459. Repeated Substring Pattern

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

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

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

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

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

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

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

Solution 2

TypeScript Code
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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;
}