24. Trapping Rain Water
hardAsked at PostmanCompute how much rain water can be trapped between bars of given heights.
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.length1 <= n <= 2 * 10^40 <= height[i] <= 10^5
Examples
Example 1
height = [0,1,0,2,1,0,1,3,2,1,2,1]6Example 2
height = [4,2,0,3,2,5]9Approaches
1. Per-column scan
For each column compute the max bar to its left and right, water = min(L,R) - h[i].
- Time
- O(n^2)
- Space
- O(1)
for each i: scan left max, right max; add max(0, min(L,R)-h[i])Tradeoff:
2. Two-pointer with running maxima
Move two pointers inward, always processing the side whose running max is lower since it bounds the water level.
- Time
- O(n)
- Space
- O(1)
function trap(h) {
let lo = 0, hi = h.length - 1, lMax = 0, rMax = 0, total = 0;
while (lo < hi) {
if (h[lo] < h[hi]) {
if (h[lo] >= lMax) lMax = h[lo];
else total += lMax - h[lo];
lo++;
} else {
if (h[hi] >= rMax) rMax = h[hi];
else total += rMax - h[hi];
hi--;
}
}
return total;
}Tradeoff:
Postman-specific tips
Postman expects the invariant out loud — the side with the lower running max can't trap more than that max — same kind of monotonic-bound reasoning that powers their request-throttle backoff math.
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