Java

/**

 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.

 */

public class Repeated_String_Match {

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

            return -1;
        }
    }
}

// OJ: https://leetcode.com/problems/repeated-string-match/
// Time: O(B * (A + B))
// Space: O(A + B)
// Ref: https://leetcode.com/problems/repeated-string-match/solution/
class Solution {
public:
    int repeatedStringMatch(string a, string b) {
        int k = 1;
        string s = a;
        for (; s.size() < b.size(); ++k) s += a;
        if (s.find(b) != string::npos) return k;
        return (s + a).find(b) != string::npos ? k + 1 : -1;
    }
};

class Solution(object):
    def repeatedStringMatch(self, A, B):
        """
        :type A: str
        :type B: str
        :rtype: int
        """
        na, nb = len(A), len(B)
        times = nb / na + 3
        for i in range(1, times):
            if B in A * i:
                return i
        return -1

Java

class Solution {
    public int repeatedStringMatch(String A, String B) {
        int lengthA = A.length(), lengthB = B.length();
        StringBuffer sb = new StringBuffer();
        int maxLength = lengthA * 2 + lengthB;
        int repeatTimes = 0;
        while (sb.length() <= maxLength) {
            sb.append(A);
            repeatTimes++;
            if (sb.toString().indexOf(B) >= 0)
                return repeatTimes;
        }
        return -1;
    }
}

// OJ: https://leetcode.com/problems/repeated-string-match/
// Time: O(B * (A + B))
// Space: O(A + B)
// Ref: https://leetcode.com/problems/repeated-string-match/solution/
class Solution {
public:
    int repeatedStringMatch(string a, string b) {
        int k = 1;
        string s = a;
        for (; s.size() < b.size(); ++k) s += a;
        if (s.find(b) != string::npos) return k;
        return (s + a).find(b) != string::npos ? k + 1 : -1;
    }
};

class Solution(object):
    def repeatedStringMatch(self, A, B):
        """
        :type A: str
        :type B: str
        :rtype: int
        """
        na, nb = len(A), len(B)
        times = nb / na + 3
        for i in range(1, times):
            if B in A * i:
                return i
        return -1

686 - Repeated String Match

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