23. Trapping Rain Water
hardAsked at Electronic ArtsCalculate how much water can be trapped between bars of varying heights, a spatial simulation problem EA asks to test terrain modeling and two-pointer mastery.
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. Precompute left/right max arrays
For each position, the water is min(leftMax, rightMax) - height. Precompute both maxima arrays in two passes.
- Time
- O(n)
- Space
- O(n)
function trap(height) {
const n = height.length;
const lMax = new Array(n), rMax = new Array(n);
lMax[0] = height[0];
for (let i = 1; i < n; i++) lMax[i] = Math.max(lMax[i-1], height[i]);
rMax[n-1] = height[n-1];
for (let i = n-2; i >= 0; i--) rMax[i] = Math.max(rMax[i+1], height[i]);
let water = 0;
for (let i = 0; i < n; i++) water += Math.min(lMax[i], rMax[i]) - height[i];
return water;
}Tradeoff:
2. Two pointers O(1) space
Use two pointers converging inward, tracking running left and right maxima. Process the shorter side at each step — water there is determined by the shorter boundary. This O(1) space approach is the gold-standard EA solution.
- Time
- O(n)
- Space
- O(1)
function trap(height) {
let lo = 0, hi = height.length - 1;
let lMax = 0, rMax = 0, water = 0;
while (lo < hi) {
if (height[lo] <= height[hi]) {
if (height[lo] >= lMax) lMax = height[lo];
else water += lMax - height[lo];
lo++;
} else {
if (height[hi] >= rMax) rMax = height[hi];
else water += rMax - height[hi];
hi--;
}
}
return water;
}Tradeoff:
Electronic Arts-specific tips
EA interviews cover game development data structures, pathfinding (A*, BFS on grids), spatial algorithms, and simulation problems. Grid/matrix manipulation and BFS on 2D arrays are very common.
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