Count of pairs of strings which differ in exactly one position

Given an array arr[] of strings of equal lengths. The task is to calculate the total number of pairs of strings which differ in exactly one position.


Input: arr[] = {“abc”, “abd”, “bbd”}
Output: 2
(abc, abd) and (abd, bbd) are the only valid pairs.

Input: arr[] = {“def”, “deg”, “dmf”, “xef”, “dxg”}
Output: 4

Method 1: For every possible pair, check if both the strings differ in exactly a single index position with a single traversal of the strings.

Method 2: Two string can be compared in the following way in order to check whether they differ in a single index position:

Let str1 = “abc” and str2 = “adc”
For str1, add “#bc”, “a#c” and “ab#” to a set.
Now for str2, generate the string in the similar manner and if any of the generated string
is already present in the set then both the strings differ in exactly 1 index position.
For example, “a#c” is one of the generated strings.
Note that “#” is used because it will not be a part of any of the original strings.

Below is the implementation of the above approach:





# Python3 implementation of the approach
# Function to return the count of same pairs
def pair_count(d):
    return sum((i*(i-1))//2 for i in d.values())
# Function to return total number of strings 
# which satisfy required condition
def Difference(array, m):
    # Dictionary changed will store strings 
    # with wild cards
    # Dictionary same will store strings 
    # that are equal
    changed, same = {}, {}
    # Iterating for all strings in the given array
    for s in array:
        # If we found the string then increment by 1 
        # Else it will get default value 0
        same[s]= same.get(s, 0)+1
        # Iterating on a single string
        for i in range(m):
            # Adding special symbol to the string
            t = s[:i]+'#'+s[i + 1:]
            # Incrementing the string if found 
            # Else it will get default value 0
            changed[t]= changed.get(t, 0)+1
    # Return counted pairs - equal pairs
    return pair_count(changed) - pair_count(same)*m
# Driver code
if __name__=="__main__":
    n, m = 3, 3
    array =["abc", "abd", "bbd"]
    print(Difference(array, m))




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.