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Check if reversing a sub array make the array sorted

• Difficulty Level : Medium
• Last Updated : 13 Apr, 2021

Given an array of distinct n integers. The task is to check whether reversing one sub-array make the array sorted or not. If the array is already sorted or by reversing a subarray once make it sorted, 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 (Simple : O(n2)
A simple solution is to consider every subarray one by one. Try reversing every subarray and check if reversing the subarray makes the whole array sorted. If yes, return true. If reversing any subarray doesn’t make the array sorted, then return false.

Method 2 (Sorting : O(nlogn)):
The idea is to compare the given array with the sorted array. Make a copy of the given array and sort it. Now, find the first index and last index which do not match with sorted array. If no such indices are found, print “Yes”. Else check if the elements between the indices are in decreasing order.
Below is the implementation of 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(nlogn).

Method 3 (Linear : O(n)):
Observe, answer will be “Yes” when the array is sorted or when the array consist of three parts. First part is increasing subarray, then decreasing subarray and then again increasing subarray. So, we need to check that array contain increasing elements then some decreasing elements and then increasing elements. In all other case, answer will be “No”.
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`

PHP

 ` 1 && ``\$arr``[``\$j``] < ``\$arr``[``\$i` `- 2])``            ``return` `false;``        ``\$j``++;``    ``}` `    ``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 && ``\$k` `< ``\$n``)``    ``{``        ``if` `(``\$arr``[``\$k``] < ``\$arr``[``\$k` `- 1])``            ``return` `false;``        ``\$k``++;``    ``}``    ``return` `true;``}` `// Driver Code``\$arr` `= ``array``(1, 3, 4, 10, 9, 8);``\$n` `= sizeof(``\$arr``);``if``(checkReverse(``\$arr``, ``\$n``))``    ``echo` `"Yes"``;``else``    ``echo` `"No"``;` `// This code is contributed``// by Akanksha Rai(Abby_akku)``?>`

Javascript

 ``

Output:

`Yes`

Time Complexity: O(n).
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