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

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}Â
Output: TrueÂ
Explanation:Â
arr[0] ? Right rotation once modifies arr[0] to 132 (= 11 Ă— 12).Â
arr[1] ? Right rotation once modifies arr[0] to 240 (= 15 Ă— 16).Â
arr[2] ? Right rotation twice modifies arr[2] to 462 (= 21 Ă— 22).Â
arr[3] ? Right rotation twice modifies arr[3] to 992 (= 31 Ă— 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:Â

`True`

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

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