What is the time complexity of House Robber III?

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

What data structures are used to solve House Robber III?

This problem uses Dynamic Programming, Tree, Depth-First Search, Binary Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving House Robber III?

Insight 1: The problem can be solved using dynamic programming on a tree. At each node, we have two choices: rob the current house or don't rob it. Insight 2: The recursive structure of the binary tree lends itself well to a recursive solution that utilizes overlapping subproblems, making dynamic programming an ideal approach.

Which companies ask LeetCode 337. House Robber III in interviews?

This problem is frequently asked at PhonePe, oyo.

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House Robber III - LeetCode 337 Solution

House Robber III - Complete Solution Guide

House Robber III is LeetCode problem 337, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called root . Besides the root , each house has one and only one parent house. After a tour, the smart thief realized that all houses in this place form a binary tree. It will automatically contact the police if two directly-linked houses were broken into on the same night . Given the root of the binary tree, return the maximum amount of money the thief can rob without alerting the police . E

Detailed Explanation

The problem describes a scenario where a thief is attempting to rob houses arranged in a binary tree structure. Each house has a value representing the amount of money it contains. The constraint is that the thief cannot rob two directly connected houses (parent and child) without triggering an alarm. The task is to determine the maximum amount of money the thief can rob without alerting the police, given the root of the binary tree.

Solution Approach

The solution uses a recursive approach with memoization (implicitly through the return values of the recursive calls) to implement dynamic programming. For each node, the `_solve` function (or its equivalent in other languages) returns a tuple (or array) containing two values: the maximum amount of money that can be robbed including the current node, and the maximum amount of money that can be robbed excluding the current node. The `rob` function calls `_solve` on the root and returns the maximum of the two values returned by `_solve`.

Step-by-Step Algorithm

Step 1: Define a recursive helper function `_solve` (or equivalent) that takes a node as input and returns a tuple (or array) of two integers: `(rob_this, not_rob_this)`. `rob_this` represents the maximum amount that can be robbed if we rob the current node, and `not_rob_this` represents the maximum amount that can be robbed if we don't rob the current node.
Step 2: Base case: If the node is null, return (0, 0), meaning we can rob nothing in either case.
Step 3: Recursively call `_solve` on the left and right children of the current node to get the results for the left and right subtrees: `left_rob, left_not_rob` and `right_rob, right_not_rob`.
Step 4: Calculate `rob_this`: If we rob the current node, we cannot rob its children. Therefore, `rob_this = node.val + left_not_rob + right_not_rob`.
Step 5: Calculate `not_rob_this`: If we don't rob the current node, we can either rob or not rob its children. Therefore, `not_rob_this = max(left_rob, left_not_rob) + max(right_rob, right_not_rob)`.
Step 6: Return the tuple `(rob_this, not_rob_this)`.
Step 7: In the `rob` function, call `_solve` on the root and return `max(rob_this, not_rob_this)`.

Key Insights

Insight 1: The problem can be solved using dynamic programming on a tree. At each node, we have two choices: rob the current house or don't rob it.
Insight 2: The recursive structure of the binary tree lends itself well to a recursive solution that utilizes overlapping subproblems, making dynamic programming an ideal approach.
Insight 3: We need to maintain two values for each node: the maximum amount we can rob if we rob the current node, and the maximum amount we can rob if we don't rob the current node.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Dynamic Programming, Tree, Depth-First Search, Binary Tree.

Companies

Asked at: PhonePe, oyo.

House Robber III | LeetCode 337

Medium

Dynamic Programming Tree Depth-First Search Binary Tree

Time: O(n)Space: O(n)

Solution Code

Available in:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int rob(TreeNode root) {
        int[] result = _solve(root);
        return Math.max(result[0], result[1]);
    }

    private int[] _solve(TreeNode node) {
        if (node == null) {
            return new int[] {0, 0};
        }

        int[] left = _solve(node.left);
        int[] right = _solve(node.right);

        int rob_this = node.val + left[1] + right[1];
        int not_rob_this = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);

        return new int[] {rob_this, not_rob_this};
    }
}

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Problem Info

Description

Problem 337: House Robber III - Detailed explanation and solution approaches.

Topics

Dynamic Programming Tree Depth-First Search Binary Tree

Companies

PhonePeoyo

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:337

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

7 steps

1
Step 1: Define a recursive helper function `_solve` (or equivalent) that takes a node as input and returns a tuple (or array) of two integers: `(rob_this, not_rob_this)`. `rob_this` represents the maximum amount that can be robbed if we rob the current node, and `not_rob_this` represents the maximum amount that can be robbed if we don't rob the current node.
2
Step 2: Base case: If the node is null, return (0, 0), meaning we can rob nothing in either case.
3
Step 3: Recursively call `_solve` on the left and right children of the current node to get the results for the left and right subtrees: `left_rob, left_not_rob` and `right_rob, right_not_rob`.
4
Step 4: Calculate `rob_this`: If we rob the current node, we cannot rob its children. Therefore, `rob_this = node.val + left_not_rob + right_not_rob`.
5
Step 5: Calculate `not_rob_this`: If we don't rob the current node, we can either rob or not rob its children. Therefore, `not_rob_this = max(left_rob, left_not_rob) + max(right_rob, right_not_rob)`.
6
Step 6: Return the tuple `(rob_this, not_rob_this)`.
7
Step 7: In the `rob` function, call `_solve` on the root and return `max(rob_this, not_rob_this)`.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Each node is visited once. At each node, we perform constant-time operations. Therefore, the overall time complexity is O(n). The space complexity is determined by the maximum depth of the recursive call stack, which is O(n) in the worst case where the tree is skewed. If the tree is balanced, the call stack space would be O(log 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

Memoization with HashMap

Instead of relying on the recursive return values for implicit memoization, an explicit hash map can be used to store the results of the _solve function for each node to avoid redundant calculations. This is particularly useful if the same subtrees are encountered multiple times.

Time: O(n)Space: O(n)

Trade-offs: While the time complexity remains the same, explicit memoization can sometimes provide a slight performance improvement, especially if the call stack depth becomes a bottleneck. It uses more explicit memory for the hash map though.

✨

Key Insights

1Insight 1: The problem can be solved using dynamic programming on a tree. At each node, we have two choices: rob the current house or don't rob it.
2Insight 2: The recursive structure of the binary tree lends itself well to a recursive solution that utilizes overlapping subproblems, making dynamic programming an ideal approach.
3Insight 3: We need to maintain two values for each node: the maximum amount we can rob if we rob the current node, and the maximum amount we can rob if we don't rob the current node.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is null). The base case of the recursion handles this by returning (0, 0), which correctly results in 0 being returned as the maximum robbed amount.
Edge case 2: Single node tree. The algorithm correctly returns the value of that node.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly assuming that always robbing the parent node will lead to the maximum amount. This is incorrect because the children of the parent might have larger values which would contribute to a higher maximum amount.

✗Common mistake 2: Not considering the case where not robbing the current node might yield a greater overall profit due to the possibility of robbing its children.

✅

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: Resource allocation in a hierarchical organization where allocating resources to a manager may prevent direct reports from receiving resources, and the goal is to maximize the overall resource utilization.

🏦Real-world scenario 2: Optimizing network coverage where installing a repeater at a certain location interferes with the coverage of adjacent locations, and the goal is to maximize the overall coverage area.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

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: November 5, 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 13

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

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "House Robber III"?

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

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