docs(content): add new problem pages, improve binary‑search docs, fix image syntax, update appendix badge

apps/docs/content/(01-linear-structures)/two-pointers/42-trapping-rain-water.mdx

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+---
+title: '42. Trapping Rain Water'
+description: Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Two Pointers", slug: "two-pointers" }
+  - { name: "Dynamic Programming", slug: "dynamic-programming" }
+  - { name: "Stack", slug: "stack" }
+  - { name: "Monotonic Stack", slug: "monotonic-stack" }
+issue: "https://github.com/prdlk/leetcode/issues/211"
+sidebar:
+  label: 'Trapping Rain Water'
+  badge: 'Hard'
+---
+
+### Example 1:
+- Input: `height = [0,1,0,2,1,0,1,3,2,1,2,1]`
+- Output: `6`
+- Explanation: The above elevation map (black section) is represented by array `[0,1,0,2,1,0,1,3,2,1,2,1]`. In this case, `6` units of rain water (blue section) are being trapped.
+
+### Example 2:
+- Input: `height = [4,2,0,3,2,5]`
+- Output: `9`
+
+### Constraints:
+
+- `n == height.length`
+- `1 <= n <= 2 * 10^4`
+- `0 <= height[i] <= 10^5`
+
+## Solution
+
+```py
+class Solution:
+    def trap(self, height: list[int]) -> int:
+        left, right = 0, len(height) - 1
+        leftMax, rightMax = 0, 0
+        total = 0
+
+        while left < right:
+            leftMax = max(leftMax, height[left])
+            rightMax = max(rightMax, height[right])
+
+            if height[left] < height[right]:
+                total += leftMax - height[left]
+                left += 1
+            else:
+                total += rightMax - height[right]
+                right -= 1
+
+        return total
+```