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++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the maximum sum // with M flips void 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 program int 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 implementation of the above approach import java.util.*;
import java.lang.*;
import java.lang.Math;
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
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 |
# Python 3 program for the above approach # Function to find the maximum sum # with M flips def 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 program if __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# implementation of the above approach using 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 |
<script> // Javascript program for the above approach // Function to find the maximum sum // with M flips function 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> |
19
Time Complexity: O(NLogN)
Auxiliary Space: O(N)