# Check if reversing a sub array make the array sorted

• Difficulty Level : Medium
• Last Updated : 22 Jun, 2022

Given an array of n distinct integers. The task is to check whether reversing any one sub-array can make the array sorted or not. If the array is already sorted or can be made sorted by reversing any one subarray, print “Yes“, else print “No“.
Examples:

```Input : arr [] = {1, 2, 5, 4, 3}
Output : Yes
By reversing the subarray {5, 4, 3},
the array will be sorted.

Input : arr [] = { 1, 2, 4, 5, 3 }
Output : No```

Method 1: Brute force (O(n3))
Consider every subarray and check if reversing the subarray makes the whole array sorted. If yes, return True. If reversing any of the subarrays doesn’t make the array sorted, then return False. Considering every subarray will take O(n2), and for each subarray, checking whether the whole array will get sorted after reversing the subarray in consideration will take O(n). Thus overall complexity would be O(n3).

Method 2: Sorting ( O(n*log(n) ))
The idea is to compare the given array with its sorted version. Make a copy of the given array and sort it. Now, find the first index and last index in the given array which does not match with the sorted array. If no such indices are found (given array was already sorted), return True. Else check if the elements between the found indices are in decreasing order, if Yes then return True else return False

if Below is the implementation of the above approach:

## C++

 `// C++ program to check whether reversing a``// sub array make the array sorted or not``#include``using` `namespace` `std;` `// Return true, if reversing the subarray will``// sort the array, else return false.``bool` `checkReverse(``int` `arr[], ``int` `n)``{``    ``// Copying the array.``    ``int` `temp[n];``    ``for` `(``int` `i = 0; i < n; i++)``        ``temp[i] = arr[i];` `    ``// Sort the copied array.``    ``sort(temp, temp + n);` `    ``// Finding the first mismatch.``    ``int` `front;``    ``for` `(front = 0; front < n; front++)``        ``if` `(temp[front] != arr[front])``            ``break``;` `    ``// Finding the last mismatch.``    ``int` `back;``    ``for` `(back = n - 1; back >= 0; back--)``        ``if` `(temp[back] != arr[back])``            ``break``;` `    ``// If whole array is sorted``    ``if` `(front >= back)``        ``return` `true``;` `    ``// Checking subarray is decreasing or not.``    ``do``    ``{``        ``front++;``        ``if` `(arr[front - 1] < arr[front])``            ``return` `false``;``    ``} ``while` `(front != back);` `    ``return` `true``;``}` `// Driven Program``int` `main()``{``    ``int` `arr[] = { 1, 2, 5, 4, 3 };``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);` `    ``checkReverse(arr, n)? (cout << ``"Yes"` `<< endl):``                          ``(cout << ``"No"` `<< endl);``    ``return` `0;``}`

## Java

 `// Java program to check whether reversing a``// sub array make the array sorted or not` `import` `java.util.Arrays;` `class` `GFG {` `// Return true, if reversing the subarray will``// sort the array, else return false.``    ``static` `boolean` `checkReverse(``int` `arr[], ``int` `n) {``        ``// Copying the array.``        ``int` `temp[] = ``new` `int``[n];``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``temp[i] = arr[i];``        ``}` `        ``// Sort the copied array.``        ``Arrays.sort(temp);` `        ``// Finding the first mismatch.``        ``int` `front;``        ``for` `(front = ``0``; front < n; front++) {``            ``if` `(temp[front] != arr[front]) {``                ``break``;``            ``}``        ``}` `        ``// Finding the last mismatch.``        ``int` `back;``        ``for` `(back = n - ``1``; back >= ``0``; back--) {``            ``if` `(temp[back] != arr[back]) {``                ``break``;``            ``}``        ``}` `        ``// If whole array is sorted``        ``if` `(front >= back) {``            ``return` `true``;``        ``}` `        ``// Checking subarray is decreasing or not.``        ``do` `{``            ``front++;``            ``if` `(arr[front - ``1``] < arr[front]) {``                ``return` `false``;``            ``}``        ``} ``while` `(front != back);` `        ``return` `true``;``    ``}` `// Driven Program``    ``public` `static` `void` `main(String[] args) {` `        ``int` `arr[] = {``1``, ``2``, ``5``, ``4``, ``3``};``        ``int` `n = arr.length;` `        ``if` `(checkReverse(arr, n)) {``            ``System.out.print(``"Yes"``);``        ``} ``else` `{``            ``System.out.print(``"No"``);``        ``}``    ``}` `}``//This code contributed by 29AjayKumar`

