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feat: add solutions to lc problem: No.1876 #3594

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126 changes: 101 additions & 25 deletions ...tion/1800-1899/1876.Substrings of Size Three with Distinct Characters/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -64,7 +64,17 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：滑动窗口

我们可以维护一个滑动窗口，使得窗口内的字符不重复。初始时，我们用一个长度为 $26$ 的二进制整数 $\textit{mask}$ 表示窗口内的字符，其中第 $i$ 位为 $1$ 表示字符 $i$ 在窗口内出现过，否则表示字符 $i$ 在窗口内没有出现过。

然后，我们遍历字符串 $s$，对于每一个位置 $r$，如果 $\textit{s}[r]$ 在窗口内出现过，我们需要将窗口的左边界 $l$ 右移，直到窗口内不再有重复的字符。在这之后，我们将 $\textit{s}[r]$ 加入窗口内，此时如果窗口的长度大于等于 $3$，那么我们就找到了一个以 $\textit{s}[r]$ 结尾的长度为 $3$ 的好子字符串。

遍历结束后，我们就找到了所有的好子字符串的数量。

时间复杂度 $O(n)$，其中 $n$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$。

> 该解法可以拓展到长度为 $k$ 的好子字符串的数量。

<!-- tabs:start -->

Expand All @@ -73,44 +83,97 @@ tags:
```python
class Solution:
def countGoodSubstrings(self, s: str) -> int:
count, n = 0, len(s)
for i in range(n - 2):
count += s[i] != s[i + 1] and s[i] != s[i + 2] and s[i + 1] != s[i + 2]
return count
ans = mask = l = 0
for r, x in enumerate(map(lambda c: ord(c) - 97, s)):
while mask >> x & 1:
y = ord(s[l]) - 97
mask ^= 1 << y
l += 1
mask |= 1 << x
ans += int(r - l + 1 >= 3)
return ans
```

#### Java

```java
class Solution {
public int countGoodSubstrings(String s) {
int count = 0, n = s.length();
for (int i = 0; i < n - 2; ++i) {
char a = s.charAt(i), b = s.charAt(i + 1), c = s.charAt(i + 2);
if (a != b && a != c && b != c) {
++count;
int ans = 0;
int n = s.length();
for (int l = 0, r = 0, mask = 0; r < n; ++r) {
int x = s.charAt(r) - 'a';
while ((mask >> x & 1) == 1) {
int y = s.charAt(l++) - 'a';
mask ^= 1 << y;
}
mask |= 1 << x;
ans += r - l + 1 >= 3 ? 1 : 0;
}
return count;
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
int countGoodSubstrings(string s) {
int ans = 0;
int n = s.length();
for (int l = 0, r = 0, mask = 0; r < n; ++r) {
int x = s[r] - 'a';
while ((mask >> x & 1) == 1) {
int y = s[l++] - 'a';
mask ^= 1 << y;
}
mask |= 1 << x;
ans += r - l + 1 >= 3 ? 1 : 0;
}
return ans;
}
};
```

#### Go

```go
func countGoodSubstrings(s string) (ans int) {
mask, l := 0, 0
for r, c := range s {
x := int(c - 'a')
for (mask>>x)&1 == 1 {
y := int(s[l] - 'a')
l++
mask ^= 1 << y
}
mask |= 1 << x
if r-l+1 >= 3 {
ans++
}
}
return
}
```

#### TypeScript

```ts
function countGoodSubstrings(s: string): number {
const n: number = s.length;
let count: number = 0;
for (let i: number = 0; i < n - 2; ++i) {
let a: string = s.charAt(i),
b: string = s.charAt(i + 1),
c: string = s.charAt(i + 2);
if (a != b && a != c && b != c) {
++count;
let ans = 0;
const n = s.length;
for (let l = 0, r = 0, mask = 0; r < n; ++r) {
const x = s.charCodeAt(r) - 'a'.charCodeAt(0);
while ((mask >> x) & 1) {
const y = s.charCodeAt(l++) - 'a'.charCodeAt(0);
mask ^= 1 << y;
}
mask |= 1 << x;
ans += r - l + 1 >= 3 ? 1 : 0;
}
return count;
return ans;
}
```

