Number of pairs with Pandigital Concatenation

A pair of strings when concatenated is said to be a ‘Pandigital Concatenation’ if their concatenation consists of all digits from (0 – 9) in any order at least once.The task is, given N strings, compute the number of pairs resulting in a ‘Pandigital Concatenation’. 

Examples: 

Input  : num[] = {"123567", "098234", "14765", "19804"}
Output : 3
The pairs, 1st and 2nd giving 
(123567098234),1st and 4rd giving(12356719804) and 
2nd and 3rd giving (09823414765),
on concatenation result in Pandigital Concatenations. 

Input : num[] =  {"56789", "098345", "1234"}
Output : 0
None of the pairs on concatenation result in Pandigital 
Concatenations.

Method 1 (Brute Force): A possible brute-force solution is to form all possible concatenations by forming all pairs in O(n² and using a frequency array for digits (0 – 9), we check if each digit exists at least once in each concatenation formed for every pair. 

Implementation:

3

Time Complexity : The time complexity of the given program is O(n^2 * k), where n is the number of strings in the input vector and k is the length of the longest string in the vector. This is because the program has nested loops that iterate over all pairs of strings in the input vector, and the isPanDigital function has a loop that iterates over each character in the concatenated string, which takes O(k) time. Therefore, the overall time complexity is O(n^2 * k).

Space Complexity: The space complexity of the program is O(1), as it uses a constant amount of additional space regardless of the size of the input vector or the length of the strings. This is because it uses a fixed-size boolean array of size 10 to keep track of the presence of digits in a string.

Method 2 (Efficient): 

Now we look for something better than the brute-force discussed above. Careful analysis suggests that, for every digit 0 – 9 to be present we have a mask as 1111111111 (i.e. all numbers 0-9 exist in the array of numbers

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 digits
exists at-least once thus for all such Pandigital 
Concatenations, this relationship should hold.
So we can represent 11...11 as a valid mask for
pandigital concatenations.

So now the approach is to represent every string as a mask of 10 bits where the i^th bit is set if the i^th digit exists in the string. 

E.g., "11405" can be represented as
Digits -           0  1  2  3  4  5  6  7  8  9
                   |  |  |  |  |  |  |  |  |  |
Mask for 11405 -   1  1  0  0  1  1  0  0  0  0

The approach though may look complete is still not efficient as we still have to iterate over all pairs and check if the OR of these two strings results in the mask of a valid Pandigital Concatenation. 

If we analyze the possible masks of all possible strings we can understand that every single string would be only comprised of digits 0 – 9, so every number can at max contain all digits 0 to 9 at least once thus the mask of such a number would be 1111111111 (1023 in decimal). Thus, in the decimal system all masks exit in (0 – 1023]. 

Now we just have to maintain a frequency array to store the number of times a mask exists in the array of strings.

Let two masks be i and j with frequencies freq_i and freq_j respectively,
If (i OR j) = Mask_{pandigital concatenation} 
Then, 
Number of Pairs = freq_i * freq_j  

Implementation:

3

Complexity : O(N * |s| + 1023 * 1023) where |s| gives length of strings in the array.