Description

Given a string s, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together.

Example 1:

Input: s = "abab"
Output: true
Explanation: It is the substring "ab" twice.

Example 2:

Input: s = "aba"
Output: false

Example 3:

Input: s = "abcabcabcabc"
Output: true
Explanation: It is the substring "abc" four times or the substring "abcabc" twice.

Constraints:

1 <= s.length <= 10^4
s consists of lowercase English letters.

Solution

Iteration

Search from the left, when current sub-string has the same elements as the whole string, we start to check if the length of the sub-string could be divided by the whole string without residuals. Keep checking until the result is no longer great than 1.

Time complexity: $o(s.length)$
Space complexity: $o(s.length)$

Math

If a string is periodical, then it must exist in s+s[1:-1]

Code

Iteration

class Solution:
    def repeatedSubstringPattern(self, s: str) -> bool:
        uniq_chs = set(s)
        sub_string = ''
        for i in range(len(s)):
            sub_string += s[i]
            if not(uniq_chs - set(sub_string)):
                times = len(s) // len(sub_string)
                residuals = len(s) % len(sub_string)
                if times <= 1:
                    return False
                if residuals == 0 and s == sub_string * times:
                    return True
        return False

Math

class Solution:
    def repeatedSubstringPattern(self, s: str) -> bool:
        return s in (s + s)[1:-1]