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42. Trapping Rain Water

hardAsked at Netflix

Given heights, compute how much water trapped between bars after rain. Netflix asks this on senior-IC loops to see whether you can articulate the two-pointer invariant (water at i depends only on min(maxLeft, maxRight)) without falling into O(n^2).

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Netflix loops.

  • Glassdoor (2026-Q1)— Netflix senior IC onsite reports list this as the algorithmic-thinking hard.
  • Blind (2025-10)— Recurring in Netflix L5+ writeups as the hard problem at the end of the loop.

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

Explanation: Six units of water are trapped between the bars.

Example 2

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

Approaches

1. Brute-force per index

For each index i, scan left and right to find maxLeft and maxRight; add min(maxLeft, maxRight) - height[i].

Time
O(n^2)
Space
O(1)
function trapBrute(height) {
  let total = 0;
  for (let i = 0; i < height.length; i++) {
    let maxLeft = 0, maxRight = 0;
    for (let j = 0; j <= i; j++) maxLeft = Math.max(maxLeft, height[j]);
    for (let j = i; j < height.length; j++) maxRight = Math.max(maxRight, height[j]);
    total += Math.min(maxLeft, maxRight) - height[i];
  }
  return total;
}

Tradeoff: Captures the right invariant (water at i = min(maxLeft, maxRight) - height[i]) but recomputes maxes from scratch each iteration.

2. Prefix max + suffix max arrays

Precompute maxLeft[i] and maxRight[i] in two linear scans, then sum min(left, right) - height[i] across i.

Time
O(n)
Space
O(n)
function trapDP(height) {
  const n = height.length;
  if (n === 0) return 0;
  const maxLeft = new Array(n).fill(0);
  const maxRight = new Array(n).fill(0);
  maxLeft[0] = height[0];
  for (let i = 1; i < n; i++) maxLeft[i] = Math.max(maxLeft[i - 1], height[i]);
  maxRight[n - 1] = height[n - 1];
  for (let i = n - 2; i >= 0; i--) maxRight[i] = Math.max(maxRight[i + 1], height[i]);
  let total = 0;
  for (let i = 0; i < n; i++) total += Math.min(maxLeft[i], maxRight[i]) - height[i];
  return total;
}

Tradeoff: Linear time at O(n) extra space — a clean improvement over brute-force. Mention this before going to two-pointer.

3. Two-pointer (optimal)

Track maxLeft and maxRight while two pointers walk inward. Whichever side has the smaller max defines that index's water.

Time
O(n)
Space
O(1)
function trap(height) {
  let l = 0, r = height.length - 1;
  let maxLeft = 0, maxRight = 0;
  let total = 0;
  while (l < r) {
    if (height[l] < height[r]) {
      if (height[l] >= maxLeft) maxLeft = height[l];
      else total += maxLeft - height[l];
      l++;
    } else {
      if (height[r] >= maxRight) maxRight = height[r];
      else total += maxRight - height[r];
      r--;
    }
  }
  return total;
}

Tradeoff: O(1) extra space. The key insight: if height[l] < height[r], then maxLeft fully constrains what water can rest at l — we don't need to know the exact maxRight, only that some bar to the right is at least height[l].

Netflix-specific tips

Netflix interviewers want the two-pointer invariant articulated before the code. Say: 'At each step, the smaller side limits the water height — so we can safely advance that pointer.' Netflix loops also frequently follow up with: 'What if the heights were a stream and you could only do one pass?' — the two-pointer is already one pass; the prefix-max version isn't.

Common mistakes

  • Treating the problem as 'find the global max and subtract' — wrong, water depends on the local min of maxes, not the global max.
  • Forgetting that water[i] = min(maxLeft, maxRight) - height[i], not just maxLeft - height[i].
  • Off-by-one errors in the two-pointer version, especially around the termination condition l < r vs l <= r.

Follow-up questions

An interviewer at Netflix may pivot to one of these next:

  • What if it's a 2D grid (Trapping Rain Water II)?
  • Could you do it with a monotonic stack? (Yes — different framing, same O(n) complexity.)
  • What if the elevation changed over time (online problem)?

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Output

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FAQ

Why is two-pointer better than prefix-max arrays?

Same time complexity but O(1) space instead of O(n). For very large arrays, that's the difference between fitting in cache and not.

What's the intuition for why two-pointer works?

If height[l] < height[r], we know there's at least one bar of height >= height[l] somewhere on the right (specifically at r). So the water at l can be safely computed using only maxLeft — we don't need maxRight's exact value.

Free learning resources

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