# 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) {
}
}
}
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:
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 {
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
}