Example with multiple pits
Input
height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output
6
The elevation map traps 6 units of water in total.
Full lesson preview
Solve trapping rain water using two pointers
Compute how much water can be trapped after raining given an elevation map using an optimal two-pointer method.
Python practice30 minTwo Pointers & Sliding WindowAdvancedLast updated March 26, 2026
Problem statement
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.
Implement an algorithm using two pointers that runs in linear time and uses constant extra space.
Task
Apply the two-pointer technique to calculate trapped rain water in O(n) time and O(1) extra space.
Examples
Input format
Call trap_rain_water(height) where height is a list of non-negative integers.
Output format
Return an integer indicating total trapped water.
Constraints
0 <= len(height) <= 10^5, 0 <= height[i] <= 10^5. Use O(1) extra space and O(n) time.
Samples
Sample 1
Input
[4,2,0,3,2,5]
Output
9
Classic example with total trapped water equal to 9.