The Question

Trapping Rain Water

Given an array of $n$ non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining. For example, given [0,1,0,2,1,0,1,3,2,1,2,1], the output should be 6. Constraints: - $1 \le n \le 2 \times 10^4$ - $0 \le height[i] \le 10^5$

Java

Two Pointers

Questions & Insights

Clarifying Questions

Input Constraints: What is the maximum size of the elevation array? (Assuming

N \le 10^5

for an

O(N)

solution).

Value Range: Can the heights be negative? (Assuming non-negative integers representing elevation).

Memory Constraints: Is there a specific auxiliary space limit? (Assuming

O(1)

space is the target for an optimal solution).

Edge Cases: How should we handle arrays with fewer than 3 elements? (Implicitly, these cannot trap water, so the result should be 0).

Assumptions:

The input is a non-empty array of non-negative integers.

The width of each bar is 1.

O(N)

time and

O(1)

space solution is preferred over

O(N)

space (DP or Stack-based) approaches.

Thinking Process

Observation: The amount of water trapped at any index

i

s determined by the minimum of the maximum height to its left and the maximum height to its right, minus the height at . Mathematically:

Water[i] = \max(0, \min(\text{left\_max}_i, \text{right\_max}_i) - \text{height}[i])

Two Pointers Logic: Instead of pre-calculating the left and right max arrays, we can use two pointers (left and right) starting at the ends of the array. We maintain left_max and right_max variables.

Movement Strategy: Since the amount of water is limited by the shorter of the two boundaries, we always move the pointer pointing to the smaller height. This ensures that the left_max or right_max we are currently using is indeed the limiting factor for the current index.

Iteration: Move the pointer, update the respective side's maximum, and if the current height is less than that maximum, add the difference to the total volume.

Implementation Breakdown

Problem Set

Functional Requirements:

Input: int[] height.

Output: int representing total units of trapped rain water.

Constraints:

n == height.length

1 \le n \le 2 \times 10^4

0 \le height[i] \le 10^5

Approach

Algorithm: Two Pointers.

Data Structure: None (Primitive variables for tracking state).

Complexity:

Time:

O(n)

where

n

is the length of the array (single pass).

Space:

O(1)

(only a few integer pointers and variables).

Implementation

Wrap Up

Advanced Topics

Parallelization: This problem can be solved using a Divide and Conquer approach for very large datasets across multiple machines. You find the global maximum, split the array, and calculate trapped water on the left and right slopes independently.

Streaming/Iterative: If the data is a stream (e.g., coming from a sensor), we can't easily find the "right max" without knowing future values. This would require buffering or a different probabilistic approach if real-time estimation is needed.

Readability vs. Performance: While the Two-Pointer approach is

O(1)

space, the Dynamic Programming approach (storing left_max and right_max in arrays) is often more intuitive for beginners to grasp the "min of maxes" logic.