README_EN.md

1316. Distinct Echo Substrings

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Description

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.

Solutions

Python3

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)

Java

class Solution {
    private long[] h;
    private long[] p;

    public int distinctEchoSubstrings(String text) {
        int n = text.length();
        int base = 131;
        h = new long[n + 10];
        p = new long[n + 10];
        p[0] = 1;
        for (int i = 0; i < n; ++i) {
            int t = text.charAt(i) - 'a' + 1;
            h[i + 1] = h[i] * base + t;
            p[i + 1] = p[i] * base;
        }
        Set<Long> vis = new HashSet<>();
        for (int i = 0; i < n - 1; ++i) {
            for (int j = i + 1; j < n; j += 2) {
                int k = (i + j) >> 1;
                long a = get(i + 1, k + 1);
                long b = get(k + 2, j + 1);
                if (a == b) {
                    vis.add(a);
                }
            }
        }
        return vis.size();
    }

    private long get(int i, int j) {
        return h[j] - h[i - 1] * p[j - i + 1];
    }
}

C++

typedef unsigned long long ull;

class Solution {
public:
    int distinctEchoSubstrings(string text) {
        int n = text.size();
        int base = 131;
        vector<ull> p(n + 10);
        vector<ull> h(n + 10);
        p[0] = 1;
        for (int i = 0; i < n; ++i) {
            int t = text[i] - 'a' + 1;
            p[i + 1] = p[i] * base;
            h[i + 1] = h[i] * base + t;
        }
        unordered_set<ull> vis;
        for (int i = 0; i < n - 1; ++i) {
            for (int j = i + 1; j < n; j += 2) {
                int k = (i + j) >> 1;
                ull a = get(i + 1, k + 1, p, h);
                ull b = get(k + 2, j + 1, p, h);
                if (a == b) vis.insert(a);
            }
        }
        return vis.size();
    }

    ull get(int l, int r, vector<ull>& p, vector<ull>& h) {
        return h[r] - h[l - 1] * p[r - l + 1];
    }
};

Go

func distinctEchoSubstrings(text string) int {
	n := len(text)
	base := 131
	h := make([]int, n+10)
	p := make([]int, n+10)
	p[0] = 1
	for i, c := range text {
		t := int(c-'a') + 1
		p[i+1] = p[i] * base
		h[i+1] = h[i]*base + t
	}
	get := func(l, r int) int {
		return h[r] - h[l-1]*p[r-l+1]
	}
	vis := map[int]bool{}
	for i := 0; i < n-1; i++ {
		for j := i + 1; j < n; j += 2 {
			k := (i + j) >> 1
			a, b := get(i+1, k+1), get(k+2, j+1)
			if a == b {
				vis[a] = true
			}
		}
	}
	return len(vis)
}

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