# Minimum elements to be removed from the ends to make the array sorted

Last Updated : 13 Apr, 2023

Given an array arr[] of length N, the task is to remove the minimum number of elements from the ends of the array to make the array non-decreasing. Elements can only be removed from the left or the right end.

Examples:

Input: arr[] = {1, 2, 4, 1, 5}
Output:
We can’t make the array sorted after one removal.
But if we remove 2 elements from the right end, the
array becomes {1, 2, 4} which is sorted.
Input: arr[] = {3, 2, 1}
Output:

Approach: A very simple solution to this problem is to find the length of the longest non-decreasing subarray of the given array. Let’s say the length is L. So, the count of elements that need to be removed will be N – L. The length of the longest non-decreasing subarray can be easily found using the approach discussed in this article.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the minimum number` `// of elements to be removed from the ends` `// of the array to make it sorted` `int` `findMin(``int``* arr, ``int` `n)` `{`   `    ``// To store the final answer` `    ``int` `ans = 1;`   `    ``// Two pointer loop` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``int` `j = i + 1;`   `        ``// While the array is increasing increment j` `        ``while` `(j < n and arr[j] >= arr[j - 1])` `            ``j++;`   `        ``// Updating the ans` `        ``ans = max(ans, j - i);`   `        ``// Updating the left pointer` `        ``i = j - 1;` `    ``}`   `    ``// Returning the final answer` `    ``return` `n - ans;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 3, 2, 1 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);`   `    ``cout << findMin(arr, n);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{` `    `  `    ``// Function to return the minimum number ` `    ``// of elements to be removed from the ends ` `    ``// of the array to make it sorted ` `    ``static` `int` `findMin(``int` `arr[], ``int` `n) ` `    ``{ ` `    `  `        ``// To store the final answer ` `        ``int` `ans = ``1``; ` `    `  `        ``// Two pointer loop ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``int` `j = i + ``1``; ` `    `  `            ``// While the array is increasing increment j ` `            ``while` `(j < n && arr[j] >= arr[j - ``1``]) ` `                ``j++; ` `    `  `            ``// Updating the ans ` `            ``ans = Math.max(ans, j - i); ` `    `  `            ``// Updating the left pointer ` `            ``i = j - ``1``; ` `        ``} ` `    `  `        ``// Returning the final answer ` `        ``return` `n - ans; ` `    ``} ` `    `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)` `    ``{ ` `        ``int` `arr[] = { ``3``, ``2``, ``1` `}; ` `        ``int` `n = arr.length; ` `        ``System.out.println(findMin(arr, n)); ` `    ``} ` `}`   `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the approach`   `# Function to return the minimum number` `# of elements to be removed from the ends` `# of the array to make it sorted` `def` `findMin(arr, n):`   `    ``# To store the final answer` `    ``ans ``=` `1`   `    ``# Two pointer loop` `    ``for` `i ``in` `range``(n):` `        ``j ``=` `i ``+` `1`   `        ``# While the array is increasing increment j` `        ``while` `(j < n ``and` `arr[j] >``=` `arr[j ``-` `1``]):` `            ``j ``+``=` `1`   `        ``# Updating the ans` `        ``ans ``=` `max``(ans, j ``-` `i)`   `        ``# Updating the left pointer` `        ``i ``=` `j ``-` `1`   `    ``# Returning the final answer` `    ``return` `n ``-` `ans`   `# Driver code` `arr ``=` `[``3``, ``2``, ``1``]` `n ``=` `len``(arr)`   `print``(findMin(arr, n))`   `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach ` `using` `System;`   `class` `GFG ` `{` `    `  `// Function to return the minimum number ` `// of elements to be removed from the ends ` `// of the array to make it sorted ` `static` `int` `findMin(``int` `[]arr, ``int` `n) ` `{ `   `    ``// To store the readonly answer ` `    ``int` `ans = 1; `   `    ``// Two pointer loop ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``int` `j = i + 1; `   `        ``// While the array is increasing increment j ` `        ``while` `(j < n && arr[j] >= arr[j - 1]) ` `            ``j++; `   `        ``// Updating the ans ` `        ``ans = Math.Max(ans, j - i); `   `        ``// Updating the left pointer ` `        ``i = j - 1; ` `    ``} `   `    ``// Returning the readonly answer ` `    ``return` `n - ans; ` `} `   `// Driver code ` `public` `static` `void` `Main(String[] args)` `{ ` `    ``int` `[]arr = { 3, 2, 1 }; ` `    ``int` `n = arr.Length; ` `    ``Console.WriteLine(findMin(arr, n)); ` `} ` `}`   `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`2`

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

Method #2: Using Dynamic Programming

The idea is to use dynamic programming to keep track of the length of the increasing subsequence ending at each index.

Step-by-Step Algorithm:

1. Initialize ans to 1 and create a new array dp of length n, where dp[i] will store the length of the longest increasing subsequence ending at index i.
2. Set dp[0] to 1 since the longest increasing subsequence ending at index 0 has length 1.
3. Loop through the array from index 1 to n-1:
a) Set dp[i] to 1.
b) If arr[i] is greater than or equal to arr[i-1], add the length of the increasing subsequence ending at i-1 to dp[i]. This means that we’re extending the increasing subsequence ending at i-1 to include the current element arr[i].
c) Update ans to be the maximum value between ans and dp[i].
4. Delete the dp array to free up memory.
5. Return n – ans as the minimum number of elements that need to be removed to make the array sorted.

