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true |
Medium |
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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 * 1041 <= count <= 3 * 104sconsists only of lowercase English letters.
We can enumerate the number of types of letters in the substring within the range of
Next, we take the current substring length as the size of the window, count the number of types of letters in the window size that are equal to
The time complexity is
class Solution:
def equalCountSubstrings(self, s: str, count: int) -> int:
ans = 0
for i in range(1, 27):
k = i * count
if k > len(s):
break
cnt = Counter()
t = 0
for j, c in enumerate(s):
cnt[c] += 1
t += cnt[c] == count
t -= cnt[c] == count + 1
if j >= k:
cnt[s[j - k]] -= 1
t += cnt[s[j - k]] == count
t -= cnt[s[j - k]] == count - 1
ans += i == t
return ansclass Solution {
public int equalCountSubstrings(String s, int count) {
int ans = 0;
int[] cnt = new int[26];
int n = s.length();
for (int i = 1; i < 27 && i * count <= n; ++i) {
int k = i * count;
Arrays.fill(cnt, 0);
int t = 0;
for (int j = 0; j < n; ++j) {
int a = s.charAt(j) - 'a';
++cnt[a];
t += cnt[a] == count ? 1 : 0;
t -= cnt[a] == count + 1 ? 1 : 0;
if (j - k >= 0) {
int b = s.charAt(j - k) - 'a';
--cnt[b];
t += cnt[b] == count ? 1 : 0;
t -= cnt[b] == count - 1 ? 1 : 0;
}
ans += i == t ? 1 : 0;
}
}
return ans;
}
}class Solution {
public:
int equalCountSubstrings(string s, int count) {
int ans = 0;
int n = s.size();
int cnt[26];
for (int i = 1; i < 27 && i * count <= n; ++i) {
int k = i * count;
memset(cnt, 0, sizeof(cnt));
int t = 0;
for (int j = 0; j < n; ++j) {
int a = s[j] - 'a';
t += ++cnt[a] == count;
t -= cnt[a] == count + 1;
if (j >= k) {
int b = s[j - k] - 'a';
t += --cnt[b] == count;
t -= cnt[b] == count - 1;
}
ans += i == t;
}
}
return ans;
}
};func equalCountSubstrings(s string, count int) (ans int) {
n := len(s)
for i := 1; i < 27 && i*count <= n; i++ {
k := i * count
cnt := [26]int{}
t := 0
for j, c := range s {
a := c - 'a'
cnt[a]++
if cnt[a] == count {
t++
} else if cnt[a] == count+1 {
t--
}
if j >= k {
b := s[j-k] - 'a'
cnt[b]--
if cnt[b] == count {
t++
} else if cnt[b] == count-1 {
t--
}
}
if i == t {
ans++
}
}
}
return
}function equalCountSubstrings(s: string, count: number): number {
const n = s.length;
let ans = 0;
for (let i = 1; i < 27 && i * count <= n; ++i) {
const k = i * count;
const cnt: number[] = Array(26).fill(0);
let t = 0;
for (let j = 0; j < n; ++j) {
const a = s.charCodeAt(j) - 97;
t += ++cnt[a] === count ? 1 : 0;
t -= cnt[a] === count + 1 ? 1 : 0;
if (j >= k) {
const b = s.charCodeAt(j - k) - 97;
t += --cnt[b] === count ? 1 : 0;
t -= cnt[b] === count - 1 ? 1 : 0;
}
ans += i === t ? 1 : 0;
}
}
return ans;
}/**
* @param {string} s
* @param {number} count
* @return {number}
*/
var equalCountSubstrings = function (s, count) {
const n = s.length;
let ans = 0;
for (let i = 1; i < 27 && i * count <= n; ++i) {
const k = i * count;
const cnt = Array(26).fill(0);
let t = 0;
for (let j = 0; j < n; ++j) {
const a = s.charCodeAt(j) - 97;
t += ++cnt[a] === count ? 1 : 0;
t -= cnt[a] === count + 1 ? 1 : 0;
if (j >= k) {
const b = s.charCodeAt(j - k) - 97;
t += --cnt[b] === count ? 1 : 0;
t -= cnt[b] === count - 1 ? 1 : 0;
}
ans += i === t ? 1 : 0;
}
}
return ans;
};