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feat: add solutions to lc problem: No.2067 #2570

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Apr 11, 2024
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feat: add solutions to lc problem: No.2067 #2570

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da5a858
feat: add solutions to lc problem: No.2067
yanglbme Apr 11, 2024
f84a1e6
fix: solution
yanglbme Apr 11, 2024
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180 changes: 92 additions & 88 deletions solution/2000-2099/2067.Number of Equal Count Substrings/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -63,9 +63,9 @@

### 方法一：枚举 + 滑动窗口

我们可以在 $[1..26]$ 范围内枚举子串的字母种类数 $x$，那么子串长度就是 $count \times x$。
我们可以在 $[1..26]$ 范围内枚举子串的字母种类数 $i$，那么子串长度就是 $i \times count$。

接下来，我们将当前子串长度作为窗口的大小，统计窗口大小中有多少种字母的个数为 $count$，记录在 $y$ 中。如果此时 $x = y$，说明当前窗口中的字母个数都为 $count$，那么就可以将答案加一。
接下来，我们将当前子串长度作为窗口的大小，统计窗口大小中有多少种字母的个数为 $count$，记录在 $t$ 中。如果此时 $i = t$，说明当前窗口中的字母个数都为 $count$，那么就可以将答案加一。

时间复杂度 $O(n \times C)$，空间复杂度 $O(C)$。其中 $n$ 为字符串 $s$ 的长度，而 $C$ 为字母的种类数，本题中 $C = 26$。

Expand All @@ -75,57 +75,46 @@
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):
for i in range(1, 27):
k = i * count
if k > len(s):
break
cnt = Counter()
y = 0
for i, c in enumerate(s):
t = 0
for j, 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
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 ans
```

```java
class Solution {
public int equalCountSubstrings(String s, int count) {
int ans = 0;
int[] cnt = new int[26];
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';
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];
if (cnt[a] == count) {
++y;
}
if (cnt[a] == count + 1) {
--y;
}
int j = i - m;
if (j >= 0) {
int b = s.charAt(j) - '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];
if (cnt[b] == count) {
++y;
}
if (cnt[b] == count - 1) {
--y;
}
}
if (x == y) {
++ans;
t += cnt[b] == count ? 1 : 0;
t -= cnt[b] == count - 1 ? 1 : 0;
}
ans += i == t ? 1 : 0;
}
}
return ans;
Expand All @@ -140,23 +129,20 @@ public:
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;
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 += x == y;
ans += i == t;
}
}
return ans;
Expand All @@ -167,31 +153,28 @@ public:
```go
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
for i := 1; i < 27 && i*count <= n; i++ {
k := i * count
cnt := [26]int{}
for i, c := range s {
t := 0
for j, c := range s {
a := c - 'a'
cnt[a]++
if cnt[a] == count {
y++
t++
} else if cnt[a] == count+1 {
t--
}
if cnt[a] == count+1 {
y--
}
j := i - m
if j >= 0 {
b := s[j] - 'a'
if j >= k {
b := s[j-k] - 'a'
cnt[b]--
if cnt[b] == count {
y++
}
if cnt[b] == count-1 {
y--
t++
} else if cnt[b] == count-1 {
t--
}
}
if x == y {
if i == t {
ans++
}
}
Expand All @@ -200,32 +183,53 @@ func equalCountSubstrings(s string, count int) (ans int) {
}
```

```ts
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;
}
```

```js
/**
* @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;
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 += x == y;
ans += i === t ? 1 : 0;
}
}
return ans;
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