Skip to content

feat: update solutions to lc problem: No.1456 #2554

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
merged 1 commit into from
Apr 8, 2024
Merged

feat: update solutions to lc problem: No.1456 #2554

Show file tree
Hide file tree
Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
100 changes: 48 additions & 52 deletions ...400-1499/1456.Maximum Number of Vowels in a Substring of Given Length/README.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -64,43 +64,45 @@

### 方法一：滑动窗口

找出所有窗口为 $k$ 中的元音字母数目，并记录最大值。
我们首先统计前 $k$ 个字符中元音字母的个数，记为 $cnt$，初始化答案 $ans$ 为 $cnt$。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。
然后我们从 $k$ 开始遍历字符串，每次遍历时，我们将当前字符加入窗口，如果当前字符是元音字母，则 $cnt$ 加一；将窗口第一个字符移出窗口，如果移除的字符是元音字母，则 $cnt$ 减一。然后，我们更新答案 $ans = \max(ans, cnt)$。

遍历结束后，返回答案即可。

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

<!-- tabs:start -->

```python
class Solution:
def maxVowels(self, s: str, k: int) -> int:
vowels = set('aeiou')
t = sum(c in vowels for c in s[:k])
ans = t
vowels = set("aeiou")
ans = cnt = sum(c in vowels for c in s[:k])
for i in range(k, len(s)):
t += s[i] in vowels
t -= s[i - k] in vowels
ans = max(ans, t)
cnt += int(s[i] in vowels) - int(s[i - k] in vowels)
ans = max(ans, cnt)
return ans
```

```java
class Solution {
public int maxVowels(String s, int k) {
int t = 0, n = s.length();
int cnt = 0;
for (int i = 0; i < k; ++i) {
if (isVowel(s.charAt(i))) {
++t;
++cnt;
}
}
int ans = t;
for (int i = k; i < n; ++i) {
int ans = cnt;
for (int i = k; i < s.length(); ++i) {
if (isVowel(s.charAt(i))) {
++t;
++cnt;
}
if (isVowel(s.charAt(i - k))) {
--t;
--cnt;
}
ans = Math.max(ans, t);
ans = Math.max(ans, cnt);
}
return ans;
}
Expand All @@ -115,20 +117,17 @@ class Solution {
class Solution {
public:
int maxVowels(string s, int k) {
int t = 0, n = s.size();
for (int i = 0; i < k; ++i) t += isVowel(s[i]);
int ans = t;
for (int i = k; i < n; ++i) {
t += isVowel(s[i]);
t -= isVowel(s[i - k]);
ans = max(ans, t);
auto isVowel = [](char c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
};
int cnt = count_if(s.begin(), s.begin() + k, isVowel);
int ans = cnt;
for (int i = k; i < s.size(); ++i) {
cnt += isVowel(s[i]) - isVowel(s[i - k]);
ans = max(ans, cnt);
}
return ans;
}

bool isVowel(char c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
};
```

Expand All @@ -137,46 +136,44 @@ func maxVowels(s string, k int) int {
isVowel := func(c byte) bool {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'
}
t := 0
cnt := 0
for i := 0; i < k; i++ {
if isVowel(s[i]) {
t++
cnt++
}
}
ans := t
ans := cnt
for i := k; i < len(s); i++ {
if isVowel(s[i]) {
t++
}
if isVowel(s[i-k]) {
t--
cnt--
}
if isVowel(s[i]) {
cnt++
}
ans = max(ans, t)
ans = max(ans, cnt)
}
return ans
}
```

```ts
function maxVowels(s: string, k: number): number {
function isVowel(c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
let t = 0;
for (let i = 0; i < k; ++i) {
if (isVowel(s.charAt(i))) {
t++;
const isVowel = (c: string) => ['a', 'e', 'i', 'o', 'u'].includes(c);
let cnt = 0;
for (let i = 0; i < k; i++) {
if (isVowel(s[i])) {
cnt++;
}
}
let ans = t;
for (let i = k; i < s.length; ++i) {
if (isVowel(s.charAt(i))) {
t++;
let ans = cnt;
for (let i = k; i < s.length; i++) {
if (isVowel(s[i])) {
cnt++;
}
if (isVowel(s.charAt(i - k))) {
t--;
if (isVowel(s[i - k])) {
cnt--;
}
ans = Math.max(ans, t);
ans = Math.max(ans, cnt);
}
return ans;
}
Expand All @@ -194,23 +191,22 @@ class Solution {
}
function maxVowels($s, $k) {
$cnt = 0;
$rs = 0;
for ($i = 0; $i < $k; $i++) {
if ($this->isVowel($s[$i])) {
$cnt++;
}
}
$rs = $cnt;
$ans = $cnt;
for ($j = $k; $j < strlen($s); $j++) {
if ($this->isVowel($s[$j - $k])) {
$cnt--;
}
if ($this->isVowel($s[$j])) {
$cnt++;
}
$rs = max($rs, $cnt);
$ans = max($ans, $cnt);
}
return $rs;
return $ans;
}
}
```
Expand Down
102 changes: 51 additions & 51 deletions ...-1499/1456.Maximum Number of Vowels in a Substring of Given Length/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -46,41 +46,47 @@

## Solutions

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

First, we count the number of vowels in the first $k$ characters, denoted as $cnt$, and initialize the answer $ans$ as $cnt$.

Then we start traversing the string from $k$. For each iteration, we add the current character to the window. If the current character is a vowel, we increment $cnt$. We remove the first character from the window. If the removed character is a vowel, we decrement $cnt$. Then, we update the answer $ans = \max(ans, cnt)$.

After the traversal, we return the answer.

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

<!-- tabs:start -->

