Python3 Program to Check if it is possible to sort the array after rotating it

Last Updated : 05 Sep, 2024

Given an array of size N, the task is to determine whether its possible to sort the array or not by just one shuffle. In one shuffle, we can shift some contiguous elements from the end of the array and place it in the front of the array.
For eg:

A = {2, 3, 1, 2}, we can shift {1, 2} from the end of the array to the front of the array to sort it.
A = {1, 2, 3, 2} since we cannot sort it in one shuffle hence it's not possible to sort the array.

Examples:

Input: arr[] = {1, 2, 3, 4} 
Output: Possible 
Since this array is already sorted hence no need for shuffle.

Input: arr[] = {6, 8, 1, 2, 5}
Output: Possible
Place last three elements at the front 
in the same order i.e. {1, 2, 5, 6, 8}

Approach:

Check if the array is already sorted or not. If yes return true.
Else start traversing the array elements until the current element is smaller than next element. Store that index where arr[i] > arr[i+1].
Traverse from that point and check if from that index elements are in increasing order or not.
If above both conditions satisfied then check if last element is smaller than or equal to the first element of given array.
Print "Possible" if above three conditions satisfied else print "Not possible" if any of the above 3 conditions failed.

Below is the implementation of above approach:

Python 3

# Python 3 implementation of 
# above approach
def is_sorted(a):
    all(a[i] <= a[i + 1] 
    for i in range(len(a) - 1))
    
# Function to check if 
# it is possible
def isPossible(a, n):

    # step 1
    if (is_sorted(a)) :
        print("Possible")
    
    else :

        # break where a[i] > a[i+1]
        flag = True
        for i in range(n - 1) :
            if (a[i] > a[i + 1]) :
                break
            
        # break point + 1
        i += 1

        # check whether the sequence is
        # further increasing or not
        for k in range(i, n - 1) :
            if (a[k] > a[k + 1]) :
                flag = False
                break

        # If not increasing after 
        # break point
        if (not flag):
            return False

        else :

            # last element <= First element
            if (a[n - 1] <= a[0]):
                return True

            else:
                return False

# Driver code
if __name__ == "__main__":

    arr = [ 3, 1, 2, 2, 3 ]
    n = len(arr)

    if (isPossible(arr, n)):
        print("Possible")

    else:
        print("Not Possible")

# This code is contributed
# by ChitraNayal