# Python3 Program to Modify given array to a non-decreasing array by rotation

Given an array arr[] of size N (consisting of duplicates), the task is to check if the given array can be converted to a non-decreasing array by rotating it. If it’s not possible to do so, then print “No“. Otherwise, print “Yes“.

Examples:

Input: arr[] = {3, 4, 5, 1, 2}
Output: Yes
Explanation: After 2 right rotations, the array arr[] modifies to {1, 2, 3, 4, 5}

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

Approach: The idea is based on the fact that a maximum of N distinct arrays can be obtained by rotating the given array and check for each individual rotated array, whether it is non-decreasing or not. Follow the steps below to solve the problem:

• Initialize a vector, say v, and copy all the elements of the original array into it.
• Sort the vector v.
• Traverse the original array and perform the following steps:
• Rotate by 1 in each iteration.
• If the array becomes equal to vector v, print “Yes“. Otherwise, print “No“.

Below is the implementation of the above approach:

## Python3

 `# Python 3 program for the above approach`  `# Function to check if a``# non-decreasing array can be obtained``# by rotating the original array``def` `rotateArray(arr, N):``  ` `    ``# Stores copy of original array``    ``v ``=` `arr` `    ``# Sort the given vector``    ``v.sort(reverse ``=` `False``)` `    ``# Traverse the array``    ``for` `i ``in` `range``(``1``, N ``+` `1``, ``1``):``      ` `        ``# Rotate the array by 1``        ``x ``=` `arr[N ``-` `1``]``        ``i ``=` `N ``-` `1``        ``while``(i > ``0``):``            ``arr[i] ``=` `arr[i ``-` `1``]``            ``arr[``0``] ``=` `x``            ``i ``-``=` `1``            ` `        ``# If array is sorted``        ``if` `(arr ``=``=` `v):``            ``print``(``"YES"``)``            ``return` `    ``# If it is not possible to``    ``# sort the array``    ``print``(``"NO"``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``  ` `    ``# Given array``    ``arr ``=`  `[``3``, ``4``, ``5``, ``1``, ``2``]` `    ``# Size of the array``    ``N ``=` `len``(arr)` `    ``# Function call to check if it is possible``    ``# to make array non-decreasing by rotating``    ``rotateArray(arr, N)``    ` `    ``# This code is contributed by ipg2016107.`

Output
`YES`

Time Complexity: O(N2)
Auxiliary Space: O(N)

Method 2: Identify the index of the smallest element in the array using the index method, and then creates a rotated version of the array where the smallest element is the first element. It then checks if the rotated array is non-decreasing by traversing through it and comparing each element with the next element.

• First, we identify the index of the smallest element in the array using the index method.
• We then create a rotated version of the array where the smallest element is the first element. We do this by slicing the original array into two parts and concatenating them in reverse order.
• Finally, we traverse through the rotated array and check if it is non-decreasing. If we find an element that is greater than the next element, we return “NO”. Otherwise, we return “YES”.

Implementation:

## Python3

 `def` `check_rotation(arr):``    ``n ``=` `len``(arr)``    ` `    ``# Identify the index of the smallest element in the array``    ``idx ``=` `arr.index(``min``(arr))``    ` `    ``# Rotate the array to make the smallest element the first element``    ``rotated_arr ``=` `arr[idx:] ``+` `arr[:idx]``    ` `    ``# Check if the rotated array is non-decreasing``    ``for` `i ``in` `range``(n``-``1``):``        ``if` `rotated_arr[i] > rotated_arr[i``+``1``]:``            ``return` `"NO"``    ` `    ``return` `"YES"` `# Example usage``arr ``=` `[``3``, ``4``, ``5``, ``1``, ``2``]``result ``=` `check_rotation(arr)``print``(result) ``# Output: NO`

Output
`YES`

Time complexity: O(n)
Auxiliary Space: O(n)

Please refer complete article on Modify given array to a non-decreasing array by rotation for more details!

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