```python
class Solution:
def maxVowels(self, s: str, k: int) -> int:
vowels = set('aeiou')
t = sum(c in vowels for c in s[:k])
ans = t
vowels = set("aeiou")
ans = cnt = sum(c in vowels for c in s[:k])
for i in range(k, len(s)):
t += s[i] in vowels
t -= s[i - k] in vowels
ans = max(ans, t)
cnt += int(s[i] in vowels) - int(s[i - k] in vowels)
ans = max(ans, cnt)
return ans
```

```java
class Solution {
public int maxVowels(String s, int k) {
int t = 0, n = s.length();
int cnt = 0;
for (int i = 0; i < k; ++i) {
if (isVowel(s.charAt(i))) {
++t;
++cnt;
}
}
int ans = t;
for (int i = k; i < n; ++i) {
int ans = cnt;
for (int i = k; i < s.length(); ++i) {
if (isVowel(s.charAt(i))) {
++t;
++cnt;
}
if (isVowel(s.charAt(i - k))) {
--t;
--cnt;
}
ans = Math.max(ans, t);
ans = Math.max(ans, cnt);
}
return ans;
}
Expand All @@ -95,20 +101,17 @@ class Solution {
class Solution {
public:
int maxVowels(string s, int k) {
int t = 0, n = s.size();
for (int i = 0; i < k; ++i) t += isVowel(s[i]);
int ans = t;
for (int i = k; i < n; ++i) {
t += isVowel(s[i]);
t -= isVowel(s[i - k]);
ans = max(ans, t);
auto isVowel = [](char c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
};
int cnt = count_if(s.begin(), s.begin() + k, isVowel);
int ans = cnt;
for (int i = k; i < s.size(); ++i) {
cnt += isVowel(s[i]) - isVowel(s[i - k]);
ans = max(ans, cnt);
}
return ans;
}

bool isVowel(char c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
};
```

Expand All @@ -117,46 +120,44 @@ func maxVowels(s string, k int) int {
isVowel := func(c byte) bool {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'
}
t := 0
cnt := 0
for i := 0; i < k; i++ {
if isVowel(s[i]) {
t++
cnt++
}
}
ans := t
ans := cnt
for i := k; i < len(s); i++ {
if isVowel(s[i]) {
t++
}
if isVowel(s[i-k]) {
t--
cnt--
}
if isVowel(s[i]) {
cnt++
}
ans = max(ans, t)
ans = max(ans, cnt)
}
return ans
}
```

```ts
function maxVowels(s: string, k: number): number {
function isVowel(c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
let t = 0;
for (let i = 0; i < k; ++i) {
if (isVowel(s.charAt(i))) {
t++;
const isVowel = (c: string) => ['a', 'e', 'i', 'o', 'u'].includes(c);
let cnt = 0;
for (let i = 0; i < k; i++) {
if (isVowel(s[i])) {
cnt++;
}
}
let ans = t;
for (let i = k; i < s.length; ++i) {
if (isVowel(s.charAt(i))) {
t++;
let ans = cnt;
for (let i = k; i < s.length; i++) {
if (isVowel(s[i])) {
cnt++;
}
if (isVowel(s.charAt(i - k))) {
t--;
if (isVowel(s[i - k])) {
cnt--;
}
ans = Math.max(ans, t);
ans = Math.max(ans, cnt);
}
return ans;
}
Expand All @@ -174,23 +175,22 @@ class Solution {
}
function maxVowels($s, $k) {
$cnt = 0;
$rs = 0;
for ($i = 0; $i < $k; $i++) {
if ($this->isVowel($s[$i])) {
$cnt++;
}
}
$rs = $cnt;
$ans = $cnt;
for ($j = $k; $j < strlen($s); $j++) {
if ($this->isVowel($s[$j - $k])) {
$cnt--;
}
if ($this->isVowel($s[$j])) {
$cnt++;
}
$rs = max($rs, $cnt);
$ans = max($ans, $cnt);
}
return $rs;
return $ans;
}
}
```
Expand Down
19 changes: 8 additions & 11 deletions solution/1400-1499/1456.Maximum Number of Vowels in a Substring of Given Length/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,18 +1,15 @@
class Solution {
public:
int maxVowels(string s, int k) {
int t = 0, n = s.size();
for (int i = 0; i < k; ++i) t += isVowel(s[i]);
int ans = t;
for (int i = k; i < n; ++i) {
t += isVowel(s[i]);
t -= isVowel(s[i - k]);
ans = max(ans, t);
auto isVowel = [](char c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
};
int cnt = count_if(s.begin(), s.begin() + k, isVowel);
int ans = cnt;
for (int i = k; i < s.size(); ++i) {
cnt += isVowel(s[i]) - isVowel(s[i - k]);
ans = max(ans, cnt);
}
return ans;
}

bool isVowel(char c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
};
Loading