How do you approach House Robber?

Think about the decision at each house: rob it or skip it.

What is the time complexity of House Robber?

Time complexity: O(n). Space complexity: O(1).

What are common mistakes on House Robber?

Iterating the dimensions in an order that reads a cell before it has been written. Getting the base case wrong. It carries the whole recurrence, and the error stays invisible until a large input. Rolling down to one dimension without reversing the inner loop, which lets a 0/1 item be taken twice.

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House Robber

MediumAlgorithms

Description

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. The only constraint stopping you from robbing each of them is that adjacent houses have security systems connected - it will automatically contact the police if **two adjacent houses** were broken into on the same night. Given an integer array `nums` representing the amount of money of each house, return the **maximum amount of money** you can rob tonight **without alerting the police**.

Example 1

Input: nums = [1,2,3,1]

Output: 4

Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). Total = 1 + 3 = 4.

Example 2

Input: nums = [2,7,9,3,1]

Output: 12

Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1). Total = 2 + 9 + 1 = 12.

Constraints

- 1 <= nums.length <= 100 - 0 <= nums[i] <= 400

House Robber: approach, complexity and pitfalls

A medium dynamic programming problem, graded against 7 test cases (4 of them hidden).

The pattern: Dynamic Programming

Overlapping subproblems solved once and reused - the pattern candidates fear most.

Reach for it when you see: "Number of ways", "minimum/maximum cost", or a recursion that recomputes the same state.

More Dynamic Programmingproblems →

If you are stuck

Show the 3-step hint ladder

Think about the decision at each house: rob it or skip it.
If you rob house i, you can't rob house i-1. Your profit is nums[i] + max profit from house i-2.
Use DP: dp[i] = max(dp[i-1], dp[i-2] + nums[i]).

The optimal approach

Show the approach: Dynamic Programming

At each house, choose the maximum of: (1) skip this house (take dp[i-1]) or (2) rob this house (take nums[i] + dp[i-2]). Only need two variables for space optimization.

Time: O(n)
Space: O(1)

The full reference solution in every supported language stays in the editor above - reveal it there once you have had a real attempt.

What the tests check

Read off this problem's own test suite, so these are the cases a submission actually has to survive.

Single element. A one-element case is included. Two-pointer and window solutions frequently return the wrong thing when the two indices start out equal.
Duplicate values. Repeated values appear in the input, so any logic that assumes distinct elements - or that skips over equal neighbours - is exercised.
Zero. Zero appears in the input, which is the value most likely to break a truthiness check used in place of an explicit comparison.

Common Dynamic Programming mistakes

These apply to the pattern as a whole, not just this problem.

Iterating the dimensions in an order that reads a cell before it has been written.
Getting the base case wrong. It carries the whole recurrence, and the error stays invisible until a large input.
Rolling down to one dimension without reversing the inner loop, which lets a 0/1 item be taken twice.