# Remove k corner elements to maximize remaining sum

Given an array, the task is to remove total k elements from corners to maximize the sum of remaining elements. For example, if we k = 5 and if we remove 2 elements from left corner, then we need to remove 3 elements from right corner.

Examples:

Input : arr = [11, 49, 100, 20, 86, 29, 72], k = 4
Output : 206
Explanation :: We remove 29 and 72 from right corner. We also remove 11 and 49 from left corner to get the maximum sum as 206 for remaining elements.

Input : arr[] = [1, 2, 3, 4, 5, 6, 1], k = 3
Output : 12
Explanation :: We remove two elements from left corner (1 and 2) and one element from right corner (1).

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach :
1) Initialize result as negative infinity.
2) Compute total sum.
3) Run a loop for x = 1 to k
…..Remove ‘x’ elements from left side and k – i elements from right side.
…..If the remaining elements have sum more than result, update the result.

Time Complexity : O(n * k)

Efficient Approach (Using Window Sliding Technique)

1) Find the total sum of array.
2) Find sum of first n-k elements and initialize this as current sum and also initialize this as result.
3) Run a loop for i = n-k to n-1
….curr_sum = curr_sum – arr[i – n + k] + arr[i]
….res = max(res, curr_sum)

In step 3, we mainly run sliding window. We remove an element from left side and add an element from right side.

Below is the c++ implementation of above problem statement.

## CPP

 `#include ` `using` `namespace` `std; ` `int` `calculate(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// calculate the total sum of all elements ` `    ``// present in the array.. ` `    ``int` `total_sum = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``total_sum += arr[i]; ` ` `  `    ``// now calculate the sum of all elements ` `    ``// excluding the last k elements.. ` `    ``int` `curr_sum = 0; ` `    ``for` `(``int` `i = 0; i < n - k; i++) ` `        ``curr_sum += arr[i]; ` ` `  `    ``// now here its time to use sliding window ` `    ``// concept, remove the first element from ` `    ``// the current window and add the new element ` `    ``// in it in order to get the sum of all n-k size ` `    ``// of elements in arr. ` `    ``// Calculate the minimum sum of elements of ` `    ``// size n-k and stored it into the result ` `    ``int` `res = curr_sum; ` `    ``for` `(``int` `i = n - k; i < n; i++) { ` `        ``curr_sum = curr_sum - arr[i - n + k] + arr[i]; ` `        ``res = max(res, curr_sum); ` `    ``} ` ` `  `    ``// Now return result (sum of remaining n-k elements) ` `    ``return` `res; ` `} ` ` `  `// main function ` `int` `main() ` `{ ` `    ``int` `arr[] = { 11, 49, 100, 20, 86, 29, 72 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `k = 4; ` `    ``cout << ``"Maximum sum of remaining elements "` `         ``<< calculate(arr, n, k) << ``"\n"``; ` `    ``return` `0; ` `} `

## Python3

 `def` `calculate(arr, n, k): ` `     `  `    ``# calculate the total sum of all elements ` `    ``# present in the array.. ` `    ``total_sum ``=` `0` `    ``for` `i ``in` `arr: ` `        ``total_sum ``+``=` `i ` ` `  `    ``# now calculate the sum of all elements ` `    ``# excluding the last k elements.. ` `    ``curr_sum ``=` `0` `    ``for` `i ``in` `range``(n ``-` `k): ` `        ``curr_sum ``+``=` `arr[i] ` ` `  `    ``# now here its time to use sliding window ` `    ``# concept, remove the first element from ` `    ``# the current window and add the new element ` `    ``# in it in order to get the sum of all n-k size ` `    ``# of elements in arr. ` `    ``# Calculate the minimum sum of elements of ` `    ``# size n-k and stored it into the result ` `    ``res ``=` `curr_sum ` `    ``for` `i ``in` `range``(n ``-` `k, n): ` `        ``curr_sum ``=` `curr_sum ``-` `arr[i ``-` `n ``+` `k] ``+` `arr[i] ` `        ``res ``=` `max``(res, curr_sum) ` ` `  `    ``# Now return result (sum of remaining n-k elements) ` `    ``return` `res ` ` `  `# main function ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr``=``[``11``, ``49``, ``100``, ``20``, ``86``, ``29``, ``72``] ` `    ``n ``=` `len``(arr) ` `    ``k ``=` `4` `    ``print``(``"Maximum sum of remaining elements "``,calculate(arr, n, k)) ` ` `  `# This code is contributed by mohit kumar 29     `

Output:

```Maximum sum of remaining elements 206
```

Time Complexity: O(1)
Auxiliary Space : O(1)

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