Given two strings A and B, find the minimum number of times A has to be repeated such that B is a substring of it. If no such solution, return -1.
For example, with A = "abcd" and B = "cdabcdab".
Return 3, because by repeating A three times (“abcdabcdabcd”), B is a substring of it; and B is not a substring of A repeated two times ("abcdabcd").
Note:
The length of A and B will be between 1 and 10000.
class Solution:
def repeatedStringMatch(self, A, B):
"""
:type A: str
:type B: str
:rtype: int
"""
if len(set(A)) < len(set(B)):
return -1
for i in range(1,len(B)+1):
t = A*i
if t.find(B)!=-1:
return i
return -1