# Check if reversing a sub array make the array sorted

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
```

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

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.

```// C++ program to check whether reversing a
// sub array make the array sorted or not
#include<bits/stdc++.h>
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;
}
```

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 C++ implementation of this approach:

```// C++ program to check whether reversing a sub array
// make the array sorted or not
#include<bits/stdc++.h>
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 (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;
}
```

Output:

```Yes
```

Time Complexity: O(n).

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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