25. Trapping Rain Water

hardAsked at Tesla

Calculate how much water a histogram traps after rain — Tesla's thermal management team uses a structurally identical algorithm to model coolant pooling in battery-pack channel geometries, identifying where fluid accumulates before designing drainage paths.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-26

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

Explanation: The elevation map traps 6 units of water (visualize the gaps between bars).

Example 2

Input

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

Output

Approaches

1. Precompute left/right max arrays

Build leftMax[i] and rightMax[i] arrays. Water at position i = min(leftMax[i], rightMax[i]) - height[i]. Two extra O(n) arrays.

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-pointer O(1) space

Maintain left/right pointers and running max on each side. Whichever side has the smaller max, compute its water contribution and advance that pointer inward.

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]) {
      if (height[left] >= leftMax) {
        leftMax = height[left];
      } else {
        water += leftMax - height[left];
      }
      left++;
    } else {
      if (height[right] >= rightMax) {
        rightMax = height[right];
      } else {
        water += rightMax - height[right];
      }
      right--;
    }
  }
  return water;
}

Tradeoff:

Tesla-specific tips

Tesla asks this in hardware-adjacent roles because the thermal and structural simulation teams think in exactly these terms. The two-pointer solution is the expected answer at the senior level — derive it by explaining why you can commit to a water value on the side with the smaller max (the other side is guaranteed to be at least as tall). Many candidates stumble on why it's safe to advance; proving that invariant out loud is what separates a strong pass from a borderline one.

Solve it now

Free. No sign-up. Python and JavaScript run instantly in your browser.

Output

Press Run or Cmd+Enter to execute

Practice these live with InterviewChamp.AI

Drill Trapping Rain Water and other Tesla interview questions under real-loop conditions with instant feedback on your reasoning, complexity claims, and code.

Practice these live with InterviewChamp.AI →