Expand All @@ -123,13 +186,26 @@ class Solution {
* @return Integer
*/
function countGoodSubstrings($s) {
$cnt = 0;
for ($i = 0; $i < strlen($s) - 2; $i++) {
if ($s[$i] != $s[$i + 1] && $s[$i] != $s[$i + 2] && $s[$i + 1] != $s[$i + 2]) {
$cnt++;
$ans = 0;
$n = strlen($s);
$l = 0;
$r = 0;
$mask = 0;

while ($r < $n) {
$x = ord($s[$r]) - ord('a');
while (($mask >> $x) & 1) {
$y = ord($s[$l++]) - ord('a');
$mask ^= 1 << $y;
}
$mask |= 1 << $x;
if ($r - $l + 1 >= 3) {
$ans++;
}
$r++;
}
return $cnt++;

return $ans;
}
}
```
Expand Down
128 changes: 102 additions & 26 deletions ...n/1800-1899/1876.Substrings of Size Three with Distinct Characters/README_EN.md
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Expand Up @@ -35,7 +35,7 @@ tags:
<pre>
<strong>Input:</strong> s = &quot;xyzzaz&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> There are 4 substrings of size 3: &quot;xyz&quot;, &quot;yzz&quot;, &quot;zza&quot;, and &quot;zaz&quot;.
<strong>Explanation:</strong> There are 4 substrings of size 3: &quot;xyz&quot;, &quot;yzz&quot;, &quot;zza&quot;, and &quot;zaz&quot;.
The only good substring of length 3 is &quot;xyz&quot;.
</pre>

Expand All @@ -62,7 +62,17 @@ The good substrings are &quot;abc&quot;, &quot;bca&quot;, &quot;cab&quot;, and &

<!-- solution:start -->

### Solution 1
### Solution 1: Sliding Window

We can maintain a sliding window such that the characters within the window are not repeated. Initially, we use a binary integer $\textit{mask}$ of length $26$ to represent the characters within the window, where the $i$-th bit being $1$ indicates that character $i$ has appeared in the window, otherwise it indicates that character $i$ has not appeared in the window.

Then, we traverse the string $s$. For each position $r$, if $\textit{s}[r]$ has appeared in the window, we need to move the left boundary $l$ of the window to the right until there are no repeated characters in the window. After this, we add $\textit{s}[r]$ to the window. At this point, if the length of the window is greater than or equal to $3$, then we have found a good substring of length $3$ ending at $\textit{s}[r]$.

After the traversal, we have found the number of all good substrings.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

> This solution can be extended to find the number of good substrings of length $k$.

<!-- tabs:start -->

Expand All @@ -71,44 +81,97 @@ The good substrings are &quot;abc&quot;, &quot;bca&quot;, &quot;cab&quot;, and &
```python
class Solution:
def countGoodSubstrings(self, s: str) -> int:
count, n = 0, len(s)
for i in range(n - 2):
count += s[i] != s[i + 1] and s[i] != s[i + 2] and s[i + 1] != s[i + 2]
return count
ans = mask = l = 0
for r, x in enumerate(map(lambda c: ord(c) - 97, s)):
while mask >> x & 1:
y = ord(s[l]) - 97
mask ^= 1 << y
l += 1
mask |= 1 << x
ans += int(r - l + 1 >= 3)
return ans
```

#### Java

