1316. Distinct Echo Substrings

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

Solution 1

Python3JavaC++GoRust

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)

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];
    }
}

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];
    }
};

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)
}

use std::collections::HashSet;

const BASE: u64 = 131;

impl Solution {
    #[allow(dead_code)]
    pub fn distinct_echo_substrings(text: String) -> i32 {
        let n = text.len();
        let mut vis: HashSet<u64> = HashSet::new();
        let mut base_vec: Vec<u64> = vec![1; n + 1];
        let mut hash_vec: Vec<u64> = vec![0; n + 1];

        // Initialize the base vector & hash vector
        for i in 0..n {
            let cur_char = ((text.chars().nth(i).unwrap() as u8) - ('a' as u8) + 1) as u64;
            // Update base vector
            base_vec[i + 1] = base_vec[i] * BASE;
            // Update hash vector
            hash_vec[i + 1] = hash_vec[i] * BASE + cur_char;
        }

        // Traverse the text to find the result pair, using rolling hash
        for i in 0..n - 1 {
            for j in i + 1..n {
                // Prevent overflow
                let k = i + (j - i) / 2;
                let left = Self::get_hash(i + 1, k + 1, &base_vec, &hash_vec);
                let right = Self::get_hash(k + 2, j + 1, &base_vec, &hash_vec);
                if left == right {
                    vis.insert(left);
                }
            }
        }

        vis.len() as i32
    }

    #[allow(dead_code)]
    fn get_hash(start: usize, end: usize, base_vec: &Vec<u64>, hash_vec: &Vec<u64>) -> u64 {
        hash_vec[end] - hash_vec[start - 1] * base_vec[end - start + 1]
    }
}

Comments