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42. Trapping Rain Water

hardAsked at Bloomberg

Calculating trapped water between histogram bars maps directly onto Bloomberg's time-series gap analysis — measuring how much volume pools between market microstructure events — and rewards the two-pointer insight that eliminates extra passes.

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

Explanation: The elevation map traps 6 units of rainwater.

Example 2

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

Approaches

1. Precomputed prefix/suffix maxima

For each index, water trapped = min(maxLeft, maxRight) - height[i]. Precompute both arrays in two passes, then a third pass accumulates the total.

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

Tradeoff:

2. Two pointers (optimal)

Use left and right pointers converging inward. Whichever side has the smaller max boundary determines the trapped water at that index — no auxiliary arrays needed.

Time
O(n)
Space
O(1)
function trap(height) {
  let left = 0, right = height.length - 1;
  let maxL = 0, maxR = 0, water = 0;
  while (left < right) {
    if (height[left] < height[right]) {
      if (height[left] >= maxL) maxL = height[left];
      else water += maxL - height[left];
      left++;
    } else {
      if (height[right] >= maxR) maxR = height[right];
      else water += maxR - height[right];
      right--;
    }
  }
  return water;
}

Tradeoff:

Bloomberg-specific tips

Bloomberg interviewers love this problem because the two-pointer solution demonstrates the same reasoning Bloomberg quants use for intraday liquidity windows — the side with the smaller maximum boundary is the binding constraint, so you process it first. Articulate that invariant clearly before writing a line of code; it signals you understand the why, not just the how.

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Output

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