✨ (solution): Problem 42. Trapping Rain Water

Add Solution of 42. Trapping Rain Water

Author: xchux

.github/workflows/code-check.yml

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+name: Code Quality Check
+
+on:
+  push:
+    branches: [ main, master ]
+  pull_request:
+    branches: [ main, master ]
+
+jobs:
+  python-check:
+    name: Python Code Check
+    runs-on: ubuntu-latest
+    
+    steps:
+    - uses: actions/checkout@v4
+    
+    - name: Set up Python
+      uses: actions/setup-python@v5
+      with:
+        python-version: '3.x'
+        
+    - name: Install dependencies
+      run: |
+        python -m pip install --upgrade pip
+        pip install ruff
+        
+    - name: Run Ruff
+      run: |
+        # Run ruff on Python files in solutions directory
+        ruff check ./solutions 

pyproject.toml

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+[tool.ruff]
+# Enable pycodestyle (`E`) and Pyflakes (`F`) codes by default.
+select = ["E", "F"]
+ignore = []
+
+# Allow autofix for all enabled rules (when `--fix`) is provided.
+fixable = ["A", "B", "C", "D", "E", "F"]
+unfixable = []
+
+# Exclude a variety of commonly ignored directories.
+exclude = [
+    ".bzr",
+    ".direnv",
+    ".eggs",
+    ".git",
+    ".git-rewrite",
+    ".hg",
+    ".mypy_cache",
+    ".md",
+    ".nox",
+    ".pants.d",
+    ".pytype",
+    ".ruff_cache",
+    ".svn",
+    ".tox",
+    ".venv",
+    "__pypackages__",
+    "_build",
+    "buck-out",
+    "build",
+    "dist",
+    "node_modules",
+    "venv",
+]
+
+# Same as Black.
+line-length = 88
+
+# Allow unused variables when underscore-prefixed.
+dummy-variable-rgx = "^(_+|(_+[a-zA-Z0-9_]*[a-zA-Z0-9]+?))$"
+
+# Assume Python 3.12
+target-version = "py312"

solutions/42. Trapping Rain Water/README.md

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+---
+comments: true
+difficulty: easy
+# Follow `Topics` tags
+tags:
+    - Array
+    - Two Pointers
+    - Dynamic Programming
+    - Stack
+    - Monotonic Stack
+---
+
+# [42. Trapping Rain Water](https://leetcode.com/problems/trapping-rain-water/description/)
+
+## Description
+
+Given an elevation map with width of 1 which representing by `n` non-negative integers. Calculate amount of water that can be trapped after rainfall.
+
+**Example 1:**
+```
+Input: height = [0,1,0,0,1,0,1,3,2,0,1]
+Output: 4
+Explanation: [1,0,0,1] can trapped 2 units, [1,0,1] can trapped 1 units, [2,0,1] can trapped 1 units.
+```
+
+**Example 2:**
+```
+Input: height = [4,1,2,4,5]
+Output: 5
+Explanation: [4,1,2,4] can trapped 5 units.
+```
+
+
+**Constraints:**
+`n == height.length`
+`1 <= n <= 2 * 10^4`
+`0 <= height[i] <= 10^5`
+
+## Solution
+
+Shrink from the border. If the outside is higher than the inside, rain will accumulate. Therefore, the difference between the higher and lower bars represents the amount of retained rainwater.
+
+
+```java
+class Solution {
+    public int trap(int[] height) {
+        int left = 0, right = height.length - 1;
+        int leftMax = height[left], rightMax = height[right];
+        int water = 0;
+
+        while (left < right) {
+            if (leftMax < rightMax) {
+                left++;
+                leftMax = Math.max(leftMax, height[left]);
+                water += leftMax - height[left];
+            } else {
+                right--;
+                rightMax = Math.max(rightMax, height[right]);
+                water += rightMax - height[right];
+            }
+        }
+        return water;
+    }
+}
+```
+
+```python
+class Solution:
+    def trap(self, height: list[int]) -> int:
+        if not height:
+            return 0
+
+        left = 0
+        right = len(height) - 1
+        left_max = height[left]
+        right_max = height[right]
+        water = 0
+
+        while left < right:
+            if left_max < right_max:
+                left += 1
+                left_max = max(left_max, height[left])
+                water += left_max - height[left]
+            else:
+                right -= 1
+                right_max = max(right_max, height[right])
+                water += right_max - height[right]
+        return water
+```
+
+## Complexity
+
+- Time complexity: $$O(n)$$
+<!-- Add time complexity here, e.g. $$O(n)$$ -->
+
+- Space complexity: $$O(1)$$
+<!-- Add space complexity here, e.g. $$O(n)$$ -->

solutions/42. Trapping Rain Water/Solution.java

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+class Solution {
+    public int trap(int[] height) {
+        int left = 0, right = height.length - 1;
+        int leftMax = height[left], rightMax = height[right];
+        int water = 0;
+
+        while (left < right) {
+            if (leftMax < rightMax) {
+                left++;
+                leftMax = Math.max(leftMax, height[left]);
+                water += leftMax - height[left];
+            } else {
+                right--;
+                rightMax = Math.max(rightMax, height[right]);
+                water += rightMax - height[right];
+            }
+        }
+        return water;
+    }
+}

solutions/42. Trapping Rain Water/Solution.py

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+class Solution:
+    def trap(self, height: list[int]) -> int:
+        if not height:
+            return 0
+
+        left = 0
+        right = len(height) - 1
+        left_max = height[left]
+        right_max = height[right]
+        water = 0
+
+        while left < right:
+            if left_max < right_max:
+                left += 1
+                left_max = max(left_max, height[left])
+                water += left_max - height[left]
+            else:
+                right -= 1
+                right_max = max(right_max, height[right])
+                water += right_max - height[right]
+        return water