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Remove minimum elements from ends of array so that sum decreases by at least K | O(N)
• Difficulty Level : Medium
• Last Updated : 13 May, 2021

Given an array arr[] consisting of N elements, the task is to remove minimum number of elements from the ends of the array such that the total sum of the array decreases by at least K. Note that K will always be less than or equal to the sum of all the elements of the array.
Examples:

Input: arr[] = {1, 11, 5, 5}, K = 11
Output:
After removing two elements form the left
end, the sum decreases by 1 + 11 = 12.
Input: arr[] = {1, 2, 3}, K = 6
Output:

Approach: A dynamic programming based approach has already been discussed in another post. In this article, an approach using the two-pointer technique will be discussed. It can be observed that the task is to find the longest sub-array with the sum of its elements at most sum(arr) – K where sum(arr) is the sum of all the elements of the array arr[]
Let the length of such subarray be L. Thus, the minimum number of elements to be removed from the array will be equal to N – L. To find the length of longest such subarray, refer this article.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the count of minimum``// elements to be removed from the ends``// of the array such that the sum of the``// array decrease by at least K``int` `minCount(``int``* arr, ``int` `n, ``int` `k)``{` `    ``// To store the final answer``    ``int` `ans = 0;` `    ``// Maximum possible sum required``    ``int` `sum = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``sum += arr[i];``    ``sum -= k;` `    ``// Left point``    ``int` `l = 0;` `    ``// Right pointer``    ``int` `r = 0;` `    ``// Total current sum``    ``int` `tot = 0;` `    ``// Two pointer loop``    ``while` `(l < n) {` `        ``// If the sum fits``        ``if` `(tot <= sum) {` `            ``// Update the answer``            ``ans = max(ans, r - l);``            ``if` `(r == n)``                ``break``;` `            ``// Update the total sum``            ``tot += arr[r++];``        ``}` `        ``else` `{` `            ``// Increment the left pointer``            ``tot -= arr[l++];``        ``}``    ``}` `    ``return` `(n - ans);``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 11, 5, 5 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);``    ``int` `k = 11;` `    ``cout << minCount(arr, n, k);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``    ` `    ``// Function to return the count of minimum``    ``// elements to be removed from the ends``    ``// of the array such that the sum of the``    ``// array decrease by at least K``    ``static` `int` `minCount(``int` `arr[],``                        ``int` `n, ``int` `k)``    ``{``    ` `        ``// To store the final answer``        ``int` `ans = ``0``;``    ` `        ``// Maximum possible sum required``        ``int` `sum = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``sum += arr[i];``        ``sum -= k;``    ` `        ``// Left point``        ``int` `l = ``0``;``    ` `        ``// Right pointer``        ``int` `r = ``0``;``    ` `        ``// Total current sum``        ``int` `tot = ``0``;``    ` `        ``// Two pointer loop``        ``while` `(l < n)``        ``{``    ` `            ``// If the sum fits``            ``if` `(tot <= sum)``            ``{``    ` `                ``// Update the answer``                ``ans = Math.max(ans, r - l);``                ``if` `(r == n)``                    ``break``;``    ` `                ``// Update the total sum``                ``tot += arr[r++];``            ``}``            ``else``            ``{``    ` `                ``// Increment the left pointer``                ``tot -= arr[l++];``            ``}``        ``}``        ``return` `(n - ans);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `arr[] = { ``1``, ``11``, ``5``, ``5` `};``        ``int` `n = arr.length;``        ``int` `k = ``11``;``    ` `        ``System.out.println(minCount(arr, n, k));``    ``}``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count of minimum``# elements to be removed from the ends``# of the array such that the sum of the``# array decrease by at least K``def` `minCount(arr, n, k) :` `    ``# To store the final answer``    ``ans ``=` `0``;` `    ``# Maximum possible sum required``    ``sum` `=` `0``;``    ``for` `i ``in` `range``(n) :``        ``sum` `+``=` `arr[i];``    ``sum` `-``=` `k;` `    ``# Left point``    ``l ``=` `0``;` `    ``# Right pointer``    ``r ``=` `0``;` `    ``# Total current sum``    ``tot ``=` `0``;` `    ``# Two pointer loop``    ``while` `(l < n) :` `        ``# If the sum fits``        ``if` `(tot <``=` `sum``) :` `            ``# Update the answer``            ``ans ``=` `max``(ans, r ``-` `l);``            ``if` `(r ``=``=` `n) :``                ``break``;` `            ``# Update the total sum``            ``tot ``+``=` `arr[r];``            ``r ``+``=` `1``    ` `        ``else` `:` `            ``# Increment the left pointer``            ``tot ``-``=` `arr[l];``            ``l ``+``=` `1``    ` `    ``return` `(n ``-` `ans);` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``1``, ``11``, ``5``, ``5` `];``    ``n ``=` `len``(arr);``    ``k ``=` `11``;` `    ``print``(minCount(arr, n, k));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `    ``// Function to return the count of minimum``    ``// elements to be removed from the ends``    ``// of the array such that the sum of the``    ``// array decrease by at least K``    ``static` `int` `minCount(``int` `[]arr,``                        ``int` `n, ``int` `k)``    ``{``    ` `        ``// To store the final answer``        ``int` `ans = 0;``    ` `        ``// Maximum possible sum required``        ``int` `sum = 0;``        ``for` `(``int` `i = 0; i < n; i++)``            ``sum += arr[i];``        ``sum -= k;``    ` `        ``// Left point``        ``int` `l = 0;``    ` `        ``// Right pointer``        ``int` `r = 0;``    ` `        ``// Total current sum``        ``int` `tot = 0;``    ` `        ``// Two pointer loop``        ``while` `(l < n)``        ``{``    ` `            ``// If the sum fits``            ``if` `(tot <= sum)``            ``{``    ` `                ``// Update the answer``                ``ans = Math.Max(ans, r - l);``                ``if` `(r == n)``                    ``break``;``    ` `                ``// Update the total sum``                ``tot += arr[r++];``            ``}``            ``else``            ``{``    ` `                ``// Increment the left pointer``                ``tot -= arr[l++];``            ``}``        ``}``        ``return` `(n - ans);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = { 1, 11, 5, 5 };``        ``int` `n = arr.Length;``        ``int` `k = 11;``    ` `        ``Console.WriteLine(minCount(arr, n, k));``    ``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``
Output:
`2`

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