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Formatted question description: https://leetcode.ca/all/1316.html

1316. Distinct Echo Substrings (Hard)

Return the number of distinct non-empty substrings of text that can be written as the concatenation of some string with itself (i.e. it can be written as a + a where a is some string).

 

Example 1:

Input: text = "abcabcabc"
Output: 3
Explanation: The 3 substrings are "abcabc", "bcabca" and "cabcab".

Example 2:

Input: text = "leetcodeleetcode"
Output: 2
Explanation: The 2 substrings are "ee" and "leetcodeleetcode".

 

Constraints:

  • 1 <= text.length <= 2000
  • text has only lowercase English letters.

Related Topics:
String, Rolling Hash

Solution 1. Rolling Hash

// OJ: https://leetcode.com/problems/distinct-echo-substrings/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int distinctEchoSubstrings(string s) {
        int N = s.size(), mod = 1e8 + 7, cnt = 0;
        vector<long> v(N);
        for (int len = 1; len <= N / 2; ++len) {
            long hash = s[0] - 'a', p = 1;
            unordered_set<string> seen;
            for (int i = 1; i < N; ++i) {
                if (i < len) {
                    hash = (hash * 26 + s[i] - 'a') % mod;
                    p = (p * 26) % mod;
                } else {
                    hash = ((hash - (s[i - len] - 'a') * p) * 26 + s[i] - 'a') % mod;
                    if (hash < 0) hash += mod;
                    if (i - 2 * len + 1 >= 0 && v[i - len] == hash) {
                        auto a = s.substr(i - 2 * len + 1, len);
                        if (seen.count(a)) continue;
                        seen.insert(a);
                        if (a == s.substr(i - len + 1, len)) ++cnt;
                    }
                }
                v[i] = hash;
            }
        }
        return cnt;
    }
};

  • class Solution {
    public int distinctEchoSubstrings(String text) {
        Set<String> substringsSet = new HashSet<String>();
        int length = text.length();
        for (int i = 0; i < length; i++) {
            int maxLength = length - i;
            for (int j = 2; j <= maxLength; j += 2) {
                String substr1 = text.substring(i, i + j / 2), substr2 = text.substring(i + j / 2, i + j);
                if (substr1.equals(substr2))
                    substringsSet.add(substr1 + substr2);
            }
        }
        return substringsSet.size();
    }
}

  • // OJ: https://leetcode.com/problems/distinct-echo-substrings/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int distinctEchoSubstrings(string s) {
        int N = s.size(), mod = 1e8 + 7, cnt = 0;
        vector<long> v(N);
        for (int len = 1; len <= N / 2; ++len) {
            long hash = s[0] - 'a', p = 1;
            unordered_set<string> seen;
            for (int i = 1; i < N; ++i) {
                if (i < len) {
                    hash = (hash * 26 + s[i] - 'a') % mod;
                    p = (p * 26) % mod;
                } else {
                    hash = ((hash - (s[i - len] - 'a') * p) * 26 + s[i] - 'a') % mod;
                    if (hash < 0) hash += mod;
                    if (i - 2 * len + 1 >= 0 && v[i - len] == hash) {
                        auto a = s.substr(i - 2 * len + 1, len);
                        if (seen.count(a)) continue;
                        seen.insert(a);
                        if (a == s.substr(i - len + 1, len)) ++cnt;
                    }
                }
                v[i] = hash;
            }
        }
        return cnt;
    }
};

  • class Solution:
    def distinctEchoSubstrings(self, text: str) -> int:
        def get(l, r):
            return (h[r] - h[l - 1] * p[r - l + 1]) % mod

        n = len(text)
        base = 131
        mod = int(1e9) + 7
        h = [0] * (n + 10)
        p = [1] * (n + 10)
        for i, c in enumerate(text):
            t = ord(c) - ord('a') + 1
            h[i + 1] = (h[i] * base) % mod + t
            p[i + 1] = (p[i] * base) % mod
        vis = set()
        for i in range(n - 1):
            for j in range(i + 1, n, 2):
                k = (i + j) >> 1
                a = get(i + 1, k + 1)
                b = get(k + 2, j + 1)
                if a == b:
                    vis.add(a)
        return len(vis)

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