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}
Output: 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 the 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 maximize the sum value while (i < j) { // If the left end element // is smaller than the right end if (arr[i] < arr[j]) { // remove the left end element sum -= arr[i]; i++; } // If the right end element // is smaller than the 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
import java.util.Arrays; public class MaxRemainingSum { // Function to find the maximum sum of the array // by removing end elements static int maxRemainingSum( int [] arr, int n) { int sum = Arrays.stream(arr).sum(); int i = 0 ; int j = n - 1 ; int result = 0 ; // Traverse and remove the // minimum value from an end // to maximize the sum value while (i < j) { // If the left end element // is smaller than the right end if (arr[i] < arr[j]) { // remove the left end element sum -= arr[i]; i++; } 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)); } } |
Python3
def max_remaining_sum(arr, n): total_sum = sum (arr) i = 0 j = n - 1 result = 0 # Traverse and remove the # minimum value from an end # to maximize the sum value while i < j: # If the left end element # is smaller than the right end if arr[i] < arr[j]: # remove the left end element total_sum - = arr[i] i + = 1 else : # remove the right end element total_sum - = arr[j] j - = 1 # Add the remaining element # sum in the result result + = total_sum # Return the maximum sum of sum return result # Driver code arr = [ 3 , 1 , 7 , 2 , 1 ] N = len (arr) print (max_remaining_sum(arr, N)) |
C#
using System; class Program { static int MaxRemainingSum( int [] arr, int n) { int sum = 0; // compute the sum of the whole array for ( int i = 0; i < n; i++) sum += arr[i]; int left = 0; int right = n - 1; int result = 0; // Traverse and remove the // minimum value from an end // to maximize the sum value while (left < right) { // If the left end element // is smaller than the right end if (arr[left] < arr[right]) { // remove the left end element sum -= arr[left]; left++; } else { // remove the right end element sum -= arr[right]; right--; } // Add the remaining element // sum in the result result += sum; } // Return the maximum sum of sum return result; } static void Main() { int [] arr = {3, 1, 7, 2, 1 }; int N = arr.Length; Console.WriteLine(MaxRemainingSum(arr, N)); } } |
Javascript
function maxRemainingSum(arr) { let sum = 0; // Compute the sum of the whole array for (let i = 0; i < arr.length; i++) sum += arr[i]; let left = 0; let right = arr.length - 1; let result = 0; // Traverse and remove the minimum value from an end // to maximize the sum value while (left < right) { // If the left end element is smaller than the right end if (arr[left] < arr[right]) { // Remove the left end element sum -= arr[left]; left++; } else { // Remove the right end element sum -= arr[right]; right--; } // Add the remaining element sum in the result result += sum; } // Return the maximum sum of sum return result; } // Test the function const arr = [3, 1, 7, 2, 1 ]; console.log(maxRemainingSum(arr)); |
39
Time complexity: O(N), where N is the size of the given array.
Auxiliary space: O(1), as constant space is used.
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