimum index
            if (arr[min] > arr[i]):
                min = i
 
        # Perform remaining operation
        # on the smallest element
        arr[min] *= -1
 
    # Return the sum of the updated array
    return sumArr(arr, n)
 
# Driver code
if __name__ == '__main__':
    arr = [-5, 4, 1, 3, 2]
    n = len(arr)
    k = 4
 
    print(maxSum(arr, n, k))
 
# This code is contributed by
# Surendra_Gangwar

C#




// C# implementation of the above approach
using System;
using System.Linq;
 
class GFG
{
 
// Utility function to return the sum
// of the array elements
static int sumArr(int [] arr, int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
 
    return sum;
}
 
// Function to return the maximized sum
// of the array after performing
// the given operation exactly k times
static int maxSum(int [] arr, int n, int k)
{
    // Sort the array elements
    Array.Sort(arr);
 
    int i = 0;
     
    // Change signs of the negative elements
    // starting from the smallest
    while (i < n && k > 0 && arr[i] < 0)
    {
        arr[i] *= -1;
        k--;
        i++;
    }
 
    // If a single operation has to be
    // performed then it must be performed
    // on the smallest positive element
    if (k % 2 == 1)
    {
 
        // To store the index of the
        // minimum element
        int min = 0;
        for (i = 1; i < n; i++)
 
            // Update the minimum index
            if (arr[min] > arr[i])
                min = i;
 
        // Perform remaining operation
        // on the smallest element
        arr[min] *= -1;
    }
 
    // Return the sum of the updated array
    return sumArr(arr, n);
}
 
// Driver code
static void Main()
{
    int []arr= { -5, 4, 1, 3, 2 };
    int n = arr.Length;
    int k = 4;
 
    Console.WriteLine(maxSum(arr, n, k));
}
}
 
// This code is contributed by mohit kumar 29

PHP




<?php
// PHP implementation of the approach
 
// Utility function to return the sum
// of the array elements
function sumArr($arr, $n)
{
    $sum = 0;
    for ($i = 0; $i < $n; $i++)
        $sum += $arr[$i];
 
    return $sum;
}
 
// Function to return the maximized sum
// of the array after performing
// the given operation exactly k times
function maxSum($arr, $n, $k)
{
    // Sort the array elements
    sort($arr);
 
    $i = 0;
    // Change signs of the negative elements
    // starting from the smallest
    while ($i < $n && $k > 0 &&
                $arr[$i] < 0)
    {
        $arr[$i] *= -1;
        $k--;
        $i++;
    }
 
    // If a single operation has to be
    // performed then it must be performed
    // on the smallest positive element
    if ($k % 2 == 1)
    {
 
        // To store the index of the
        // minimum element
        $min = 0;
        for ($i = 1; $i < $n; $i++)
 
            // Update the minimum index
            if ($arr[$min] > $arr[$i])
                $min = $i;
 
        // Perform remaining operation
        // on the smallest element
        $arr[$min] *= -1;
    }
 
    // Return the sum of the updated array
    return sumArr($arr, $n);
}
 
// Driver code
$arr = array( -5, 4, 1, 3, 2 );
$n = sizeof($arr);
$k = 4;
 
echo maxSum($arr, $n, $k), "\n";
 
// This code is contributed by ajit.
?>

Javascript




<script>
    // Javascript implementation of the above approach
     
    // Utility function to return the sum
    // of the array elements
    function sumArr(arr, n)
    {
        let sum = 0;
        for (let i = 0; i < n; i++)
            sum += arr[i];
 
        return sum;
    }
 
    // Function to return the maximized sum
    // of the array after performing
    // the given operation exactly k times
    function maxSum(arr, n, k)
    {
        // Sort the array elements
        arr.sort(function(a, b){return a - b});
 
        let i = 0;
 
        // Change signs of the negative elements
        // starting from the smallest
        while (i < n && k > 0 && arr[i] < 0)
        {
            arr[i] *= -1;
            k--;
            i++;
        }
 
        // If a single operation has to be
        // performed then it must be performed
        // on the smallest positive element
        if (k % 2 == 1)
        {
 
            // To store the index of the
            // minimum element
            let min = 0;
            for (i = 1; i < n; i++)
 
                // Update the minimum index
                if (arr[min] > arr[i])
                    min = i;
 
            // Perform remaining operation
            // on the smallest element
            arr[min] *= -1;
        }
 
        // Return the sum of the updated array
        return sumArr(arr, n);
    }
     
    let arr= [ -5, 4, 1, 3, 2 ];
    let n = arr.length;
    let k = 4;
   
    document.write(maxSum(arr, n, k));
 
</script>
Output: 
13

 




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