## Python3

 `# Python3 program to check whether``# reversing a sub array make the``# array sorted or not` `# Return true, if reversing the``# subarray will sort the array,``# else return false.``def` `checkReverse(arr, n):` `    ``# Copying the array``    ``temp ``=` `[``0``] ``*` `n``    ``for` `i ``in` `range``(n):``        ``temp[i] ``=` `arr[i]` `    ``# Sort the copied array.``    ``temp.sort()` `    ``# Finding the first mismatch.``    ``for` `front ``in` `range``(n):``        ``if` `temp[front] !``=` `arr[front]:``            ``break` `    ``# Finding the last mismatch.``    ``for` `back ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``):``        ``if` `temp[back] !``=` `arr[back]:``            ``break` `    ``#If whole array is sorted``    ``if` `front >``=` `back:``        ``return` `True``    ``while` `front !``=` `back:``        ``front ``+``=` `1``        ``if` `arr[front ``-` `1``] < arr[front]:``            ``return` `False``    ``return` `True` `# Driver code``arr ``=` `[``1``, ``2``, ``5``, ``4``, ``3``]``n ``=` `len``(arr)``if` `checkReverse(arr, n) ``=``=` `True``:``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)` `# This code is contributed``# by Shrikant13`

## C#

 `// C# program to check whether reversing a``// sub array make the array sorted or not``using` `System;` `class` `GFG``{` `// Return true, if reversing the``// subarray will sort the array,``// else return false.``static` `bool` `checkReverse(``int` `[]arr, ``int` `n)``{``    ``// Copying the array.``    ``int` `[]temp = ``new` `int``[n];``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``temp[i] = arr[i];``    ``}` `    ``// Sort the copied array.``    ``Array.Sort(temp);` `    ``// Finding the first mismatch.``    ``int` `front;``    ``for` `(front = 0; front < n; front++)``    ``{``        ``if` `(temp[front] != arr[front])``        ``{``            ``break``;``        ``}``    ``}` `    ``// Finding the last mismatch.``    ``int` `back;``    ``for` `(back = n - 1; back >= 0; back--)``    ``{``        ``if` `(temp[back] != arr[back])``        ``{``            ``break``;``        ``}``    ``}` `    ``// If whole array is sorted``    ``if` `(front >= back)``    ``{``        ``return` `true``;``    ``}` `    ``// Checking subarray is decreasing``    ``// or not.``    ``do``    ``{``        ``front++;``        ``if` `(arr[front - 1] < arr[front])``        ``{``            ``return` `false``;``        ``}``    ``} ``while` `(front != back);` `    ``return` `true``;``}` `// Driven Program``public` `static` `void` `Main()``{``    ``int` `[]arr = {1, 2, 5, 4, 3};``    ``int` `n = arr.Length;` `    ``if` `(checkReverse(arr, n))``    ``{``        ``Console.Write(``"Yes"``);``    ``}``    ``else``    ``{``        ``Console.Write(``"No"``);``    ``}``}``}` `// This code is contributed``// by PrinciRaj`

## PHP

 `= 0; ``\$back``--)``        ``if` `(``\$temp``[``\$back``] != ``\$arr``[``\$back``])``            ``break``;` `    ``// If whole array is sorted``    ``if` `(``\$front` `>= ``\$back``)``        ``return` `true;` `    ``// Checking subarray is decreasing or not.``    ``do``    ``{``        ``\$front``++;``        ``if` `(``\$arr``[``\$front` `- 1] < ``\$arr``[``\$front``])``            ``return` `false;``    ``} ``while` `(``\$front` `!= ``\$back``);` `    ``return` `true;``}` `// Driver Code``\$arr` `= ``array``( 1, 2, 5, 4, 3 );``\$n` `= sizeof(``\$arr``);` `if``(checkReverse(``\$arr``, ``\$n``))``    ``echo` `"Yes"` `. ``"\n"``;``else``    ``echo` `"No"` `. ``"\n"``;` `// This code is contributed``// by Akanksha Rai``?>`

## Javascript

 ``

Output

```Yes
```

Time Complexity: O(n*log(n) ).