```java
class Solution {
public int countGoodSubstrings(String s) {
int count = 0, n = s.length();
for (int i = 0; i < n - 2; ++i) {
char a = s.charAt(i), b = s.charAt(i + 1), c = s.charAt(i + 2);
if (a != b && a != c && b != c) {
++count;
int ans = 0;
int n = s.length();
for (int l = 0, r = 0, mask = 0; r < n; ++r) {
int x = s.charAt(r) - 'a';
while ((mask >> x & 1) == 1) {
int y = s.charAt(l++) - 'a';
mask ^= 1 << y;
}
mask |= 1 << x;
ans += r - l + 1 >= 3 ? 1 : 0;
}
return count;
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
int countGoodSubstrings(string s) {
int ans = 0;
int n = s.length();
for (int l = 0, r = 0, mask = 0; r < n; ++r) {
int x = s[r] - 'a';
while ((mask >> x & 1) == 1) {
int y = s[l++] - 'a';
mask ^= 1 << y;
}
mask |= 1 << x;
ans += r - l + 1 >= 3 ? 1 : 0;
}
return ans;
}
};
```

#### Go

```go
func countGoodSubstrings(s string) (ans int) {
mask, l := 0, 0
for r, c := range s {
x := int(c - 'a')
for (mask>>x)&1 == 1 {
y := int(s[l] - 'a')
l++
mask ^= 1 << y
}
mask |= 1 << x
if r-l+1 >= 3 {
ans++
}
}
return
}
```

#### TypeScript

```ts
function countGoodSubstrings(s: string): number {
const n: number = s.length;
let count: number = 0;
for (let i: number = 0; i < n - 2; ++i) {
let a: string = s.charAt(i),
b: string = s.charAt(i + 1),
c: string = s.charAt(i + 2);
if (a != b && a != c && b != c) {
++count;
let ans = 0;
const n = s.length;
for (let l = 0, r = 0, mask = 0; r < n; ++r) {
const x = s.charCodeAt(r) - 'a'.charCodeAt(0);
while ((mask >> x) & 1) {
const y = s.charCodeAt(l++) - 'a'.charCodeAt(0);
mask ^= 1 << y;
}
mask |= 1 << x;
ans += r - l + 1 >= 3 ? 1 : 0;
}
return count;
return ans;
}
```

Expand All @@ -121,13 +184,26 @@ class Solution {
* @return Integer
*/
function countGoodSubstrings($s) {
$cnt = 0;
for ($i = 0; $i < strlen($s) - 2; $i++) {
if ($s[$i] != $s[$i + 1] && $s[$i] != $s[$i + 2] && $s[$i + 1] != $s[$i + 2]) {
$cnt++;
$ans = 0;
$n = strlen($s);
$l = 0;
$r = 0;
$mask = 0;

while ($r < $n) {
$x = ord($s[$r]) - ord('a');
while (($mask >> $x) & 1) {
$y = ord($s[$l++]) - ord('a');
$mask ^= 1 << $y;
}
$mask |= 1 << $x;
if ($r - $l + 1 >= 3) {
$ans++;
}
$r++;
}
return $cnt++;

return $ans;
}
}
```
Expand Down
17 changes: 17 additions & 0 deletions solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -0,0 +1,17 @@
class Solution {
public:
int countGoodSubstrings(string s) {
int ans = 0;
int n = s.length();
for (int l = 0, r = 0, mask = 0; r < n; ++r) {
int x = s[r] - 'a';
while ((mask >> x & 1) == 1) {
int y = s[l++] - 'a';
mask ^= 1 << y;
}
mask |= 1 << x;
ans += r - l + 1 >= 3 ? 1 : 0;
}
return ans;
}
};
16 changes: 16 additions & 0 deletions solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.go
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Original file line number Diff line number Diff line change
@@ -0,0 +1,16 @@
func countGoodSubstrings(s string) (ans int) {
mask, l := 0, 0
for r, c := range s {
x := int(c - 'a')
for (mask>>x)&1 == 1 {
y := int(s[l] - 'a')
l++
mask ^= 1 << y
}
mask |= 1 << x
if r-l+1 >= 3 {
ans++
}
}
return
}
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