## C++

 `#include ` `using` `namespace` `std;`   `int` `findMin(``int``* arr, ``int` `n)` `{` `    ``int` `ans = 1;` `    ``int``* dp = ``new` `int``[n];` `    ``dp[0] = 1;`   `    ``for` `(``int` `i = 1; i < n; i++) {` `        ``dp[i] = 1;` `        ``if` `(arr[i] >= arr[i - 1])` `            ``dp[i] += dp[i - 1];` `        ``ans = max(ans, dp[i]);` `    ``}`   `    ``delete``[] dp;` `    ``return` `n - ans;` `}`   `int` `main()` `{` `    ``int` `arr[] = { 3, 2, 1 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);`   `    ``cout << findMin(arr, n);`   `    ``return` `0;` `}`

## Java

 `import` `java.util.*;`   `public` `class` `GFG {` `    ``// Function to find the minimum number of elements to` `    ``// remove from an array` `    ``public` `static` `int` `findMin(``int``[] arr, ``int` `n)` `    ``{` `        ``int` `ans = ``1``;` `        ``int``[] dp = ``new` `int``[n];` `        ``dp[``0``] = ``1``;`   `        ``// Calculate the length of the longest` `        ``// non-decreasing subsequence ending at each index` `        ``// Dynamic Programming approach to find the length` `        ``// of` `        ``// Longest Increasing Subsequence` `        ``for` `(``int` `i = ``1``; i < n; i++) {` `            ``dp[i] = ``1``;` `            ``if` `(arr[i] >= arr[i - ``1``])` `                ``dp[i] += dp[i - ``1``];` `            ``ans = Math.max(ans, dp[i]);` `        ``}` `        ``// Return the minimum number of elements to remove` `        ``return` `n - ans;` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int``[] arr = { ``3``, ``2``, ``1` `};` `        ``int` `n = arr.length;` `        ``// Call the FindMin function and print the result` `        ``System.out.println(findMin(arr, n));` `    ``}` `}`

## C#

 `using` `System;`   `class` `Program {` `    ``// Function to find the minimum number of elements to` `    ``// remove from an array` `    ``static` `int` `FindMin(``int``[] arr, ``int` `n)` `    ``{` `        ``int` `ans = 1;` `        ``int``[] dp = ``new` `int``[n];` `        ``dp[0] = 1;`   `        ``// Calculate the length of the longest` `        ``// non-decreasing subsequence ending at each index` `        ``for` `(``int` `i = 1; i < n; i++) {` `            ``dp[i] = 1;` `            ``if` `(arr[i] >= arr[i - 1])` `                ``dp[i] += dp[i - 1];` `            ``ans = Math.Max(ans, dp[i]);` `        ``}`   `        ``// Free the dynamically allocated memory for the dp` `        ``// array` `        ``Array.Clear(dp, 0, dp.Length);`   `        ``// Return the minimum number of elements to remove` `        ``return` `n - ans;` `    ``}`   `    ``static` `void` `Main(``string``[] args)` `    ``{` `        ``int``[] arr = { 3, 2, 1 };` `        ``int` `n = arr.Length;`   `        ``// Call the FindMin function and print the result` `        ``Console.WriteLine(FindMin(arr, n));` `    ``}` `}`

## Python3

 `def` `findMin(arr, n):` `    ``ans ``=` `1` `    ``dp ``=` `[``1``] ``*` `n`   `    ``for` `i ``in` `range``(``1``, n):` `        ``if` `arr[i] >``=` `arr[i``-``1``]:` `            ``dp[i] ``+``=` `dp[i``-``1``]` `        ``ans ``=` `max``(ans, dp[i])`   `    ``return` `n ``-` `ans`   `arr ``=` `[``3``, ``2``, ``1``]` `n ``=` `len``(arr)`   `print``(findMin(arr, n))`

## Javascript

 `function` `findMin(arr, n) {` `    ``let ans = 1;` `    ``let dp = ``new` `Array(n);` `    ``dp[0] = 1;`   `    ``for` `(let i = 1; i < n; i++) {` `        ``dp[i] = 1;` `        ``if` `(arr[i] >= arr[i - 1]) dp[i] += dp[i - 1];` `        ``ans = Math.max(ans, dp[i]);` `    ``}`   `    ``return` `n - ans;` `}`   `let arr = [3, 2, 1];` `let n = arr.length;`   `console.log(findMin(arr, n));`   `// This code is contributed by sarojmcy2e`

Output

`2`

Complexity Analysis :

Time Complexity: O(n) where n is the length

This is because the algorithm loops through the array once, so the time complexity is O(n).

Auxiliary Space: O(1)

This is because we uses an additional array of size n, so the space complexity is O(n). However, since we delete the dp array before returning, the actual space used by the algorithm is O(1) in terms of memory that is not reused.

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