Count pairs in an array which have at least one digit common

Given an array of N numbers. Find out the number of pairs i and j such that i < j and A_i and A_j have at least one digit common (For e.g. (11, 19) have 1 digit common but (36, 48) have no digit common)

Examples: 

Input: A[] = { 10, 12, 24 } 
Output: 2 
Explanation: Two valid pairs are (10, 12) and (12, 24) which have atleast one digit common 

Method 1 (Brute Force):

A naive approach to solve this problem is just by running two nested loops and consider all possible pairs. We can check if the two numbers have at least one common digit, by extracting every digit of the first number and try to find it in the extracted digits of second number. The task would become much easier we simply convert them into strings.

Below is the implementation of the above approach:

2

Time Complexity: O(N²) where N is the size of array.
Space Complexity : O( 1 )

Method 2 (Bit Masking):

An efficient approach of solving this problem is creating a bit mask for every digit present in a particular number. Thus, for every digit to be present in a number if we have a mask of 1111111111. 

Digits - 0 1 2 3 4 5 6 7 8 9
 | | | | | | | | | |
Mask - 1 1 1 1 1 1 1 1 1 1 

Here 1 denotes that the corresponding ith digit is set. 
For e.g. 1235 can be represented as
Digits - 0 1 2 3 4 5 6 7 8 9
 | | | | | | | | | |
Mask for 1235 - 0 1 1 1 0 1 0 0 0 0

Now we just have to extract every digit of a number and make the corresponding bit set (1 << i^th digit) and store the whole number as a mask. Careful analysis suggests that the maximum value of the mask is 1023 in decimal (which contains all the digits from 0 - 9). Since the same set of digits can exist in more than one number, we need to maintain a frequency array to store the count of mask value. 

Let the frequencies of two masks i and j be freq_i and freq_j respectively, 
If(i AND j) return true, means i_th and j_th mask contains atleast one common set bit which in turn implies that the numbers from which these masks have been built also contain a common digit 
then, 
increment the answer 
ans += freq_i * freq_j [ if i != j ] 
ans += (freq_i * (freq_i - 1)) / 2 [ if j == i ]  

Below is the implementation of this efficient approach:

2

Time Complexity: O(N + 1024 * 1024), where N is the size of the array.

Space Complexity : O( 1 )