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.
class Solution:
def equalCountSubstrings(self, s: str, count: int) -> int:
n = len(s)
if count > n:
return 0
counter = [[0] * 26 for _ in range(n + 1)]
def check(i, j):
c1 = counter[i]
c2 = counter[j + 1]
for k in range(26):
if c2[k] == 0 or c1[k] == c2[k]:
continue
if c2[k] - c1[k] != count:
return False
return True
ans = 0
for i, c in enumerate(s):
idx = ord(c) - ord('a')
for j in range(26):
counter[i + 1][j] = counter[i][j]
counter[i + 1][idx] = counter[i][idx] + 1
l = 0
for _ in range(26):
l += count
j = i - l + 1
if j < 0:
break
ans += check(j, i)
return ansclass Solution {
public int equalCountSubstrings(String s, int count) {
int n = s.length();
if (count > n) {
return 0;
}
int[][] counter = new int[n + 1][26];
int ans = 0;
for (int i = 0; i < n; ++i) {
int idx = s.charAt(i) - 'a';
for (int j = 0; j < 26; ++j) {
counter[i + 1][j] = counter[i][j];
}
counter[i + 1][idx] = counter[i][idx] + 1;
int l = 0;
for (int k = 0; k < 26; ++k) {
l += count;
int j = i - l + 1;
if (j < 0) {
break;
}
ans += check(j, i, count, counter) ? 1 : 0;
}
}
return ans;
}
private boolean check(int i, int j, int count, int[][] counter) {
int[] c1 = counter[i];
int[] c2 = counter[j + 1];
for (int k = 0; k < 26; ++k) {
if (c2[k] == 0 || c1[k] == c2[k]) {
continue;
}
if (c2[k] - c1[k] != count) {
return false;
}
}
return true;
}
}class Solution {
public:
int equalCountSubstrings(string s, int count) {
int n = s.size();
if (count > n) return 0;
vector<vector<int>> counter(n + 1, vector<int>(26));
int ans = 0;
for (int i = 0; i < n; ++i)
{
int idx = s[i] - 'a';
for (int j = 0; j < 26; ++j) counter[i + 1][j] = counter[i][j];
counter[i + 1][idx] = counter[i][idx] + 1;
int l = 0;
for (int k = 0; k < 26; ++k)
{
l += count;
int j = i - l + 1;
if (j < 0) break;
ans += check(j, i, count, counter);
}
}
return ans;
}
bool check(int i, int j, int count, vector<vector<int>>& counter) {
auto& c1 = counter[i];
auto& c2 = counter[j + 1];
for (int k = 0; k < 26; ++k)
{
if (c2[k] == 0 || c1[k] == c2[k]) continue;
if (c2[k] - c1[k] != count) return false;
}
return true;
}
};func equalCountSubstrings(s string, count int) int {
n := len(s)
if count > n {
return 0
}
counter := make([][]int, n+1)
for i := range counter {
counter[i] = make([]int, 26)
}
ans := 0
check := func(i, j int) bool {
c1, c2 := counter[i], counter[j+1]
for k := 0; k < 26; k++ {
if c2[k] == 0 || c1[k] == c2[k] {
continue
}
if c2[k]-c1[k] != count {
return false
}
}
return true
}
for i, c := range s {
idx := c - 'a'
for j := 0; j < 26; j++ {
counter[i+1][j] = counter[i][j]
}
counter[i+1][idx] = counter[i][idx] + 1
l := 0
for k := 0; k < 26; k++ {
l += count
j := i - l + 1
if j < 0 {
break
}
if check(j, i) {
ans++
}
}
}
return ans
}