How Do You Solve the Trapping Rain Water Problem in Coding Interviews?

One common question in technical interviews, especially for roles related to algorithms and data structures, is the "Trapping Rain Water" problem. This problem tests your understanding of arrays, two-pointer techniques, and your ability to optimize solutions for efficiency. In this article, we'll go over what the problem involves, an efficient approach to solving it, and a clean code example to help you prepare.

Understanding the Problem

Imagine it’s a rainy day, and the landscape is represented by an array of non-negative integers, where each element indicates the height of a vertical bar or wall. The goal is to determine how much water can be trapped between these barriers after it rains.

For example, given the height array:

[0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1]

Visualized, it might look like this:

HtmlCopy code

The total trapped water in this scenario is 6 units.

Why Is This Challenging?

The initial approach might be to check each position to see how much water it can hold based on the highest walls to its left and right. But a naive implementation can be inefficient, especially for large arrays, with a time complexity of O(n²). The challenge is to optimize this to run in linear time, O(n), while still using only constant extra space.

An Efficient Approach: Using Two Pointers

One effective way to solve this problem involves the two-pointer method, which is both simple and optimal. Here's how it works:

1. Set two pointers: one at the beginning (left) and one at the end (right) of the array.
2. Keep track of the maximum height encountered from the left (left_max) and from the right (right_max).
3. Proceed inward from both ends:
   • If the height at left is less than at right, then:
     • If height[left] is greater than left_max, update left_max.
     • Else, this position can trap left_max - height[left] units of water.
     • Move left pointer to the right.
   • Else:
     • If height[right] is greater than right_max, update right_max.
     • Else, this position can trap right_max - height[right] units of water.
     • Move right pointer to the left.
4. Continue this process until left and right pointers cross.

This method ensures that at each step, the trapped water calculation is based on known maximum boundaries, preventing redundant comparisons.

Sample Implementation in Python

Here’s a clean, easy-to-understand code example implementing this approach:

PythonCopy code

Explaining the Code

• The function starts by checking for an empty input array.
• Two pointers (left and right) are initialized at either end, with left_max and right_max to keep track of the highest walls seen so far.
• The while loop continues until the two pointers meet.
• Inside the loop, comparisons decide which pointer to move:
  • When the height at left is less than at right, focus on the left side.
  • Otherwise, focus on the right side.
• When a new maximum height is found, update left_max or right_max.
• If the current height is less than the maximum height seen so far, accumulate the difference (this is the trapped water).

Why This Solution Works

This approach makes a single pass through the array, resulting in a linear time complexity (O(n)). It maintains constant extra space because only a few variables are used regardless of input size. The logic leverages the fact that trapped water depends on the shorter boundary, so moving the pointers inward based on comparisons ensures all cases are efficiently covered.

The "Trapping Rain Water" problem is a common technical interview question that tests your understanding of array traversal and optimization. The two-pointer