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• /**

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;
}
}
}

############

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;
}
}

• // 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:
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

############

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

• 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;
}


• 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;
}
}

############

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;
}
}

• // 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:
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

############

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

• 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;
}