Maximize the sum of sum of the Array by removing end elements

Given an array arr of size N, the task is to maximize the sum of sum, of the remaining elements in the array, by removing the end elements.

Example:

Input: arr[] = {2, 3}
Output: 3
Explanation:
At first we will delete 2, then sum of remaining elements = 3. Then delete 3. Therefore sum of sum = 3 + 0 = 3.
We can also delete 3 first and then 2, but in this case, the sum of sum = 2 + 0 = 2
But since 3 is larger, therefore the output is 3.

Input: arr[] = {3, 1, 7, 2, 1}
Ouput: 39
Explanation:
At first we will delete 1 from last, then the sum of remaining elements
will be 13.
Then delete 2 from last, then the sum of remaining elements will be 11.
Then delete 3 from the beginning, then the sum of remaining elements will be 8.
Then we delete 1, the remaining sum is 7 and then delete 7.
Therefore the Sum of all remaining sums is 13 + 11 + 8 + 7 = 39, which is the maximum case.

Approach: The idea is to use Greedy Algorithm to solve this problem.



  1. First to calculate the total sum of the array.
  2. Then compare the elements on both ends and subtract the minimum value among the two, from the sum. This will make the remaining sum maximum.
  3. Then, add remaining sum to the result.
  4. Repeat the above steps till all the elements have been removed from the array. Then print the resultant sum.

Below is the implementation of the above approach:

C++

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// C++ program to maximize the sum
// of sum of the Array by
// removing end elements
  
#include <iostream>
using namespace std;
  
// Function to find
// the maximum sum of sum
int maxRemainingSum(int arr[], int n)
{
    int sum = 0;
  
    // compute the sum of whole array
    for (int i = 0; i < n; i++)
        sum += arr[i];
  
    int i = 0;
    int j = n - 1;
  
    int result = 0;
  
    // Traverse and remove the
    // minimum value from an end
    // to maximum the sum value
    while (i < j) {
  
        // If the left end element
        // is smaller than right end
        if (arr[i] < arr[j]) {
  
            // remove the left end element
            sum -= arr[i];
  
            i++;
        }
  
        // If the right end element
        // is smaller than left end
        else {
  
            // remove the right end element
            sum -= arr[j];
            j--;
        }
  
        // Add the remaining element
        // sum in the result
        result += sum;
    }
  
    // Return the maximum
    // sum of sum
    return result;
}
  
// Driver code
int main()
{
    int arr[] = { 3, 1, 7, 2, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
  
    cout << maxRemainingSum(arr, N);
  
    return 0;
}

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Output:

39

Time complexity: O(N)

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