HARD NC#14 Two Pointers
42. Trapping Rain Water
📖 Problem
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.
🧠 Visual Learning Aid
1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases
JS/TS Refreshers
- •Array methods (`push`, `pop`, `shift`, `slice`)
- •Object/Map/Set usage patterns
- •Function parameter and return typing
- •In-place array updates
- •Sorted array traversal
- •Boundary condition checks
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply Two Pointers reasoning pattern
💡 Approach
- → Water at index i = min(leftMax, rightMax) - height[i]
- → Use two pointers from both ends
- → Track max height seen from left and right
- → Always move pointer with smaller max height
- → Time: O(n), Space: O(1)
🧭 Prerequisites
🛠️ Hints & Pitfalls
Hints
- •Water at index i = min(leftMax, rightMax) - height[i]
- •Use two pointers from both ends
- •Track max height seen from left and right
Common Pitfalls
- •Always move pointer with smaller max height
- •Time: O(n), Space: O(1)
🧪 Test Cases
Hidden tests on submit: 2
Test Case 1
Not run Input:
trap([0,1,0,2,1,0,1,3,2,1,2,1]); Expected:
6 Test Case 2
Not run Input:
trap([4,2,0,3,2,5]); Expected:
9 Test Case 3
Not run Input:
trap([0,0,0]); Expected:
0 📝 Code Editor
📤 Output