What is the time complexity of Trapping Rain Water II?

The optimal solution has a time complexity of O(m*n*log(m*n)).

What data structures are used to solve Trapping Rain Water II?

This problem uses Array, Breadth-First Search, Heap (Priority Queue), Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Trapping Rain Water II?

Insight 1: The water level at any cell is determined by the minimum height of its surrounding boundary cells. The cells on the boundary cannot trap any water, so they form the starting point. Insight 2: We can use a priority queue (min-heap) to process cells in increasing order of their height. This allows us to simulate the water filling process from the boundary inwards.

Which companies ask LeetCode 407. Trapping Rain Water II in interviews?

This problem is frequently asked at ByteDance, Otter.ai, Palantir Technologies, Qualcomm, X.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Trapping Rain Water II - LeetCode 407 Solution

Q: What is the space complexity of Trapping Rain Water II?

The space complexity is O(m*n).

Trapping Rain Water II - Complete Solution Guide

Trapping Rain Water II is LeetCode problem 407, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an m x n integer matrix heightMap representing the height of each unit cell in a 2D elevation map, return the volume of water it can trap after raining . Example 1: Input: heightMap = [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]] Output: 4 Explanation: After the rain, water is trapped between the blocks. We have two small ponds 1 and 3 units trapped. The total volume of water trapped is 4. Example 2: Input: heightMap = [[3,3,3,3,3],[3,2,2,2,3],[3,2,1,2,3],[3,2,2,2,3],[3,3,3,3,3]] Output: 10

Detailed Explanation

The problem asks us to calculate the amount of water that can be trapped in a given 2D elevation map after it rains. The elevation map is represented by an `m x n` integer matrix `heightMap`, where `heightMap[i][j]` represents the height of each cell. The water can be trapped between cells of varying heights, similar to how water is trapped in a valley formed by mountains. The goal is to find the total volume of water trapped after the rain fills the depressions in the map.

Solution Approach

The solution employs a priority queue based approach combined with Breadth-First Search (BFS). We start by adding all boundary cells of the `heightMap` to a min-heap. Then, we iteratively extract the cell with the minimum height from the heap. For each extracted cell, we explore its unvisited neighbors. If a neighbor's height is less than the current cell's height, water is trapped, and the total trapped water is updated. The neighbor is then added to the priority queue with the height being the maximum of the current cell's height and the neighbor's actual height. This ensures that the water level propagates correctly throughout the map.

Step-by-Step Algorithm

Step 1: Initialize a visited matrix to keep track of visited cells and a priority queue (min-heap) to store boundary cells.
Step 2: Add all boundary cells to the priority queue along with their coordinates and mark them as visited.
Step 3: Initialize a variable `total_water` to store the total trapped water.
Step 4: While the priority queue is not empty, extract the cell with the minimum height.
Step 5: For each unvisited neighbor of the extracted cell:
Step 6: Mark the neighbor as visited.
Step 7: Calculate the water trapped at the neighbor as `max(0, current_cell_height - neighbor_height)` and add it to `total_water`.
Step 8: Add the neighbor to the priority queue with height as `max(current_cell_height, neighbor_height)` to propagate the water level.
Step 9: Return the `total_water`.

Key Insights

Insight 1: The water level at any cell is determined by the minimum height of its surrounding boundary cells. The cells on the boundary cannot trap any water, so they form the starting point.
Insight 2: We can use a priority queue (min-heap) to process cells in increasing order of their height. This allows us to simulate the water filling process from the boundary inwards.
Insight 3: For each cell, we compare its height to the current water level (the height of the boundary cell). If the cell's height is lower, water gets trapped. The water level for the neighboring cells is then updated to the maximum of the current water level and their own height.

Complexity Analysis

Time Complexity: O(m*n*log(m*n))

Space Complexity: O(m*n)

Topics

This problem involves: Array, Breadth-First Search, Heap (Priority Queue), Matrix.

Companies

Asked at: ByteDance, Otter.ai, Palantir Technologies, Qualcomm, X.

