Two Pointers/Trapping Rain WaterHard

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Trapping Rain Water

Two Pointers·Hard

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.

The classic two-pointer approach maintains a running leftMax and rightMax and advances whichever side is the shorter wall, accumulating trapped water on that side.

Examples

Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]

Output: 6

The elevation map traps 6 units of rain water (the canonical staircase example).

Input: height = [4,2,0,3,2,5]

Output: 9

Input: height = []

Output: 0

Constraints

n == height.length
1 <= n <= 2 * 10^4
0 <= height[i] <= 10^5

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Tests

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Passed32 ms

First attempt — retry to track metric movement here.

Your approach

Pattern:Two Pointers
Time:O(N)
Space:O(1)

Complete approach — pattern and complexity both named.

Coach

What you got right

You correctly matched the sorted input requirement and implemented a linear-time scan using two pointers.

Where it diverged

No divergence detected. The code correctly aligns with the stated approach.

Next attempt: focus on

Try solving related sliding window problems to build familiarity with two-pointer variants.

Signals

No signals fired — clean run, you stayed in flow.