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 the left corner, then we need to remove 3 elements from the 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).
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 the result, update the result.
Time Complexity: O(n * k)
Efficient Approach (Using Window Sliding Technique)
1) Find the total sum of the array.
2) Find the sum of first n-k elements and initialize this as a 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 the above problem statement.
C++
#include <bits/stdc++.h>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 functionint main(){ int arr[] = { 11, 49, 100, 20, 86, 29, 72 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 4; cout << "Maximum sum of remaining elements " << calculate(arr, n, k) << "\n"; return 0;} |
Java
// Java program for the // above approachimport java.util.*;class GFG{static 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 = Math.max(res, curr_sum); } // Now return result (sum of // remaining n-k elements) return res;}// Driver codepublic static void main(String[] args){ int[] arr = {11, 49, 100, 20, 86, 29, 72}; int n = arr.length; int k = 4; System.out.print("Maximum sum of remaining " + "elements " + calculate(arr, n, k) + "\n");}}// This code is contributed by Chitranayal |
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 functionif __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 |
C#
using System;using System.Collections;using System.Collections.Generic; class GFG{ static 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 = Math.Max(res, curr_sum); } // Now return result (sum of // remaining n-k elements) return res;}// Driver codepublic static void Main(string[] args){ int []arr = { 11, 49, 100, 20, 86, 29, 72 }; int n = arr.Length; int k = 4; Console.Write("Maximum sum of remaining " + "elements " + calculate(arr, n, k) + "\n");}}// This code is contributed by rutvik_56 |
Maximum sum of remaining elements 206
Time Complexity: O(k)
Auxiliary Space : O(1)
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