23. Trapping Rain Water
hardAsked at IndeedCalculate how much water is trapped between elevation bars — Indeed uses this two-pointer pattern in capacity-planning and bandwidth-allocation simulations.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given n non-negative integers representing an elevation map where each bar has width 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. Prefix/suffix max arrays
Precompute left-max and right-max arrays; trapped water at each position = min(leftMax,rightMax) - height[i].
- Time
- O(n)
- Space
- O(n)
function trap(height) {
const n = height.length;
const left = new Array(n), right = new Array(n);
left[0] = height[0]; right[n-1] = height[n-1];
for (let i = 1; i < n; i++) left[i] = Math.max(left[i-1], height[i]);
for (let i = n-2; i >= 0; i--) right[i] = Math.max(right[i+1], height[i]);
let water = 0;
for (let i = 0; i < n; i++) water += Math.min(left[i], right[i]) - height[i];
return water;
}Tradeoff:
2. Two-pointer O(1) space
Maintain left and right pointers and running max heights; move the pointer with the smaller max since it determines the water level, achieving O(1) space.
- Time
- O(n)
- Space
- O(1)
function trap(height) {
let left = 0, right = height.length - 1;
let leftMax = 0, rightMax = 0, water = 0;
while (left < right) {
if (height[left] <= height[right]) {
if (height[left] >= leftMax) leftMax = height[left];
else water += leftMax - height[left];
left++;
} else {
if (height[right] >= rightMax) rightMax = height[right];
else water += rightMax - height[right];
right--;
}
}
return water;
}Tradeoff:
Indeed-specific tips
Indeed interviewers ask this to probe two-pointer mastery; walk through why the smaller-max side is safe to process before implementing — verbal justification is as important as the code.
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