25. Trapping Rain Water
hardAsked at WiseCompute trapped water given an elevation map — Wise reframes it as the locked-up liquidity between two FX pool walls.
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 left and right max
For each index, scan left and right for the max height; water at i is min(maxL, maxR) - h[i].
- Time
- O(n^2)
- Space
- O(1)
let total=0;
for (let i=0;i<height.length;i++){
let l=0,r=0;
for (let j=0;j<=i;j++) l=Math.max(l,height[j]);
for (let j=i;j<height.length;j++) r=Math.max(r,height[j]);
total += Math.min(l,r)-height[i];
}
return total;Tradeoff:
2. Two-pointer with running walls
Two pointers converge; each step the smaller side dictates the trapped amount against its running max, and that pointer advances.
- Time
- O(n)
- Space
- O(1)
function trap(height){
let l = 0, r = height.length - 1;
let maxL = 0, maxR = 0, total = 0;
while (l < r){
if (height[l] < height[r]){
if (height[l] >= maxL) maxL = height[l];
else total += maxL - height[l];
l++;
} else {
if (height[r] >= maxR) maxR = height[r];
else total += maxR - height[r];
r--;
}
}
return total;
}Tradeoff:
Wise-specific tips
Wise grades whether you reach for the converging two-pointer pattern — their cross-border liquidity scanner uses the same shape to compute pooled balance between two FX walls.
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