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23. Trapping Rain Water

hardAsked at DigitalOcean

Calculate the total water trapped by an elevation map — a classic hard problem DigitalOcean uses to probe two-pointer mastery and capacity planning thinking.

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 it can trap after raining.

Constraints

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

Examples

Example 1

Input
height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output
6

Example 2

Input
height = [4,2,0,3,2,5]
Output
9

Approaches

1. Precompute left/right max arrays

For each index, water = min(leftMax[i], rightMax[i]) - height[i]. O(n) time, O(n) space for two extra arrays.

Time
O(n)
Space
O(n)
function trap(height) {
  const n=height.length;
  const lMax=new Array(n), rMax=new Array(n);
  lMax[0]=height[0]; rMax[n-1]=height[n-1];
  for(let i=1;i<n;i++) lMax[i]=Math.max(lMax[i-1],height[i]);
  for(let i=n-2;i>=0;i--) rMax[i]=Math.max(rMax[i+1],height[i]);
  let water=0;
  for(let i=0;i<n;i++) water+=Math.min(lMax[i],rMax[i])-height[i];
  return water;
}

Tradeoff:

2. Two-pointer O(1) space

Maintain left and right pointers and track leftMax/rightMax. The side with the smaller max is the bottleneck — add water from that side and advance the pointer inward.

Time
O(n)
Space
O(1)
function trap(height) {
  let lo=0, hi=height.length-1;
  let leftMax=0, rightMax=0, water=0;
  while(lo<hi){
    if(height[lo]<height[hi]){
      height[lo]>=leftMax ? leftMax=height[lo] : water+=leftMax-height[lo];
      lo++;
    } else {
      height[hi]>=rightMax ? rightMax=height[hi] : water+=rightMax-height[hi];
      hi--;
    }
  }
  return water;
}

Tradeoff:

DigitalOcean-specific tips

DigitalOcean uses this problem to assess capacity-planning intuition — interviewers appreciate framing the 'bottleneck side' insight in terms of network throughput constraints.

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Output

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