docs(docs): add LeetCode problem markdown pages and docs site assets

apps/docs/content/(two-pointers)/125-valid-palindrome.mdx

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+---
+title: '125. Valid Palindrome'
+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
+sidebar:
+  label: 'Valid Palindrome'
+  badge: 'Easy'
+---
+
+<Badge variant="accent">Two Pointers</Badge>
+
+### Example 1:
+- Input: `s = "A man, a plan, a canal: Panama"`
+- Output: `true`
+- Explanation: `"amanaplanacanalpanama"` is a palindrome.
+
+### Example 2:
+- Input: `s = "race a car"`
+- Output: `false`
+- Explanation: `"raceacar"` is not a palindrome.
+
+### Example 3:
+- Input: `s = " "`
+- Output: `true`
+- 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.
+
+### Constraints:
+
+- `1 <= s.length <= 2 * 10^5`
+- `s` consists only of printable ASCII characters.
+
+## Solution
+
+```js
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+var isPalindrome = function(s) {
+  let normal = s.replace(/[^a-zA-Z0-9]/g, "").toLowerCase()    
+  let i = 0, j = normal.length - 1;
+
+  while (i < j) {
+    if(normal[i] !== normal[j]) {
+      return false;
+    }
+    i++;
+    j--;
+  }
+  return true;
+};
+```

apps/docs/content/(two-pointers)/344-reverse-string.mdx

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+---
+title: '344. Reverse String'
+description: Write a function that reverses a string. The input string is given as an array of characters s
+sidebar:
+  label: 'Reverse String'
+  badge: 'Easy'
+---
+
+<Badge variant="accent">Two Pointers</Badge>
+
+:::warning
+You must do this by modifying the input array in-place with O(1) extra memory.
+:::
+
+### Example 1:
+- Input: `s = ["h","e","l","l","o"]`
+- Output: `["o","l","l","e","h"]`
+
+### Example 2:
+- Input: `s = ["H","a","n","n","a","h"]`
+- Output: `["h","a","n","n","a","H"]`
+
+### Constraints:
+
+- `1 <= s.length <= 10^5`
+- `s[i]` is a printable ascii character.
+
+## Solution
+
+```js
+/**
+ * @param {character[]} s
+ * @return {void} Do not return anything, modify s in-place instead.
+ */
+var reverseString = function(s) {
+    let i = 0;
+    let j = s.length - 1;
+    
+    while (i < j) {
+        [s[i], s[j]] = [s[j], s[i]];
+        j--;
+        i++;
+    }   
+};
+```

apps/docs/content/(two-pointers)/392-is-subsequence.mdx

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+---
+title: '392. Is Subsequence'
+description: Given two strings s and t, return true if s is a subsequence of t, or false otherwise
+sidebar:
+  label: 'Is Subsequence'
+  badge: 'Easy'
+---
+
+<Badge variant="accent">Two Pointers</Badge>
+
+::::warning
+Suppose there are lots of incoming s, say s1, s2, ..., sk where k >= 10^9, and you want to check one by one to see if t has its subsequence. In this scenario, how would you change your code?
+::::
+
+### Example 1:
+- Input: `s = "abc", t = "ahbgdc"`
+- Output: `true`
+
+### Example 2:
+- Input: `s = "axc", t = "ahbgdc"`
+- Output: `false`
+
+### Constraints:
+
+- `0 <= s.length <= 100`
+- `0 <= t.length <= 10^4`
+- `s` and `t` consist only of lowercase English letters.
+
+## Solution
+
+```py
+class Solution:
+    def isSubsequence(self, s: str, t: str) -> bool:
+        if len(s) > len(t):
+            return False
+
+        i, j = 0, 0
+        while i < len(s) and j < len(t):
+            if s[i] == t[j]:
+                i += 1
+            j += 1
+
+        return i == len(s)
+```