Count Pandigital Fractions pairs in given Array

Given an array arr[], the task is to count the pairs in the array such that arr[i]/arr[j] is a Pandigital Fraction.

A fraction N/D is called a Pandigital Fraction if the fraction contains all the digits from 0 to 9.  

Examples:

Input: arr = [ 12345, 67890, 123, 4567890 ] 
Output: 3 
Explanation:
The fractions are 12345/67890, 12345/4567890, and 123/4567890 

Input: arr = [ 12345, 6789 ] 
Output: 0 

Approach: The idea is to iterate over every possible pair of the array using two nested loops and for every pair concatenate arr[i] and arr[j] into a single number and check if the concatenation of arr[i] and arr[j] is a Pandigital number in base 10 then increment count.

Below is the implementation of the above approach:

Python3

# Python3 implementation of the
# above approach

import math

# Function to concatenate
# two numbers into one
def numConcat(num1, num2):

     # Find number of digits in num2
     digits = len(str(num2))

     # Add zeroes to the end of num1
     num1 = num1 * (10**digits)

     # Add num2 to num1
     num1 += num2

     return num1

# Return true if n is pandigit
# else return false.
def checkPanDigital(n):
    n = str(n)
    b = 10

    # Checking length is
    # less than base
    if (len(n) < b):
        return 0;

    hash = [0] * b;

    # Traversing each digit
    # of the number.
    for i in range(len(n)):

        # If digit is integer
        if (n[i] >= '0' and \
            n[i] <= '9'):
            hash[ord(n[i]) - ord('0')] = 1;

        # If digit is alphabet
        elif (ord(n[i]) - ord('A') <= \
                            b - 11):
            hash[ord(n[i]) - \
                 ord('A') + 10] = 1;

    # Checking hash array, if any index is
    # unmarked.
    for i in range(b):
        if (hash[i] == 0):
            return 0;

    return 1;

# Returns true if N is a
# Pandigital Fraction Number
def isPandigitalFraction(N, D):
    join = numConcat(N, D)
    return checkPanDigital(join)

# Returns number pandigital fractions
# in the array
def countPandigitalFraction(v, n) :

    # iterate over all
    # pair of strings
    count = 0
    for i in range(0, n) :

        for j in range (i + 1,
                        n) :

            if (isPandigitalFraction(v[i],
                             v[j])) :
                count = count + 1
    return count


# Driver Code
if __name__ == "__main__":

    arr = [ 12345, 67890, 123, 4567890 ]
    n = len(arr)

    print(countPandigitalFraction(arr, n))

Output:

3

Time Complexity: O(N²)
Reference: https://mathworld.wolfram.com/PandigitalFraction.html