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docs: add sidebar labels and reorganize solution docs into category folders
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---
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title: '125. Valid Palindrome'
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description: A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers
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sidebar:
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label: 'Valid Palindrome'
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badge: 'Easy'
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---
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<Badge variant="accent">Two Pointers</Badge>
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### Example 1:
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- Input: `s = "A man, a plan, a canal: Panama"`
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- Output: `true`
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- Explanation: `"amanaplanacanalpanama"` is a palindrome.
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### Example 2:
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- Input: `s = "race a car"`
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- Output: `false`
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- Explanation: `"raceacar"` is not a palindrome.
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### Example 3:
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- Input: `s = " "`
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- Output: `true`
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- Explanation: `s` is an empty string `""` after removing non-alphanumeric characters. Since an empty string reads the same forward and backward, it is a palindrome.
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### Constraints:
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- `1 <= s.length <= 2 * 10^5`
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- `s` consists only of printable ASCII characters.
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## Solution
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```js
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/**
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* @param {string} s
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* @return {boolean}
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*/
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var isPalindrome = function(s) {
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let normal = s.replace(/[^a-zA-Z0-9]/g, "").toLowerCase()
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let i = 0, j = normal.length - 1;
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while (i < j) {
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if(normal[i] !== normal[j]) {
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return false;
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}
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i++;
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j--;
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}
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return true;
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};
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```
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---
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title: '344. Reverse String'
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description: Write a function that reverses a string. The input string is given as an array of characters s
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sidebar:
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label: 'Reverse String'
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badge: 'Easy'
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---
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<Badge variant="accent">Two Pointers</Badge>
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:::warning
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You must do this by modifying the input array in-place with O(1) extra memory.
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:::
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### Example 1:
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- Input: `s = ["h","e","l","l","o"]`
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- Output: `["o","l","l","e","h"]`
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### Example 2:
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- Input: `s = ["H","a","n","n","a","h"]`
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- Output: `["h","a","n","n","a","H"]`
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### Constraints:
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- `1 <= s.length <= 10^5`
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- `s[i]` is a printable ascii character.
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## Solution
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```js
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/**
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* @param {character[]} s
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* @return {void} Do not return anything, modify s in-place instead.
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*/
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var reverseString = function(s) {
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let i = 0;
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let j = s.length - 1;
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while (i < j) {
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[s[i], s[j]] = [s[j], s[i]];
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j--;
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i++;
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}
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};
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```
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