Question

1316. Distinct Echo Substrings

中文文档

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

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]
  }
}