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Formatted question description: https://leetcode.ca/all/2067.html

# 2067. Number of Equal Count Substrings

## Description

You are given a 0-indexed string s consisting of only lowercase English letters, and an integer count. A substring of s is said to be an equal count substring if, for each unique letter in the substring, it appears exactly count times in the substring.

Return the number of equal count substrings in s.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: s = "aaabcbbcc", count = 3
Output: 3
Explanation:
The substring that starts at index 0 and ends at index 2 is "aaa".
The letter 'a' in the substring appears exactly 3 times.
The substring that starts at index 3 and ends at index 8 is "bcbbcc".
The letters 'b' and 'c' in the substring appear exactly 3 times.
The substring that starts at index 0 and ends at index 8 is "aaabcbbcc".
The letters 'a', 'b', and 'c' in the substring appear exactly 3 times.


Example 2:

Input: s = "abcd", count = 2
Output: 0
Explanation:
The number of times each letter appears in s is less than count.
Therefore, no substrings in s are equal count substrings, so return 0.


Example 3:

Input: s = "a", count = 5
Output: 0
Explanation:
The number of times each letter appears in s is less than count.
Therefore, no substrings in s are equal count substrings, so return 0

Constraints:

• 1 <= s.length <= 3 * 104
• 1 <= count <= 3 * 104
• s consists only of lowercase English letters.

## Solutions

• class Solution {
public int equalCountSubstrings(String s, int count) {
int ans = 0;
int n = s.length();
for (int x = 1; x < 27 && count * x <= n; ++x) {
int m = count * x;
int[] cnt = new int[26];
int y = 0;
for (int i = 0; i < n; ++i) {
int a = s.charAt(i) - 'a';
++cnt[a];
if (cnt[a] == count) {
++y;
}
if (cnt[a] == count + 1) {
--y;
}
int j = i - m;
if (j >= 0) {
int b = s.charAt(j) - 'a';
--cnt[b];
if (cnt[b] == count) {
++y;
}
if (cnt[b] == count - 1) {
--y;
}
}
if (x == y) {
++ans;
}
}
}
return ans;
}
}

• class Solution {
public:
int equalCountSubstrings(string s, int count) {
int ans = 0;
int n = s.size();
int cnt[26];
for (int x = 1; x < 27 && count * x <= n; ++x) {
int m = count * x;
memset(cnt, 0, sizeof cnt);
int y = 0;
for (int i = 0; i < n; ++i) {
int a = s[i] - 'a';
++cnt[a];
y += cnt[a] == count;
y -= cnt[a] == count + 1;
int j = i - m;
if (j >= 0) {
int b = s[j] - 'a';
--cnt[b];
y += cnt[b] == count;
y -= cnt[b] == count - 1;
}
ans += x == y;
}
}
return ans;
}
};

• class Solution:
def equalCountSubstrings(self, s: str, count: int) -> int:
ans = 0
for x in range(1, 27):
m = count * x
if m > len(s):
break
cnt = Counter()
y = 0
for i, c in enumerate(s):
cnt[c] += 1
y += cnt[c] == count
y -= cnt[c] == count + 1
j = i - m
if j >= 0:
cnt[s[j]] -= 1
y += cnt[s[j]] == count
y -= cnt[s[j]] == count - 1
ans += x == y
return ans


• func equalCountSubstrings(s string, count int) (ans int) {
n := len(s)
for x := 1; x < 27 && x*count <= n; x++ {
m := x * count
y := 0
cnt := [26]int{}
for i, c := range s {
a := c - 'a'
cnt[a]++
if cnt[a] == count {
y++
}
if cnt[a] == count+1 {
y--
}
j := i - m
if j >= 0 {
b := s[j] - 'a'
cnt[b]--
if cnt[b] == count {
y++
}
if cnt[b] == count-1 {
y--
}
}
if x == y {
ans++
}
}
}
return
}

• /**
* @param {string} s
* @param {number} count
* @return {number}
*/
var equalCountSubstrings = function (s, count) {
let ans = 0;
const n = s.length;
for (let x = 1; x <= 26 && x * count <= n; ++x) {
const m = x * count;
const cnt = new Array(26).fill(0);
let y = 0;
for (let i = 0; i < n; ++i) {
const a = s.charCodeAt(i) - 'a'.charCodeAt(0);
++cnt[a];
y += cnt[a] == count;
y -= cnt[a] == count + 1;
const j = i - m;
if (j >= 0) {
const b = s.charCodeAt(j) - 'a'.charCodeAt(0);
--cnt[b];
y += cnt[b] == count;
y -= cnt[b] == count - 1;
}
ans += x == y;
}
}
return ans;
};