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1062. Longest Repeating Substring πŸ”’

Description

Given a string s, return the length of the longest repeating substrings. If no repeating substring exists, return 0.

 

Example 1:

Input: s = "abcd"
Output: 0
Explanation: There is no repeating substring.

Example 2:

Input: s = "abbaba"
Output: 2
Explanation: The longest repeating substrings are "ab" and "ba", each of which occurs twice.

Example 3:

Input: s = "aabcaabdaab"
Output: 3
Explanation: The longest repeating substring is "aab", which occurs 3 times.

 

Constraints:

  • 1 <= s.length <= 2000
  • s consists of lowercase English letters.

Solutions

Solution 1

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class Solution:
    def longestRepeatingSubstring(self, s: str) -> int:
        n = len(s)
        dp = [[0] * n for _ in range(n)]
        ans = 0
        for i in range(n):
            for j in range(i + 1, n):
                if s[i] == s[j]:
                    dp[i][j] = dp[i - 1][j - 1] + 1 if i else 1
                    ans = max(ans, dp[i][j])
        return ans

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class Solution {
    public int longestRepeatingSubstring(String s) {
        int n = s.length();
        int ans = 0;
        int[][] dp = new int[n][n];
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (s.charAt(i) == s.charAt(j)) {
                    dp[i][j] = i > 0 ? dp[i - 1][j - 1] + 1 : 1;
                    ans = Math.max(ans, dp[i][j]);
                }
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int longestRepeatingSubstring(string s) {
        int n = s.size();
        vector<vector<int>> dp(n, vector<int>(n));
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (s[i] == s[j]) {
                    dp[i][j] = i ? dp[i - 1][j - 1] + 1 : 1;
                    ans = max(ans, dp[i][j]);
                }
            }
        }
        return ans;
    }
};

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func longestRepeatingSubstring(s string) int {
    n := len(s)
    dp := make([][]int, n)
    for i := range dp {
        dp[i] = make([]int, n)
    }
    ans := 0
    for i := 0; i < n; i++ {
        for j := i + 1; j < n; j++ {
            if s[i] == s[j] {
                if i == 0 {
                    dp[i][j] = 1
                } else {
                    dp[i][j] = dp[i-1][j-1] + 1
                }
                ans = max(ans, dp[i][j])
            }
        }
    }
    return ans
}

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