Maximum sum of all elements of array after performing given operations
Given an array of integers. The task is to find the maximum sum of all the elements of the array after performing the given two operations once each.
The operations are:
1. Select some(possibly none) continuous elements from the beginning of the array and multiply by -1.
2. Select some(possibly none) continuous elements from the end of the array and multiply by -1.
Examples:
Input : arr[] = {-1, 10, -5, 10, -2}
Output : 18
After 1st operation : 1 10 -5 10 -2
After 2nd operation : 1 10 -5 10 2
Input : arr[] = {-9, -8, -7}
Output : 24
After 1st operation : 9 8 -7
After 2nd operation : 9 8 7
Approach: This problem can be solved in linear time, using the following idea:
- Let the sum of elements A1 .. An be equal to S. Then when inverting signs we get -A1, -A2 .. -An, and the sum is thereafter changed to -S, i.e. sum of elements on the segment will just change its’ sign when inverting the whole segment’s signs.
- Consider the initial problem as follows: choose a consecutive subsequence, and invert all the numbers remaining out of it.
- Find the Maximum subarray sum using Kadane’ Algorithm.
- Keep that subarray intact and multiply the rest with -1.
- Considering the sum of the whole array as S, and the largest sum contiguous subarray as S1, the total sum will be equal to -(S-S1) + S1 = 2*S1 – S. This is the required sum.
Below is the implementation of the above approach:
C++
// CPP program to find the maximum // sum after given operations #include <bits/stdc++.h> using namespace std; // Function to calculate Maximum Subarray Sum // or Kadane's Algorithm int maxSubArraySum(int a[], int size) { int max_so_far = INT_MIN, max_ending_here = 0; for (int i = 0; i < size; i++) { max_ending_here = max_ending_here + a[i]; if (max_so_far < max_ending_here) max_so_far = max_ending_here; if (max_ending_here < 0) max_ending_here = 0; } return max_so_far; } // Function to find the maximum // sum after given operations int maxSum(int a[], int n) { // To store sum of all elements int S = 0; // Maximum sum of a subarray int S1 = maxSubArraySum(a, n); // Calculate the sum of all elements for (int i = 0; i < n; i++) S += a[i]; return (2 * S1 - S); } // Driver Code int main() { int a[] = { -35, 32, -24, 0, 27, -10, 0, -19 }; // size of an array int n = sizeof(a) / sizeof(a[0]); cout << maxSum(a, n); return 0; } |
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Java
// Java program to find the maximum // sum after given operations import java.io.*; class GFG { // Function to calculate Maximum Subarray Sum // or Kadane's Algorithm static int maxSubArraySum(int a[], int size) { int max_so_far = Integer.MIN_VALUE, max_ending_here = 0; for (int i = 0; i < size; i++) { max_ending_here = max_ending_here + a[i]; if (max_so_far < max_ending_here) max_so_far = max_ending_here; if (max_ending_here < 0) max_ending_here = 0; } return max_so_far; } // Function to find the maximum // sum after given operations static int maxSum(int a[], int n) { // To store sum of all elements int S = 0; // Maximum sum of a subarray int S1 = maxSubArraySum(a, n); // Calculate the sum of all elements for (int i = 0; i < n; i++) S += a[i]; return (2 * S1 - S); } // Driver Code public static void main (String[] args) { int a[] = { -35, 32, -24, 0, 27, -10, 0, -19 }; // size of an array int n = a.length; System.out.println( maxSum(a, n)); } } // This code is contributed by inder_verma |
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Python3
# Python3 program to find the maximum # sum after given operations import sys # Function to calculate Maximum # Subarray Sum or Kadane's Algorithm def maxSubArraySum(a, size) : max_so_far = -(sys.maxsize - 1) max_ending_here = 0 for i in range(size) : max_ending_here = max_ending_here + a[i] if (max_so_far < max_ending_here) : max_so_far = max_ending_here if (max_ending_here < 0) : max_ending_here = 0 return max_so_far # Function to find the maximum # sum after given operations def maxSum(a, n) : # To store sum of all elements S = 0; # Maximum sum of a subarray S1 = maxSubArraySum(a, n) # Calculate the sum of all elements for i in range(n) : S += a[i] return (2 * S1 - S) # Driver Code if __name__ == "__main__" : a = [ -35, 32, -24, 0, 27, -10, 0, -19 ] # size of an array n = len(a) print(maxSum(a, n)) # This code is contributed by Ryuga |
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C#
// C# program to find the maximum // sum after given operations using System; class GFG { // Function to calculate Maximum Subarray Sum // or Kadane's Algorithm static int maxSubArraySum(int []a, int size) { int max_so_far = int.MinValue, max_ending_here = 0; for (int i = 0; i < size; i++) { max_ending_here = max_ending_here + a[i]; if (max_so_far < max_ending_here) max_so_far = max_ending_here; if (max_ending_here < 0) max_ending_here = 0; } return max_so_far; } // Function to find the maximum // sum after given operations static int maxSum(int []a, int n) { // To store sum of all elements int S = 0; // Maximum sum of a subarray int S1 = maxSubArraySum(a, n); // Calculate the sum of all elements for (int i = 0; i < n; i++) S += a[i]; return (2 * S1 - S); } // Driver Code public static void Main () { int []a = { -35, 32, -24, 0, 27, -10, 0, -19 }; // size of an array int n = a.Length; Console.WriteLine( maxSum(a, n)); } } // This code is contributed by inder_verma |
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PHP
<?php // PHP program to find the maximum // sum after given operations // Function to calculate Maximum Subarray // Sum or Kadane's Algorithm function maxSubArraySum($a, $size) { $max_so_far = PHP_INT_MIN; $max_ending_here = 0; for ($i = 0; $i < $size; $i++) { $max_ending_here = $max_ending_here + $a[$i]; if ($max_so_far < $max_ending_here) $max_so_far = $max_ending_here; if ($max_ending_here < 0) $max_ending_here = 0; } return $max_so_far; } // Function to find the maximum // sum after given operations function maxSum($a, $n) { // To store sum of all elements $S = 0; // Maximum sum of a subarray $S1 = maxSubArraySum($a, $n); // Calculate the sum of all elements for ($i = 0; $i < $n; $i++) $S += $a[$i]; return (2 * $S1 - $S); } // Driver Code $a = array(-35, 32, -24, 0, 27, -10, 0, -19); // size of an array $n = sizeof($a); echo( maxSum($a, $n)); // This code is contributed // by Mukul Singh |
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Output:
99
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