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The Question
Coding

Trapping Rain Water

Given an array of non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining. Example: Input: height = [0,1,0,2,1,0,1,3,2,1,2,1] Output: 6 Constraints: - n == height.length - 1 <= n <= 2 * 10^4 - 0 <= height[i] <= 10^5 Your solution should aim for linear time complexity O(n) and constant space complexity O(1).
Java
Two Pointers
Questions & Insights

Clarifying Questions

What is the maximum size of the elevation map array? (Assumption: Up to 10^5 elements).
What is the maximum value of each elevation height? (Assumption: Up to 10^5, fitting within standard integer types).
How should we handle cases with fewer than 3 bars? (Assumption: Water cannot be trapped, return 0).
Is it possible for height values to be negative? (Assumption: Heights are non-negative integers).

Thinking Process

The amount of water trapped at any index is determined by the minimum of the maximum height to its left and the maximum height to its right, minus the height at tself: Water[i] = \max(0, \min(leftMax[i], rightMax[i]) - height[i]).
A naive approach involves pre-calculating two arrays (leftMax and rightMax), which takes O(n) time and O(n) space.
We can optimize the space to O(1) using a Two-Pointer approach. By maintaining two pointers (left and right) and two variables (leftMax and rightMax), we can process the side with the smaller boundary.
If height[left] < height[right], the water level at left is bottlenecked by leftMax because we know there's a boundary at least as high as height[right] on the right side. We update leftMax, calculate water, and move left inward. We do the symmetrical logic for the right pointer.
Implementation Breakdown

Problem Set

Input: An array of non-negative integers height representing an elevation map.
Output: Total units of water trapped.
Constraints:
n == height.length
1 \le n \le 2 \cdot 10^4
0 \le height[i] \le 10^5
Time Complexity: O(n)
Space Complexity: O(1)

Approach

Algorithm: Two Pointers.
Data Structure: None (Primitive variables).
Complexity: Time O(n), Space O(1).

Implementation

Wrap Up

Advanced Topics

Monotonic Stack Approach: Another O(n) solution involves using a stack to store indices of decreasing heights. When we encounter a height taller than the stack top, we pop and calculate the bounded area. This is useful for problems where you need to process "pockets" horizontally rather than vertically.
Parallelization: This problem is inherently sequential in the two-pointer form, but if the map were massive (distributed), one could split the array, find local maximums at the boundaries, and use a map-reduce style approach to calculate global water levels.
Readability vs Performance: While O(1) space is optimal, the O(n) space Dynamic Programming approach (pre-calculating leftMax and rightMax arrays) is often more readable and easier to debug for junior developers.