Auxiliary Space: O(n).

Method 3: Linear time solution (O(n))
Observe, that the answer will be True when the array is already sorted or when the array consists of three parts. The first part is increasing subarray, then decreasing subarray, and then again increasing subarray. So, we need to check that array contains increasing elements then some decreasing elements, and then increasing elements if this is the case the answer will be True. In all other cases, the answer will be False.

Note: Simply finding the three parts does not guarantee the answer to be True eg consider

` arr [] = {10,20,30,40,4,3,2,50,60,70} `

The answer would be False in this case although we are able to find three parts. We will be handling the validity of the three parts in the code below.

Below is the implementation of this approach:

## C++

 `// C++ program to check whether reversing a sub array``// make the array sorted or not``#include``using` `namespace` `std;` `// Return true, if reversing the subarray will sort t``// he array, else return false.``bool` `checkReverse(``int` `arr[], ``int` `n)``{``    ``if` `(n == 1)``        ``return` `true``;` `    ``// Find first increasing part``    ``int` `i;``    ``for` `(i=1; i < n && arr[i-1] < arr[i]; i++);``    ``if` `(i == n)``        ``return` `true``;` `    ``// Find reversed part``    ``int` `j = i;``    ``while` `(j < n && arr[j] < arr[j-1])``    ``{``        ``if` `(i > 1 && arr[j] < arr[i-2])``            ``return` `false``;``        ``j++;``    ``}` `    ``if` `(j == n)``        ``return` `true``;` `    ``// Find last increasing part``    ``int` `k = j;` `    ``// To handle cases like {1,2,3,4,20,9,16,17}``    ``if` `(arr[k] < arr[i-1])``       ``return` `false``;` `    ``while` `(k > 1 && k < n)``    ``{``        ``if` `(arr[k] < arr[k-1])``            ``return` `false``;``        ``k++;``    ``}``    ``return` `true``;``}` `// Driven Program``int` `main()``{``    ``int` `arr[] = {1, 3, 4, 10, 9, 8};``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);``    ``checkReverse(arr, n)? cout << ``"Yes"` `: cout << ``"No"``;``    ``return` `0;``}`

## Java

 `// Java program to check whether reversing a sub array``// make the array sorted or not` `class` `GFG {` `// Return true, if reversing the subarray will sort t``// he array, else return false.``    ``static` `boolean` `checkReverse(``int` `arr[], ``int` `n) {``        ``if` `(n == ``1``) {``            ``return` `true``;``        ``}` `        ``// Find first increasing part``        ``int` `i;``        ``for` `(i = ``1``; arr[i - ``1``] < arr[i] && i < n; i++);``        ``if` `(i == n) {``            ``return` `true``;``        ``}` `        ``// Find reversed part``        ``int` `j = i;``        ``while` `(j < n && arr[j] < arr[j - ``1``]) {``            ``if` `(i > ``1` `&& arr[j] < arr[i - ``2``]) {``                ``return` `false``;``            ``}``            ``j++;``        ``}` `        ``if` `(j == n) {``            ``return` `true``;``        ``}` `        ``// Find last increasing part``        ``int` `k = j;` `        ``// To handle cases like {1,2,3,4,20,9,16,17}``        ``if` `(arr[k] < arr[i - ``1``]) {``            ``return` `false``;``        ``}` `        ``while` `(k > ``1` `&& k < n) {``            ``if` `(arr[k] < arr[k - ``1``]) {``                ``return` `false``;``            ``}``            ``k++;``        ``}``        ``return` `true``;``    ``}` `// Driven Program``    ``public` `static` `void` `main(String[] args) {` `        ``int` `arr[] = {``1``, ``3``, ``4``, ``10``, ``9``, ``8``};``        ``int` `n = arr.length;` `        ``if` `(checkReverse(arr, n)) {``            ``System.out.print(``"Yes"``);``        ``} ``else` `{``            ``System.out.print(``"No"``);``        ``}``    ``}` `}` `// This code is contributed``// by Rajput-Ji`

