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.
- First to calculate the total sum of the array.
- 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.
- Then, add remaining sum to the result.
- 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++
// 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; } |
Java
// Java program to maximize the sum // of sum of the Array by removing // end elements import java.util.*; class GFG{ // Function to find // the maximum sum of sum static 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 public static void main(String args[]) { int arr[] = { 3, 1, 7, 2, 1 }; int N = arr.length; System.out.println(maxRemainingSum(arr, N)); } } // This code is contributed by ankitkumar34 |
Python3
# Python3 program to maximize the # sum of sum of the Array by # removing end elements # Function to find the maximum # sum of sum def maxRemainingSum(arr, n): sum = 0 # Compute the sum of whole array for i in range(n): sum += arr[i] i = 0 j = n - 1 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 += 1 # If the right end element # is smaller than left end else: # Remove the right end element sum -= arr[j] j -= 1 # Add the remaining element # sum in the result result += sum; # Return the maximum # sum of sum return result # Driver code arr = [ 3, 1, 7, 2, 1 ] N = len(arr) print(maxRemainingSum(arr, N)) # This code is contributed by ankitkumar34 |
C#
// C# program to maximize the sum // of sum of the Array by removing // end elements using System; class GFG{ // Function to find // the maximum sum of sum static int maxRemainingSum(int[] arr, int n) { int i, sum = 0; // Compute the sum of whole array for(i = 0; i < n; i++) sum += arr[i]; 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 public static void Main() { int[] arr = { 3, 1, 7, 2, 1 }; int N = arr.Length; Console.Write(maxRemainingSum(arr, N)); } } // This code is contributed by chitranayal |
39
Time complexity: O(N)
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