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feat: add solutions to lc problem: No.1371 (#3868)
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No.1371.Find the Longest Substring Containing Vowels in Even Counts
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@yanglbme
yanglbme authored Dec 19, 2024
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136 changes: 88 additions & 48 deletions ...1399/1371.Find the Longest Substring Containing Vowels in Even Counts/README.md
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<!-- solution:start -->

### 方法一
### 方法一：前缀异或 + 数组或哈希表

根据题目描述，如果我们用一个数字表示字符串 $\textit{s}$ 的某个前缀中每个元音字母出现的次数的奇偶性，那么当两个前缀的这个数字相同时，这两个前缀的交集就是一个符合条件的子字符串。

我们可以用一个二进制数的低五位分别表示五个元音字母的奇偶性，其中第 $i$ 位为 $1$ 表示该元音字母在子字符串中出现了奇数次，为 $0$ 表示该元音字母在子字符串中出现了偶数次。

我们用 $\textit{mask}$ 表示这个二进制数，用一个数组或哈希表 $\textit{d}$ 记录每个 $\textit{mask}$ 第一次出现的位置。初始时，我们将 $\textit{d}[0] = -1$，表示空字符串的开始位置为 $-1$。

遍历字符串 $\textit{s}$，如果遇到元音字母，就将 $\textit{mask}$ 的对应位取反。接下来，我们判断 $\textit{mask}$ 是否在之前出现过，如果出现过，那么我们就找到了一个符合条件的子字符串，其长度为当前位置减去 $\textit{mask}$ 上一次出现的位置。否则，我们将 $\textit{mask}$ 的当前位置存入 $\textit{d}$。

遍历结束后，我们就找到了最长的符合条件的子字符串。

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

<!-- tabs:start -->

Expand All @@ -73,40 +85,39 @@ tags:
```python
class Solution:
def findTheLongestSubstring(self, s: str) -> int:
pos = [inf] * 32
pos[0] = -1
vowels = 'aeiou'
state = ans = 0
d = {0: -1}
ans = mask = 0
for i, c in enumerate(s):
for j, v in enumerate(vowels):
if c == v:
state ^= 1 << j
ans = max(ans, i - pos[state])
pos[state] = min(pos[state], i)
if c in "aeiou":
mask ^= 1 << (ord(c) - ord("a"))
if mask in d:
j = d[mask]
ans = max(ans, i - j)
else:
d[mask] = i
return ans
```

#### Java

```java
class Solution {

public int findTheLongestSubstring(String s) {
int[] pos = new int[32];
Arrays.fill(pos, Integer.MAX_VALUE);
pos[0] = -1;
String vowels = "aeiou";
int state = 0;
int ans = 0;
for (int i = 0; i < s.length(); ++i) {
char c = s.charAt(i);
int[] d = new int[32];
Arrays.fill(d, 1 << 29);
d[0] = 0;
int ans = 0, mask = 0;
for (int i = 1; i <= s.length(); ++i) {
char c = s.charAt(i - 1);
for (int j = 0; j < 5; ++j) {
if (c == vowels.charAt(j)) {
state ^= (1 << j);
mask ^= 1 << j;
break;
}
}
ans = Math.max(ans, i - pos[state]);
pos[state] = Math.min(pos[state], i);
ans = Math.max(ans, i - d[mask]);
d[mask] = Math.min(d[mask], i);
}
return ans;
}
Expand All @@ -119,16 +130,20 @@ class Solution {
class Solution {
public:
int findTheLongestSubstring(string s) {
vector<int> pos(32, INT_MAX);
pos[0] = -1;
string vowels = "aeiou";
int state = 0, ans = 0;
for (int i = 0; i < s.size(); ++i) {
for (int j = 0; j < 5; ++j)
if (s[i] == vowels[j])
state ^= (1 << j);
ans = max(ans, i - pos[state]);
pos[state] = min(pos[state], i);
vector<int> d(32, INT_MAX);
d[0] = 0;
int ans = 0, mask = 0;
for (int i = 1; i <= s.length(); ++i) {
char c = s[i - 1];
for (int j = 0; j < 5; ++j) {
if (c == vowels[j]) {
mask ^= 1 << j;
break;
}
}
ans = max(ans, i - d[mask]);
d[mask] = min(d[mask], i);
}
return ans;
}
Expand All @@ -138,24 +153,49 @@ public:
#### Go
```go
func findTheLongestSubstring(s string) int {
pos := make([]int, 32)
for i := range pos {
pos[i] = math.MaxInt32
}
pos[0] = -1
vowels := "aeiou"
state, ans := 0, 0
for i, c := range s {
for j, v := range vowels {
if c == v {
state ^= (1 << j)
}
}
ans = max(ans, i-pos[state])
pos[state] = min(pos[state], i)
}
return ans
func findTheLongestSubstring(s string) (ans int) {
vowels := "aeiou"
d := [32]int{}
for i := range d {
d[i] = 1 << 29
}
d[0] = 0
mask := 0
for i := 1; i <= len(s); i++ {
c := s[i-1]
for j := 0; j < 5; j++ {
if c == vowels[j] {
mask ^= 1 << j
break
}
}
ans = max(ans, i-d[mask])
d[mask] = min(d[mask], i)
}
return
}
```

#### TypeScript

