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28. Trapping Rain Water

hardAsked at Booking

Calculate total water trapped between elevation bars — Booking's data team uses an analogous two-pointer capacity calculation when computing the maximum available inventory across a price-bucket histogram for dynamic pricing.

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 level = min(maxLeft[i], maxRight[i]) - height[i]. Two passes to build the arrays, one pass to sum.

Time
O(n)
Space
O(n)
function trap(height) {
  const n = height.length;
  const leftMax = new Array(n).fill(0);
  const rightMax = new Array(n).fill(0);

  leftMax[0] = height[0];
  for (let i = 1; i < n; i++) leftMax[i] = Math.max(leftMax[i - 1], height[i]);

  rightMax[n - 1] = height[n - 1];
  for (let i = n - 2; i >= 0; i--) rightMax[i] = Math.max(rightMax[i + 1], height[i]);

  let water = 0;
  for (let i = 0; i < n; i++) {
    water += Math.min(leftMax[i], rightMax[i]) - height[i];
  }
  return water;
}

Tradeoff:

2. Two pointers (O(1) space)

Move the pointer on the shorter side inward; the water above it is determined by that side's max. One pass, no extra arrays.

Time
O(n)
Space
O(1)
function trap(height) {
  let left = 0, right = height.length - 1;
  let leftMax = 0, rightMax = 0;
  let water = 0;

  while (left < right) {
    if (height[left] <= height[right]) {
      leftMax = Math.max(leftMax, height[left]);
      water += leftMax - height[left];
      left++;
    } else {
      rightMax = Math.max(rightMax, height[right]);
      water += rightMax - height[right];
      right--;
    }
  }

  return water;
}

Tradeoff:

Booking-specific tips

Booking's capacity planning team thinks in histograms — frame your answer in terms of 'bounded volume between walls'. The two-pointer O(1) approach always impresses; be prepared to prove correctness by explaining why the shorter side's water level is fully determined.

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