docs(docs): add solution documentation for Container With Most Water and Is Subsequence

docs/solutions/(array)/11-container-with-most-water.mdx

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+---
+title: '11. Container With Most Water'
+description: You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the i^th line are (i, 0) and (i, height[i])
+sidebar:
+  label: 'Container With Most Water'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Two Pointers</Badge>
+
+::::warning
+Notice that you may not slant the container.
+::::
+
+### Example 1:
+- Input: `height = [1,8,6,2,5,4,8,3,7]`
+- Output: `49`
+- Explanation: The above vertical lines are represented by array `[1,8,6,2,5,4,8,3,7]`. In this case, the max area of water (blue section) the container can contain is `49`.
+
+### Example 2:
+- Input: `height = [1,1]`
+- Output: `1`
+
+### Constraints:
+
+- `n == height.length`
+- `2 <= n <= 10^5`
+- `0 <= height[i] <= 10^4`
+
+## Solution
+
+```py
+class Solution:
+    def maxArea(self, height: List[int]) -> int:  # noqa: F821
+        res = 0
+        left = 0
+        right = len(height) - 1
+
+        while left < right:
+            area = (right - left) * min(height[left], height[right])
+            res = max(res, area)
+            if height[left] < height[right]:
+                left += 1
+            else:
+                right -= 1
+
+        return res
+```

docs/solutions/(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)
+```