# Python3 Program to Check if all array elements can be converted to pronic numbers by rotating digits

• Last Updated : 27 Jan, 2022

Given an array arr[] of size N, the task is to check if it is possible to convert all of the array elements to a pronic number by rotating the digits of array elements any number of times.

Examples:

Input: {321, 402, 246, 299}
Output: True
Explanation:
arr → Right rotation once modifies arr to 132 (= 11 × 12).
arr → Right rotation once modifies arr to 240 (= 15 × 16).
arr → Right rotation twice modifies arr to 462 (= 21 × 22).
arr → Right rotation twice modifies arr to 992 (= 31 × 32).

Input: {433, 653, 402, 186}
Output: False

Approach: Follow the steps below to solve the problem:

• Traverse the array and check for each array element, whether it is possible to convert it to a pronic number.
• For each array element, apply all the possible rotations and check after each rotation, whether the generated number is pronic or not.
• If it is not possible to convert any array element to a pronic number, print “False”.
• Otherwise, print “True”.

Below is the implementation of the above approach:

## Python3

 `# Python implementation of``# the above approach`` ` `# Function to check if a number``# is a pronic number or not``def` `isPronic(n):`` ` `  ``for` `i ``in` `range``(``int``(n``*``*``(``1` `/` `2``)) ``+` `1``):``    ``if` `i ``*` `(i ``+` `1``) ``=``=` `n:``      ``return` `True`` ` `  ``return` `False`` ` `# Function to check if any permutation``# of n is a pronic number or not``def` `checkRot(n):`` ` `  ``temp ``=` `str``(n)`` ` `  ``for` `i ``in` `range``(``len``(temp)):`` ` `    ``if` `isPronic(``int``(temp)):``      ``return` `True`` ` `    ``temp ``=` `temp[``1``:]``+``temp[``0``]`` ` `  ``return` `False`` ` `# Function to check if all array``# elements can be converted to``# a pronic number or not``def` `check(arr):`` ` `  ``# Traverse the array``  ``for` `i ``in` `arr:`` ` `    ``# If current element``    ``# cannot be converted ``    ``# to a pronic number``    ``if` `not` `checkRot(i):``      ``return` `False``  ``return` `True`` ` `# Driver Code``arr ``=` `[ ``321``, ``402``, ``246``, ``299` `]``print``(check(arr))`

Output:

`True`

Time Complexity: O(N3/2)
Auxiliary Space: O(1)

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