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24. Trapping Rain Water

hardAsked at Adyen

Given an elevation map, compute how much water it traps after raining.

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

Two passes for the tallest wall to each side, then sum the gap above each bar.

Time
O(n)
Space
O(n)
const n = height.length;
const L = new Array(n), R = new Array(n);
L[0] = height[0]; for (let i = 1; i < n; i++) L[i] = Math.max(L[i-1], height[i]);
R[n-1] = height[n-1]; for (let i = n-2; i >= 0; i--) R[i] = Math.max(R[i+1], height[i]);
let total = 0;
for (let i = 0; i < n; i++) total += Math.min(L[i], R[i]) - height[i];
return total;

Tradeoff:

2. Two-pointer O(1) space

Move from the lower side and accumulate against its own running max.

Time
O(n)
Space
O(1)
function trap(h) {
  let l = 0, r = h.length - 1, lMax = 0, rMax = 0, total = 0;
  while (l < r) {
    if (h[l] < h[r]) {
      lMax = Math.max(lMax, h[l]);
      total += lMax - h[l];
      l++;
    } else {
      rMax = Math.max(rMax, h[r]);
      total += rMax - h[r];
      r--;
    }
  }
  return total;
}

Tradeoff:

Adyen-specific tips

Adyen graders favor the two-pointer O(1)-space variant — they routinely benchmark fraud-score reservoir calcs and prize constant-space passes.

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