99. Trapping Rain Water
hardAsked at OlaCompute how much water can be trapped between elevations.
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-index max
For each i, scan left/right for max heights and compute the water above i.
- Time
- O(n^2)
- Space
- O(1)
// brute per-index scan; too slow at 2*10^4Tradeoff:
2. Two pointers
Track left/right pointers and running max heights; the smaller side dictates how much can be trapped at its index.
- Time
- O(n)
- Space
- O(1)
function trap(height) {
let l = 0, r = height.length - 1;
let lMax = 0, rMax = 0, water = 0;
while (l < r) {
if (height[l] < height[r]) {
lMax = Math.max(lMax, height[l]);
water += lMax - height[l];
l++;
} else {
rMax = Math.max(rMax, height[r]);
water += rMax - height[r];
r--;
}
}
return water;
}Tradeoff:
Ola-specific tips
Ola asks this for the elegant two-pointer reasoning; tie it to estimating absorbed-demand 'pools' that build up between supply peaks on a city skyline.
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