Given an array arr[] of size N and an integer M, the task is to find the maximum sum of the array after changing the sign of any elements in the array for exactly M times. It is allowed to change the sign of the same element multiple times.
Examples:
Input: arr[ ] = {-3, 7, -1, -5, -3}, M = 4
Output: 19
Explanation:
4 operations on the array can be performed as,
Operation 1: Change the sign of arr[0] -> {3, 7, -1, -5, -3}
Operation 2: Change the sign of arr[2] -> {3, 7, 1, -5, -3}
Operation 3: Change the sign of arr[3] -> {3, 7, 1, 5, -3}
Operation 4: Change the sign of arr[4] -> {3, 7, 1, 5, 3}
The maximum sum of array obtained is 19.Input: arr[ ] = {-4, 2, 3, 1}, M = 3
Output: 10
Approach: To solve the problem, the main idea is to flip the smallest number of the array in each iteration. By doing so, the negative values will be changed to positive and the array sum will be maximized.
Follow the steps below to solve the problem:
- Initialize a min priority queue, say pq[], and push all the elements of the array arr[].
- Initialize a variable, say sum = 0, to store the maximum sum of the array.
- Iterate a while loop till M is greater than 0 and do the following:
- Pop from the priority queue and subtract it from the variable sum.
- Flip the sign of the popped element by multiplying it with -1 and add it to the sum.
- Push the new flipped element in the priority queue and subtract 1 from M.
- Finally, print the maximum sum stored in the variable sum.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum sum// with M flipsvoid findMaximumSumWithMflips( int arr[], int N, int M){ // Declare a priority queue // i.e. min heap priority_queue<int, vector<int>, greater<int> > pq; // Declare the sum as zero int sum = 0; // Push all elements of the // array in it for (int i = 0; i < N; i++) { pq.push(arr[i]); sum += arr[i]; } // Iterate for M times while (M--) { // Get the top element sum -= pq.top(); // Flip the sign of the // top element int temp = -1 * pq.top(); // Remove the top element pq.pop(); // Update the sum sum += temp; // Push the temp into // the queue pq.push(temp); } cout << sum;}// Driver programint main(){ int arr[] = { -3, 7, -1, -5, -3 }; // Size of the array int N = sizeof(arr) / sizeof(arr[0]); int M = 4; findMaximumSumWithMflips(arr, N, M); return 0;} |
Java
// Java implementation of the above approachimport java.util.*;import java.lang.*;import java.lang.Math;class GFG {// Function to find the maximum sum// with M flipsstatic void findMaximumSumWithMflips( int arr[], int N, int M){ // Declare a priority queue // i.e. min heap PriorityQueue<Integer> minHeap = new PriorityQueue<Integer>(); // Declare the sum as zero int sum = 0; // Push all elements of the // array in it for (int i = 0; i < N; i++) { minHeap.add(arr[i]); sum += arr[i]; } // Iterate for M times while (M-- >0) { // Get the top element sum -= minHeap.peek(); // Flip the sign of the // top element int temp = -1 * minHeap.peek(); // Remove the top element minHeap.remove(); // Update the sum sum += temp; // Push the temp into // the queue minHeap.add(temp); } System.out.println(sum);} // Driver Code public static void main(String[] args) { // Given input int arr[] = { -3, 7, -1, -5, -3 }; int M = 4,N=5; findMaximumSumWithMflips(arr, N, M); }}// This code is contributed by dwivediyash |
Python3
# Python 3 program for the above approach# Function to find the maximum sum# with M flipsdef findMaximumSumWithMflips(arr, N, M): # Declare a priority queue # i.e. min heap pq = [] # Declare the sum as zero sum = 0 # Push all elements of the # array in it for i in range(N): pq.append(arr[i]) sum += arr[i] pq.sort() # Iterate for M times while (M>0): # Get the top element sum -= pq[0] # Flip the sign of the # top element temp = -1 * pq[0] # Remove the top element pq = pq[1:] # Update the sum sum += temp # Push the temp into # the queue pq.append(temp) pq.sort() M -= 1 print(sum)# Driver programif __name__ == '__main__': arr = [-3, 7, -1, -5, -3] # Size of the array N = len(arr) M = 4 findMaximumSumWithMflips(arr, N, M) # This code is contributed by SURENDRA_GANGWAR. |
C#
// C# implementation of the above approachusing System;using System.Collections.Generic;public class GFG { // Function to find the maximum sum // with M flips static void findMaximumSumWithMflips(int[] arr, int N, int M) { // Declare a priority queue // i.e. min heap List<int> minHeap = new List<int>(); // Declare the sum as zero int sum = 0; // Push all elements of the // array in it for (int i = 0; i < N; i++) { minHeap.Add(arr[i]); sum += arr[i]; } minHeap.Sort(); // Iterate for M times while (M-- > 0) { // Get the top element sum -= minHeap[0]; // minHeap.RemoveAt(0); // Flip the sign of the // top element int temp = -1 * minHeap[0]; // Remove the top element minHeap.RemoveAt(0); // Update the sum sum += temp; // Push the temp into // the queue minHeap.Add(temp); minHeap.Sort(); } Console.WriteLine(sum); } // Driver Code public static void Main(String[] args) { // Given input int[] arr = { -3, 7, -1, -5, -3 }; int M = 4, N = 5; findMaximumSumWithMflips(arr, N, M); }}// This code is contributed by gauravrajput1 |
Javascript
<script>// Javascript program for the above approach// Function to find the maximum sum// with M flipsfunction findMaximumSumWithMflips(arr, N, M){ // Declare a priority queue // i.e. min heap let pq = [] // Declare the sum as zero let sum = 0 // Push all elements of the // array in it for(let i = 0; i < N; i++){ pq.push(arr[i]) sum += arr[i] pq.sort((a, b) => a - b) } // Iterate for M times while (M > 0){ // Get the top element sum -= pq[0] // Flip the sign of the // top element temp = -1 * pq[0] // Remove the top element pq.shift() // Update the sum sum += temp // Push the temp into // the queue pq.push(temp) pq.sort((a, b) => a - b) M -= 1 } document.write(sum)}// Driver program let arr = [-3, 7, -1, -5, -3] // Size of the array let N = arr.length let M = 4 findMaximumSumWithMflips(arr, N, M) // This code is contributed by gfgking. </script> |
Output:
19
Time Complexity: O(NLogN)
Auxiliary Space: O(N)
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