# Remove minimum elements from ends of array so that sum decreases by at least K | O(N)

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: 2
After removing two elements form the left
end, the sum decreases by 1 + 11 = 12.

Input: arr[] = {1, 2, 3}, K = 6
Output: 3

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

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 `

Output:

```2
```

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