686. Repeated String Match

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.

https://www.youtube.com/watch?v=H4VrKHVG5qI




https://leetcode.com/problems/repeated-string-match/discuss/108084/C++-4-lines-O(m-*-n)-or-O(1)-and-KMP-O(m-+-n)-or-O(n)


https://leetcode.com/problems/repeated-string-match/solution/

https://www.quora.com/Why-is-Rabin-Karp-string-searching-algorithm-used-for-detecting-plagiarism#

Approach #1: Ad-Hoc [Accepted]
Intuition
The question can be summarized as "What is the smallest k for which B is a substring of A * k?" We can just try every k.
Algorithm
Imagine we wrote S = A+A+A+.... If B is to be a substring of S, we only need to check whether some S[0:], S[1:], ..., S[len(A) - 1:] starts with B, as S is long enough to contain B, and S has period at most len(A).
Now, suppose q is the least number for which len(B) <= len(A * q). We only need to check whether B is a substring of A * q or A * (q+1). If we try k < q, then B has larger length than A * q and therefore can't be a substring. When k = q+1, A * k is already big enough to try all positions for B; namely, A[i:i+len(B)] == B for i = 0, 1, ..., len(A) - 1.



class Solution {
    public int repeatedStringMatch(String A, String B) {
        int res = 1;
        StringBuilder sb = new StringBuilder(A); // StringBuilder(A)
        for(; sb.length() < B.length(); res++) sb.append(A); // (; )
        if(sb.indexOf(B) >= 0) return res; 
        if(sb.append(A).indexOf(B) >= 0) return res + 1;
        return -1;
        
    }
}