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Longest Repeating Substring

Solution Explanation for Longest Repeating Substring

This problem asks to find the length of the longest repeating substring within a given string. We can efficiently solve this using dynamic programming.

Approach: Dynamic Programming

The core idea is to build a 2D table dp where dp[i][j] represents the length of the longest repeating substring ending at indices i and j (with i < j). If s[i] == s[j], it means we've potentially extended a repeating substring. The length of this extended substring is 1 plus the length of the substring ending at i-1 and j-1 (dp[i-1][j-1] + 1). If i is 0, it means this is the first occurrence of this character, so the length is simply 1. Otherwise, if s[i] != s[j], there's no extension, and dp[i][j] remains 0.

The algorithm iterates through all possible pairs of indices (i, j) with i < j, updating dp[i][j] accordingly and tracking the maximum length encountered so far.

Time and Space Complexity Analysis

  • Time Complexity: O(n^2), where n is the length of the string. This is because we iterate through all pairs of indices (i, j), resulting in a nested loop with O(n^2) iterations.

  • Space Complexity: O(n^2) to store the dp table.

Code Implementation in Different Languages

The code implementations below follow the dynamic programming approach explained above.

Python

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}
 
func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

All these code snippets implement the same dynamic programming solution, differing only in syntax and language-specific features. They all achieve a time complexity of O(n^2) and a space complexity of O(n^2).