Trapping Rain Water II | LeetCode 407

Hard

Array Breadth-First Search Heap (Priority Queue)Matrix

Time: O(m*n*log(m*n))Space: O(m*n)

Solution Code

Available in:

import java.util.PriorityQueue;

class Solution {
    public int trapRainWater(int[][] heightMap) {
        if (heightMap == null || heightMap.length == 0 || heightMap[0].length == 0) {
            return 0;
        }

        int m = heightMap.length;
        int n = heightMap[0].length;

        if (m <= 2 || n <= 2) {
            return 0;
        }

        boolean[][] visited = new boolean[m][n];
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
                    pq.offer(new int[]{heightMap[i][j], i, j});
                    visited[i][j] = true;
                }
            }
        }

        int total_water = 0;
        int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

        while (!pq.isEmpty()) {
            int[] curr = pq.poll();
            int height = curr[0];
            int r = curr[1];
            int c = curr[2];

            for (int[] dir : directions) {
                int nr = r + dir[0];
                int nc = c + dir[1];

                if (nr >= 0 && nr < m && nc >= 0 && nc < n && !visited[nr][nc]) {
                    visited[nr][nc] = true;

                    int water_trapped = Math.max(0, height - heightMap[nr][nc]);
                    total_water += water_trapped;

                    int new_height = Math.max(height, heightMap[nr][nc]);
                    pq.offer(new int[]{new_height, nr, nc});
                }
            }
        }

        return total_water;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 407: Trapping Rain Water II - Detailed explanation and solution approaches.

Topics

Array Breadth-First Search Heap (Priority Queue)Matrix

Companies

ByteDanceOtter.aiPalantir TechnologiesQualcommX

Stats

Difficulty:Hard

Languages:4

Companies:5

Problem #:407

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Initialize a visited matrix to keep track of visited cells and a priority queue (min-heap) to store boundary cells.
2
Step 2: Add all boundary cells to the priority queue along with their coordinates and mark them as visited.
3
Step 3: Initialize a variable `total_water` to store the total trapped water.
4
Step 4: While the priority queue is not empty, extract the cell with the minimum height.
5
Step 5: For each unvisited neighbor of the extracted cell:
6
Step 6: Mark the neighbor as visited.
7
Step 7: Calculate the water trapped at the neighbor as `max(0, current_cell_height - neighbor_height)` and add it to `total_water`.
8
Step 8: Add the neighbor to the priority queue with height as `max(current_cell_height, neighbor_height)` to propagate the water level.
9
Step 9: Return the `total_water`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n*log(m*n))

💾

Space Complexity

O(m*n)

📝

Why?

Time Complexity: We visit each cell at most once. For each cell, we perform a heap operation, which takes O(log(m*n)) time, where m and n are the dimensions of the matrix. Hence, the overall time complexity is O(m*n*log(m*n)). Space Complexity: We use a visited matrix of size m*n and a priority queue that can, in the worst case, contain all the elements of the matrix. Therefore, the space complexity is O(m*n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

Could potentially solve by pre-computing left/right and up/down max height arrays, but much harder to implement correctly than the heap-based approach.

Time: O(m*n)Space: O(m*n)

Trade-offs: This might improve performance constants slightly. However, it's significantly more complex to implement correctly, with a higher chance of error.

✨

Key Insights

1Insight 1: The water level at any cell is determined by the minimum height of its surrounding boundary cells. The cells on the boundary cannot trap any water, so they form the starting point.
2Insight 2: We can use a priority queue (min-heap) to process cells in increasing order of their height. This allows us to simulate the water filling process from the boundary inwards.
3Insight 3: For each cell, we compare its height to the current water level (the height of the boundary cell). If the cell's height is lower, water gets trapped. The water level for the neighboring cells is then updated to the maximum of the current water level and their own height.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Empty or invalid input heightMap (e.g., null or empty array). The code handles this by returning 0.
Edge case 2: heightMap with dimensions <= 2. No water can be trapped in such maps, so the code returns 0.
Edge case 3: All cells having the same height. The algorithm will correctly return 0, as no water will be trapped.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the boundary cells as the starting point. Starting from an inner cell will lead to incorrect water level calculations.

✗Common mistake 2: Not using a priority queue (min-heap). Using a regular queue will not guarantee that the cells are processed in increasing order of height, leading to incorrect results.

✗Common mistake 3: Failing to update the height of the neighbor cell when adding it to the priority queue. The height must be the maximum of the current cell's height and the neighbor's height to correctly propagate the water level.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Modeling flood risk assessment in geographical regions. The elevation map can be used to predict areas susceptible to flooding after heavy rainfall.

🏦Real-world scenario 2: Designing reservoirs and dams to maximize water storage capacity. The algorithm can help determine the optimal shape and size of the reservoir to trap the most water.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 18, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Trapping Rain Water II"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?