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Commit f5d8b8a

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rain84yanglbme
andauthored
feat: add ts solution to lc problem: No.409 (#3032)
* feat: add ts solution to lc problem: No.409 * Create Solution2.py * Create Solution2.java * Create Solution2.cpp * Create Solution2.go * Update Solution2.ts * Update README.md * Update README_EN.md * Update README.md --------- Co-authored-by: Libin YANG <contact@yanglibin.info>
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‎solution/0400-0499/0409.Longest Palindrome/README.md‎

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<!-- solution:end -->
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<!-- solution:start -->
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### 方法二:位运算 + 计数
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我们可以使用一个数组或哈希表 $odd$ 记录字符串 $s$ 中每个字符是否出现奇数次,用一个整型变量 $cnt$ 记录出现奇数次的字符个数。
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遍历字符串 $s,ドル对于每个字符 $c,ドル将 $odd[c]$ 取反,即 0ドル \rightarrow 1,ドル 1ドル \rightarrow 0$。如果 $odd[c]$ 由 0ドル$ 变为 1ドル,ドル则 $cnt$ 加一;如果 $odd[c]$ 由 1ドル$ 变为 0ドル,ドル则 $cnt$ 减一。
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最后,如果 $cnt$ 大于 0ドル,ドル答案为 $n - cnt + 1,ドル否则答案为 $n$。
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时间复杂度 $O(n),ドル空间复杂度 $O(|\Sigma|)$。其中,$n$ 为字符串 $s$ 的长度,而 $|\Sigma|$ 为字符集大小,在本题中 $|\Sigma| = 128$。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def longestPalindrome(self, s: str) -> int:
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odd = defaultdict(int)
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cnt = 0
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for c in s:
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odd[c] ^= 1
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cnt += 1 if odd[c] else -1
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return len(s) - cnt + 1 if cnt else len(s)
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```
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#### Java
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```java
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class Solution {
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public int longestPalindrome(String s) {
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int[] odd = new int[128];
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int n = s.length();
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int cnt = 0;
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for (int i = 0; i < n; ++i) {
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odd[s.charAt(i)] ^= 1;
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cnt += odd[s.charAt(i)] == 1 ? 1 : -1;
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}
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return cnt > 0 ? n - cnt + 1 : n;
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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int longestPalindrome(string s) {
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int odd[128]{};
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int n = s.length();
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int cnt = 0;
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for (char& c : s) {
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odd[c] ^= 1;
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cnt += odd[c] ? 1 : -1;
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}
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return cnt ? n - cnt + 1 : n;
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}
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};
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```
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#### Go
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```go
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func longestPalindrome(s string) (ans int) {
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odd := [128]int{}
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cnt := 0
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for _, c := range s {
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odd[c] ^= 1
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cnt += odd[c]
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if odd[c] == 0 {
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cnt--
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}
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}
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if cnt > 0 {
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return len(s) - cnt + 1
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}
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return len(s)
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}
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```
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#### TypeScript
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```ts
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function longestPalindrome(s: string): number {
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const odd: Record<string, number> = {};
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let cnt = 0;
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for (const c of s) {
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odd[c] ^= 1;
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cnt += odd[c] ? 1 : -1;
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}
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return cnt ? s.length - cnt + 1 : s.length;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/0400-0499/0409.Longest Palindrome/README_EN.md‎

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<!-- solution:end -->
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<!-- solution:start -->
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### Solution 2: Bit Manipulation + Counting
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We can use an array or hash table $odd$ to record whether each character in string $s$ appears an odd number of times, and an integer variable $cnt$ to record the number of characters that appear an odd number of times.
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We iterate through the string $s$. For each character $c,ドル we flip $odd[c],ドル i.e., 0ドル \rightarrow 1,ドル 1ドル \rightarrow 0$. If $odd[c]$ changes from 0ドル$ to 1ドル,ドル then we increment $cnt$ by one; if $odd[c]$ changes from 1ドル$ to 0ドル,ドル then we decrement $cnt$ by one.
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Finally, if $cnt$ is greater than 0ドル,ドル the answer is $n - cnt + 1,ドル otherwise, the answer is $n$.
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The time complexity is $O(n),ドル and the space complexity is $O(|\Sigma|)$. Where $n$ is the length of the string $s,ドル and $|\Sigma|$ is the size of the character set. In this problem, $|\Sigma| = 128$.
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def longestPalindrome(self, s: str) -> int:
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odd = defaultdict(int)
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cnt = 0
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for c in s:
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odd[c] ^= 1
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cnt += 1 if odd[c] else -1
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return len(s) - cnt + 1 if cnt else len(s)
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```
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#### Java
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```java
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class Solution {
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public int longestPalindrome(String s) {
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int[] odd = new int[128];
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int n = s.length();
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int cnt = 0;
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for (int i = 0; i < n; ++i) {
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odd[s.charAt(i)] ^= 1;
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cnt += odd[s.charAt(i)] == 1 ? 1 : -1;
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}
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return cnt > 0 ? n - cnt + 1 : n;
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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int longestPalindrome(string s) {
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int odd[128]{};
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int n = s.length();
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int cnt = 0;
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for (char& c : s) {
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odd[c] ^= 1;
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cnt += odd[c] ? 1 : -1;
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}
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return cnt ? n - cnt + 1 : n;
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}
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};
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```
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#### Go
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```go
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func longestPalindrome(s string) (ans int) {
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odd := [128]int{}
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cnt := 0
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for _, c := range s {
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odd[c] ^= 1
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cnt += odd[c]
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if odd[c] == 0 {
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cnt--
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}
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}
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if cnt > 0 {
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return len(s) - cnt + 1
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}
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return len(s)
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}
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```
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#### TypeScript
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```ts
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function longestPalindrome(s: string): number {
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const odd: Record<string, number> = {};
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let cnt = 0;
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for (const c of s) {
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odd[c] ^= 1;
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cnt += odd[c] ? 1 : -1;
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}
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return cnt ? s.length - cnt + 1 : s.length;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->
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class Solution {
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public:
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int longestPalindrome(string s) {
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int odd[128]{};
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int n = s.length();
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int cnt = 0;
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for (char& c : s) {
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odd[c] ^= 1;
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cnt += odd[c] ? 1 : -1;
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}
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return cnt ? n - cnt + 1 : n;
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}
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};
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func longestPalindrome(s string) (ans int) {
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odd := [128]int{}
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cnt := 0
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for _, c := range s {
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odd[c] ^= 1
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cnt += odd[c]
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if odd[c] == 0 {
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cnt--
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}
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}
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if cnt > 0 {
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return len(s) - cnt + 1
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}
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return len(s)
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}
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class Solution {
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public int longestPalindrome(String s) {
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int[] odd = new int[128];
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int n = s.length();
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int cnt = 0;
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for (int i = 0; i < n; ++i) {
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odd[s.charAt(i)] ^= 1;
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cnt += odd[s.charAt(i)] == 1 ? 1 : -1;
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}
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return cnt > 0 ? n - cnt + 1 : n;
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}
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}
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class Solution:
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def longestPalindrome(self, s: str) -> int:
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odd = defaultdict(int)
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cnt = 0
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for c in s:
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odd[c] ^= 1
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cnt += 1 if odd[c] else -1
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return len(s) - cnt + 1 if cnt else len(s)
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function longestPalindrome(s: string): number {
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const odd: Record<string, number> = {};
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let cnt = 0;
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for (const c of s) {
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odd[c] ^= 1;
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cnt += odd[c] ? 1 : -1;
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}
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return cnt ? s.length - cnt + 1 : s.length;
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}

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