23. Trapping Rain Water
hardAsked at ByteDanceCompute the volume of rainwater trapped between bars of varying heights — ByteDance uses it to test two-pointer DP and bottleneck-thinking under pressure.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water can be trapped after raining.
Constraints
1 <= height.length <= 2 * 10^40 <= height[i] <= 10^5
Examples
Example 1
height = [0,1,0,2,1,0,1,3,2,1,2,1]6Example 2
height = [4,2,0,3,2,5]9Approaches
1. Per-column max scan
For each column compute max-left and max-right by scanning the whole array each time.
- Time
- O(n^2)
- Space
- O(1)
// for each i, scan left for maxL and right for maxR; add min(maxL,maxR)-h[i]Tradeoff:
2. Two pointers
Walk pointers in from both ends, tracking running maxLeft and maxRight. The smaller side defines trapped water and advances inward.
- Time
- O(n)
- Space
- O(1)
function trap(height) {
let l = 0, r = height.length - 1, ml = 0, mr = 0, total = 0;
while (l < r) {
if (height[l] < height[r]) {
ml = Math.max(ml, height[l]);
total += ml - height[l];
l++;
} else {
mr = Math.max(mr, height[r]);
total += mr - height[r];
r--;
}
}
return total;
}Tradeoff:
ByteDance-specific tips
ByteDance interviewers grade hard on the bottleneck argument — why the smaller side decides trapped water — because the same reasoning shows up in their video-bandwidth backpressure controller.
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