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1062. Longest Repeating Substring

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

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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
  
  

• 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
  }
  

• function longestRepeatingSubstring(s: string): number {
      const n = s.length;
      const f: number[][] = Array.from({ length: n }).map(() => Array(n).fill(0));
      let ans = 0;
      for (let i = 1; i < n; ++i) {
          for (let j = 0; j < i; ++j) {
              if (s[i] === s[j]) {
                  f[i][j] = 1 + (f[i - 1][j - 1] || 0);
                  ans = Math.max(ans, f[i][j]);
              }
          }
      }
      return ans;
  }
  
  

• impl Solution {
      pub fn longest_repeating_substring(s: String) -> i32 {
          let n = s.len();
          let mut f = vec![vec![0; n]; n];
          let mut ans = 0;
          let s = s.as_bytes();
  
          for i in 1..n {
              for j in 0..i {
                  if s[i] == s[j] {
                      f[i][j] = if j > 0 { f[i - 1][j - 1] + 1 } else { 1 };
                      ans = ans.max(f[i][j]);
                  }
              }
          }
          ans
      }
  }
  
  

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