Question

459. Repeated Substring Pattern

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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 <= 104
• s consists of lowercase English letters.

Solutions

Solution 1

class Solution:
  def repeatedSubstringPattern(self, s: str) -> bool:
    return (s + s).index(s, 1) < len(s)

class Solution {
  public boolean repeatedSubstringPattern(String s) {
    String str = s + s;
    return str.substring(1, str.length() - 1).contains(s);
  }
}

class Solution {
public:
  bool repeatedSubstringPattern(string s) {
    return (s + s).find(s, 1) < s.size();
  }
};

func repeatedSubstringPattern(s string) bool {
  return strings.Index(s[1:]+s, s) < len(s)-1
}

function repeatedSubstringPattern(s: string): boolean {
  return (s + s).slice(1, (s.length << 1) - 1).includes(s);
}

impl Solution {
  pub fn repeated_substring_pattern(s: String) -> bool {
    (s.clone() + &s)[1..s.len() * 2 - 1].contains(&s)
  }
}

Solution 2

function repeatedSubstringPattern(s: string): boolean {
  const n = s.length;
  for (let i = 0; i < n >> 1; i++) {
    const len = i + 1;
    if (n % len !== 0) {
      continue;
    }
    const t = s.slice(0, len);
    let j: number;
    for (j = len; j < n; j += len) {
      if (s.slice(j, j + len) !== t) {
        break;
      }
    }
    if (j === n) {
      return true;
    }
  }
  return false;
}