# Sort Permutation of 1 to N by removing any element and inserting it to front or back

• Last Updated : 03 May, 2022

Given an array arr[] of size N having distinct integers from 1 to N, the task is to count the minimum number of steps required to sort the array in increasing order by removing any element and inserting it to the front or back of the array.

Examples:

Input: arr[ ] = {4, 1, 3, 2}
Output: 2
Explanation:
The given array can be sorted in increasing order by following two operations:
Operation 1: Remove 3 and add it to the back of the array. {4, 1, 3, 2} -> {4, 1, 2, 3}
Operation 2: Remove 4 and add it to the back of the array. {4, 1, 2, 3} -> {1, 2, 3, 4}

Input: arr[ ] = {4, 1, 2, 5, 3}
Output: 2

Approach: The problem can be solved by using the fact that to make the array sorted in a minimum number of steps, the minimum number of elements must be moved to the front or back. Also, the elements whose position must be changed will not lie in the increasing order initially. So the problem reduces to finding the longest increasing subarray as only those elements of the array will not be moved.

Below is the implementation of the above approach:

## C++

 `// CPP program for the above approach``#include``using` `namespace` `std;` `// Function to find the minimum steps``// to sort the array``void` `findMinStepstoSort(``int` `arr[], ``int` `N){` `  ``// Storing the positions of elements``  ``map<``int``,``int``> pos;``  ``for``(``int` `i = 0; i < N; i++)``    ``pos[arr[i]] = i;` `  ``// Initialize answer``  ``int` `ans = N-1;``  ``int` `prev = -1;``  ``int` `count = 0;` `  ``// Traversing the array``  ``for``(``int` `i = 1; i < N + 1; i++){` `    ``// If current is greater than``    ``// previous``    ``if``(pos[i] > prev)``      ``count += 1;` `    ``// else if current is less than``    ``// previous``    ``else``      ``count = 1;` `    ``// Updating previous``    ``prev = pos[i];` `    ``// Updating ans``    ``ans = min(ans, N - count);``   ``}``  ``cout<

## Java

 `// JAVA program for the above approach``import` `java.util.*;` `class` `GFG{` `// Function to find the minimum steps``// to sort the array``static` `void` `findMinStepstoSort(``int` `arr[], ``int` `N){` `  ``// Storing the positions of elements``  ``HashMap pos = ``new` `HashMap<>();``  ``for``(``int` `i = ``0``; i < N; i++)``    ``pos.put(arr[i], i);` `  ``// Initialize answer``  ``int` `ans = N - ``1``;``  ``int` `prev = -``1``;``  ``int` `count = ``0``;` `  ``// Traversing the array``  ``for``(``int` `i = ``1``; i < N + ``1``; i++){` `    ``// If current is greater than``    ``// previous``    ``if``(pos.get(i) > prev)``      ``count += ``1``;` `    ``// else if current is less than``    ``// previous``    ``else``      ``count = ``1``;` `    ``// Updating previous``    ``prev = pos.get(i);` `    ``// Updating ans``    ``ans = Math.min(ans, N - count);``   ``}``  ``System.out.print(ans);``}` `// Driver Code``public` `static` `void` `main(String[] args){` `  ``int` `N = ``5``;``  ``int` `arr[] = {``4``, ``1``, ``2``, ``5``, ``3``};` `  ``findMinStepstoSort(arr, N);``}``}` `// This code is contributed by gauravrajput1`

## Python3

 `# Python program for the above approach` `# Function to find the minimum steps``# to sort the array``def` `findMinStepstoSort(arr, N):` `  ``# Storing the positions of elements``  ``pos ``=` `{arr[i]:i ``for` `i ``in` `range``(N)}`` ` `  ``# Initialize answer``  ``ans ``=` `N``-``1``  ``prev ``=` `-``1``;count ``=` `0` `  ``# Traversing the array``  ``for` `i ``in` `range``(``1``, N ``+` `1``):` `    ``# If current is greater than``    ``# previous``    ``if` `pos[i] > prev:``      ``count ``+``=` `1` `    ``# else if current is less than``    ``# previous``    ``else``:``      ``count ``=` `1` `    ``# Updating previous``    ``prev ``=` `pos[i]` `    ``# Updating ans``    ``ans ``=` `min``(ans, N ``-` `count)``  ``print``(ans)`  `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``  ``N ``=` `5``  ``arr ``=` `[``4``, ``1``, ``2``, ``5``, ``3``]` `  ``findMinStepstoSort(arr, N)`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG``{` `// Function to find the minimum steps``// to sort the array``static` `void` `findMinStepstoSort(``int``[] arr, ``int` `N){` `  ``// Storing the positions of elements``  ``Dictionary<``int``, ``int``> pos = ``new``                ``Dictionary<``int``, ``int``>();``  ``for``(``int` `i = 0; i < N; i++)``    ``pos[arr[i]] = i;` `  ``// Initialize answer``  ``int` `ans = N - 1;``  ``int` `prev = -1;``  ``int` `count = 0;` `  ``// Traversing the array``  ``for``(``int` `i = 1; i < N + 1; i++){` `    ``// If current is greater than``    ``// previous``    ``if``(pos[i] > prev)``      ``count += 1;` `    ``// else if current is less than``    ``// previous``    ``else``      ``count = 1;` `    ``// Updating previous``    ``prev = pos[i];` `    ``// Updating ans``    ``ans = Math.Min(ans, N - count);``   ``}``  ``Console.WriteLine(ans);``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main (``string``[] args)``    ``{``        ``int` `N = 5;``        ``int``[] arr = {4, 1, 2, 5, 3};` `        ``findMinStepstoSort(arr, N);``    ``}``}` `// This code is contributed by sanjoy_62.`

## Javascript

 ``

Output

`2`

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

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