```ts
function findTheLongestSubstring(s: string): number {
const vowels = 'aeiou';
const d: number[] = Array(32).fill(1 << 29);
d[0] = 0;
let [ans, mask] = [0, 0];
for (let i = 1; i <= s.length; i++) {
const c = s[i - 1];
for (let j = 0; j < 5; j++) {
if (c === vowels[j]) {
mask ^= 1 << j;
break;
}
}
ans = Math.max(ans, i - d[mask]);
d[mask] = Math.min(d[mask], i);
}
return ans;
}
```

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136 changes: 88 additions & 48 deletions ...9/1371.Find the Longest Substring Containing Vowels in Even Counts/README_EN.md
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Expand Up @@ -62,7 +62,19 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Prefix XOR + Array or Hash Table

According to the problem description, if we use a number to represent the parity of the occurrences of each vowel in a prefix of the string $\textit{s}$, then when two prefixes have the same number, the substring between these two prefixes is a valid substring.

We can use the lower five bits of a binary number to represent the parity of the five vowels, where the $i$-th bit being $1$ means the vowel appears an odd number of times in the substring, and $0$ means it appears an even number of times.

We use $\textit{mask}$ to represent this binary number and use an array or hash table $\textit{d}$ to record the first occurrence of each $\textit{mask}$. Initially, we set $\textit{d}[0] = -1$, indicating that the start position of the empty string is $-1$.

We traverse the string $\textit{s}$, and if we encounter a vowel, we toggle the corresponding bit in $\textit{mask}$. Next, we check if $\textit{mask}$ has appeared before. If it has, we have found a valid substring, and its length is the current position minus the last occurrence of $\textit{mask}$. Otherwise, we store the current position of $\textit{mask}$ in $\textit{d}$.

After traversing the string, we will have found the longest valid substring.

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

<!-- tabs:start -->

Expand All @@ -71,40 +83,39 @@ tags:
```python
class Solution:
def findTheLongestSubstring(self, s: str) -> int:
pos = [inf] * 32
pos[0] = -1
vowels = 'aeiou'
state = ans = 0
d = {0: -1}
ans = mask = 0
for i, c in enumerate(s):
for j, v in enumerate(vowels):
if c == v:
state ^= 1 << j
ans = max(ans, i - pos[state])
pos[state] = min(pos[state], i)
if c in "aeiou":
mask ^= 1 << (ord(c) - ord("a"))
if mask in d:
j = d[mask]
ans = max(ans, i - j)
else:
d[mask] = i
return ans
```

#### Java

