docs: restructure problem paths, change section display to groups, and improve documentation content

apps/docs/content/(01-linear-structures)/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
+icon: dot
+topics:
+  - { name: "Two Pointers", slug: "two-pointers" }
+  - { name: "String", slug: "string" }
+  - { name: "Dynamic Programming", slug: "dynamic-programming" }
+video: "99RVfqklbCE"
+issue: "https://github.com/prdlk/leetcode/issues/45"
+sidebar:
+  label: 'Is Subsequence'
+  badge: 'Easy'
+---
+
+::::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.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["isSubsequence(s, t)"]) --> G{"len(s) > len(t)?"}
+  G -- yes --> X(["return False"])
+  G -- no --> I["i = 0 into s, j = 0 into t"]
+  I --> W{"i < len(s) and j < len(t)?"}
+  W -- no --> E(["return i == len(s) — every char of s was matched"])
+  W -- yes --> Q{"s[i] == t[j]?"}
+  Q -- yes --> A["i += 1 — consume the match"]
+  Q -- no --> B["j += 1 — always advance t"]
+  A --> B
+  B --> W
+```
+
+## 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)
+```