Greatest Common Divisor of Strings
To approach this problem, we first need to understand the underlying mathematical principle of greatest common divisor (gcd). Here, instead of numbers, we are dealing with strings. But the concept remains the same  we need to find the largest string that can divide both given strings.
The basic strategy is to find the smallest common divisor and iteratively check if it is also a divisor of the larger string.
Python solution:


This algorithm works by recursively reducing the problem to the gcd of the suffix of str1 (after removing str2 from the start of str1) and str2, until one string is empty. The nonempty string at that point is the gcd of strings.
The termination condition for this recursive process is when str1 doesn’t start with str2 or str2 becomes empty.
This algorithm follows the Euclidean algorithm for computing gcd of two numbers. The time complexity is O(N^2), where N is the length of the strings.
What is the base case?
The base case for this recursive solution is when the second string str2
becomes empty or when the string str1
doesn’t start with str2
.
In other words, the base cases are:
if not str1.startswith(str2)
: This checks if the larger stringstr1
starts withstr2
. If it does not, it means there’s no common divisor, and an empty string''
is returned.if len(str2) == 0
: This checks ifstr2
has been completely subtracted fromstr1
and has become empty. If it is, it means thatstr1
is the largest string that can divide both given strings, sostr1
is returned.
What is the unit of work done in each recursive call?
In each recursive call of the function, the unit of work done is checking if str1
starts with str2
and making a new recursive call with updated values for str1
and str2
.
Specifically, if str1
starts with str2
, str1
is updated by removing the starting part of str1
that is equal to str2
, and str2
remains the same. Then, a recursive call is made with the updated str1
and str2
.
If str1
does not start with str2
, a new recursive call is made with str2
and the remainder of str1
after removing str2
from the start, in order to find the greatest common divisor of the two strings.
Thus, the unit of work in each recursive call consists of string comparison and manipulation, followed by a recursive function call.
It seems like we are looking for substring matches in the solution. Am I right?
Yes, you’re correct. Specifically, we’re looking for the largest substring that is a prefix of both str1
and str2
and that can completely divide both strings. That is, both str1
and str2
are composed of one or more repetitions of this substring. This is akin to finding the Greatest Common Divisor (GCD) of two numbers, except here we’re dealing with strings.
How do we reduce the size of the problem in each recursive call?
We reduce the size of the problem in each recursive call by considering the remainder of dividing the longer string by the shorter one. This is done by taking the substring of the longer string that starts from the end of a complete occurrence of the shorter string.
If str1
is longer than str2
, we take the substring of str1
that starts from the end of a complete occurrence of str2
in str1
and recursively call the function with this new substring and str2
. If str2
is longer than str1
, we do the similar operation.
This approach is similar to the Euclidean algorithm for computing the greatest common divisor of two numbers. We continually subtract the smaller number from the larger one until we reach zero, at which point the remaining number is the GCD. In this string version, we’re continually subtracting the smaller string from the larger one by taking substrings until we reach an empty string, at which point the remaining string is the “greatest common divisor” of the original strings.