# 686. Repeated String Match

## Description

Given two strings a and b, return the minimum number of times you should repeat string a so that string b is a substring of it. If it is impossible for b​​​​​​ to be a substring of a after repeating it, return -1.

Notice: string "abc" repeated 0 times is "", repeated 1 time is "abc" and repeated 2 times is "abcabc".

Example 1:

Input: a = "abcd", b = "cdabcdab"
Output: 3
Explanation: We return 3 because by repeating a three times "abcdabcdabcd", b is a substring of it.


Example 2:

Input: a = "a", b = "aa"
Output: 2


Constraints:

• 1 <= a.length, b.length <= 104
• a and b consist of lowercase English letters.

## Solutions

• class Solution {
public int repeatedStringMatch(String a, String b) {
int m = a.length(), n = b.length();
int ans = (n + m - 1) / m;
StringBuilder t = new StringBuilder(a.repeat(ans));
for (int i = 0; i < 3; ++i) {
if (t.toString().contains(b)) {
return ans;
}
++ans;
t.append(a);
}
return -1;
}
}

• class Solution {
public:
int repeatedStringMatch(string a, string b) {
int m = a.size(), n = b.size();
int ans = (n + m - 1) / m;
string t = "";
for (int i = 0; i < ans; ++i) t += a;
for (int i = 0; i < 3; ++i) {
if (t.find(b) != -1) return ans;
++ans;
t += a;
}
return -1;
}
};

• class Solution:
def repeatedStringMatch(self, a: str, b: str) -> int:
m, n = len(a), len(b)
ans = ceil(n / m)
t = [a] * ans
for _ in range(3):
if b in ''.join(t):
return ans
ans += 1
t.append(a)
return -1


• func repeatedStringMatch(a string, b string) int {
m, n := len(a), len(b)
ans := (n + m - 1) / m
t := strings.Repeat(a, ans)
for i := 0; i < 3; i++ {
if strings.Contains(t, b) {
return ans
}
ans++
t += a
}
return -1
}

• function repeatedStringMatch(a: string, b: string): number {
const m: number = a.length,
n: number = b.length;
let ans: number = Math.ceil(n / m);
let t: string = a.repeat(ans);

for (let i = 0; i < 3; i++) {
if (t.includes(b)) {
return ans;
}
ans++;
t += a;
}

return -1;
}