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,
- If count ? k then the sign of exactly k negative numbers will be changed starting from the smallest.
- 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.
- 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 elementsint 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 timesint 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 codeint 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 approachimport java.util.Arrays;class GFG {// Utility function to return the sum// of the array elementsstatic 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 timesstatic 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 codepublic 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 elementsdef 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 timesdef 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 codeif __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 approachusing 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 elementsfunction sumArr($arr, $n){ $sum = 0; for ($i = 0; $i < $n; $i++) $sum += $arr[$i]; return $sum;}// Function to return the maximized sum// of the array after performing// the given operation exactly k timesfunction maxSum($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
Time Complexity: O(n * log n)
Auxiliary Space: O(1)
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