Maximum array sum that can be obtained after exactly k changes

• Difficulty Level : Easy
• Last Updated : 09 Jun, 2022

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:

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 â‰¥ 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 ``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

 ` 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

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Output:

`13`

Time Complexity: O(n * log n)

Auxiliary Space: O(1)

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