Suppose there are two strings A and B. We can say that A is divisible by B, when A is created by concatenating B one or more times. So if A = “abcabc”, and B = “abc”, then A is divisible by B. In this section, we will see what is the greatest common divisor of a String. So return the largest string that divides both of the strings. So if two strings are “ABABAB”, and “ABAB”, then GCD will be “AB”
To solve this, we will follow these steps −
- temp := shorter string between A and B
- m := length of temp
- x := 1
- res is an array and insert “” into the res
- while A and B has substring of size x, then add the substring into the res, and increase x by 1
- finally return the last element in the res array.
Example
Let us see the following implementation to get better understanding −
class Solution(object):
def gcdOfStrings(self, str1, str2):
if len(str1)<=len(str2):
temp = str1
else:
temp = str2
m = len(temp)
x = 1
res=[""]
while x<=m:
if m%x==0 and temp[:x] * (len(str1)//x) == str1 and temp[:x] * (len(str2)//x) == str2:
res.append(temp[:x])
x+=1
return res[-1]
ob1 = Solution()
print(ob1.gcdOfStrings("ABABAB","ABAB"))Input
"ABABAB" "ABAB"
Output
AB