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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

  • 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];
        }
    };
    
  • 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)
    
    
  • 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]
        }
    }
    
    

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