LeetCode [686] Repeated String Match

 686. Repeated String Match

Medium

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

Example 3:

Input: a = "a", b = "a"
Output: 1

Example 4:

Input: a = "abc", b = "wxyz"
Output: -1

 

Constraints:

• 1 <= a.length <= 104
• 1 <= b.length <= 104
• a and b consist of lower-case English letters.

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class Solution { public int repeatedStringMatch(String a, String b) { String as = a; for (int rep = 1; rep <= b.length() / a.length() + 2; rep++, as += a) if (as.indexOf(b) != -1) return rep; return -1; } }

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public int repeatedStringMatch(String A, String B) { int count = 0; StringBuilder sb = new StringBuilder(); while (sb.length() < B.length()) { sb.append(A); count++; } if(sb.toString().contains(B)) return count; if(sb.append(A).toString().contains(B)) return ++count; return -1; }