## Python3

 `# Python3 program to check whether reversing``# a sub array make the array sorted or not``import` `math as mt` `# Return True, if reversing the subarray``# will sort the array, else return False.``def` `checkReverse(arr, n):` `    ``if` `(n ``=``=` `1``):``        ``return` `True` `    ``# Find first increasing part``    ``i ``=` `1``    ``for` `i ``in` `range``(``1``, n):``        ``if` `arr[i ``-` `1``] < arr[i] :``            ``if` `(i ``=``=` `n):``                ``return` `True``         ` `        ``else``:``            ``break` `    ``# Find reversed part``    ``j ``=` `i``    ``while` `(j < n ``and` `arr[j] < arr[j ``-` `1``]):``     ` `        ``if` `(i > ``1` `and` `arr[j] < arr[i ``-` `2``]):``            ``return` `False``        ``j ``+``=` `1` `    ``if` `(j ``=``=` `n):``        ``return` `True` `    ``# Find last increasing part``    ``k ``=` `j` `    ``# To handle cases like 1,2,3,4,20,9,16,17``    ``if` `(arr[k] < arr[i ``-` `1``]):``        ``return` `False` `    ``while` `(k > ``1` `and` `k < n):``    ` `        ``if` `(arr[k] < arr[k ``-` `1``]):``            ``return` `False``        ``k ``+``=` `1``    ` `    ``return` `True` `# Driver Code``arr ``=` `[ ``1``, ``3``, ``4``, ``10``, ``9``, ``8``]``n ``=` `len``(arr)``if` `checkReverse(arr, n):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)``        ` `# This code is contributed by``# Mohit kumar 29`

## C#

 `// C# program to check whether reversing a``// sub array make the array sorted or not`` ` `using` `System;``public` `class` `GFG{` `// Return true, if reversing the subarray will sort t``// he array, else return false.``    ``static` `bool` `checkReverse(``int` `[]arr, ``int` `n) {``        ``if` `(n == 1) {``            ``return` `true``;``        ``}` `        ``// Find first increasing part``        ``int` `i;``        ``for` `(i = 1; arr[i - 1] < arr[i] && i < n; i++);``        ``if` `(i == n) {``            ``return` `true``;``        ``}` `        ``// Find reversed part``        ``int` `j = i;``        ``while` `(j < n && arr[j] < arr[j - 1]) {``            ``if` `(i > 1 && arr[j] < arr[i - 2]) {``                ``return` `false``;``            ``}``            ``j++;``        ``}` `        ``if` `(j == n) {``            ``return` `true``;``        ``}` `        ``// Find last increasing part``        ``int` `k = j;` `        ``// To handle cases like {1,2,3,4,20,9,16,17}``        ``if` `(arr[k] < arr[i - 1]) {``            ``return` `false``;``        ``}` `        ``while` `(k > 1 && k < n) {``            ``if` `(arr[k] < arr[k - 1]) {``                ``return` `false``;``            ``}``            ``k++;``        ``}``        ``return` `true``;``    ``}`  `// Driven Program``    ``public` `static` `void` `Main() {` `        ``int` `[]arr = {1, 3, 4, 10, 9, 8};``        ``int` `n = arr.Length;` `        ``if` `(checkReverse(arr, n)) {``            ``Console.Write(``"Yes"``);``        ``} ``else` `{``            ``Console.Write(``"No"``);``        ``}``    ``}``}``// This code is contributed``// by 29AjayKumar`

## Javascript

 ``

Output

`Yes`

Time Complexity: O(n).
Auxiliary Space: O(1).

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