```java
class Solution {

public int findTheLongestSubstring(String s) {
int[] pos = new int[32];
Arrays.fill(pos, Integer.MAX_VALUE);
pos[0] = -1;
String vowels = "aeiou";
int state = 0;
int ans = 0;
for (int i = 0; i < s.length(); ++i) {
char c = s.charAt(i);
int[] d = new int[32];
Arrays.fill(d, 1 << 29);
d[0] = 0;
int ans = 0, mask = 0;
for (int i = 1; i <= s.length(); ++i) {
char c = s.charAt(i - 1);
for (int j = 0; j < 5; ++j) {
if (c == vowels.charAt(j)) {
state ^= (1 << j);
mask ^= 1 << j;
break;
}
}
ans = Math.max(ans, i - pos[state]);
pos[state] = Math.min(pos[state], i);
ans = Math.max(ans, i - d[mask]);
d[mask] = Math.min(d[mask], i);
}
return ans;
}
Expand All @@ -117,16 +128,20 @@ class Solution {
class Solution {
public:
int findTheLongestSubstring(string s) {
vector<int> pos(32, INT_MAX);
pos[0] = -1;
string vowels = "aeiou";
int state = 0, ans = 0;
for (int i = 0; i < s.size(); ++i) {
for (int j = 0; j < 5; ++j)
if (s[i] == vowels[j])
state ^= (1 << j);
ans = max(ans, i - pos[state]);
pos[state] = min(pos[state], i);
vector<int> d(32, INT_MAX);
d[0] = 0;
int ans = 0, mask = 0;
for (int i = 1; i <= s.length(); ++i) {
char c = s[i - 1];
for (int j = 0; j < 5; ++j) {
if (c == vowels[j]) {
mask ^= 1 << j;
break;
}
}
ans = max(ans, i - d[mask]);
d[mask] = min(d[mask], i);
}
return ans;
}
Expand All @@ -136,24 +151,49 @@ public:
#### Go
```go
func findTheLongestSubstring(s string) int {
pos := make([]int, 32)
for i := range pos {
pos[i] = math.MaxInt32
}
pos[0] = -1
vowels := "aeiou"
state, ans := 0, 0
for i, c := range s {
for j, v := range vowels {
if c == v {
state ^= (1 << j)
}
}
ans = max(ans, i-pos[state])
pos[state] = min(pos[state], i)
}
return ans
func findTheLongestSubstring(s string) (ans int) {
vowels := "aeiou"
d := [32]int{}
for i := range d {
d[i] = 1 << 29
}
d[0] = 0
mask := 0
for i := 1; i <= len(s); i++ {
c := s[i-1]
for j := 0; j < 5; j++ {
if c == vowels[j] {
mask ^= 1 << j
break
}
}
ans = max(ans, i-d[mask])
d[mask] = min(d[mask], i)
}
return
}
```

#### TypeScript

```ts
function findTheLongestSubstring(s: string): number {
const vowels = 'aeiou';
const d: number[] = Array(32).fill(1 << 29);
d[0] = 0;
let [ans, mask] = [0, 0];
for (let i = 1; i <= s.length; i++) {
const c = s[i - 1];
for (let j = 0; j < 5; j++) {
if (c === vowels[j]) {
mask ^= 1 << j;
break;
}
}
ans = Math.max(ans, i - d[mask]);
d[mask] = Math.min(d[mask], i);
}
return ans;
}
```

Expand Down
24 changes: 14 additions & 10 deletions ...n/1300-1399/1371.Find the Longest Substring Containing Vowels in Even Counts/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,17 +1,21 @@
class Solution {
public:
int findTheLongestSubstring(string s) {
vector<int> pos(32, INT_MAX);
pos[0] = -1;
string vowels = "aeiou";
int state = 0, ans = 0;
for (int i = 0; i < s.size(); ++i) {
for (int j = 0; j < 5; ++j)
if (s[i] == vowels[j])
state ^= (1 << j);
ans = max(ans, i - pos[state]);
pos[state] = min(pos[state], i);
vector<int> d(32, INT_MAX);
d[0] = 0;
int ans = 0, mask = 0;
for (int i = 1; i <= s.length(); ++i) {
char c = s[i - 1];
for (int j = 0; j < 5; ++j) {
if (c == vowels[j]) {
mask ^= 1 << j;
break;
}
}
ans = max(ans, i - d[mask]);
d[mask] = min(d[mask], i);
}
return ans;
}
};
};
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