LeetCode Meditations: House Robber

Let's start with the description for House Robber:

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 and 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.

For example:

Input: nums = [1, 2, 3, 1]
Output: 4

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

Or:

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 amount you can rob = 2 + 9 + 1 = 12.

---

Let's start with the easiest options. What's the maximum amount of money we can have if we have no houses to rob?

Exactly. So, we can handle that case:

function rob(nums: number[]): number {
  if (nums.length === 0) {
    return 0;
  }
  /* ... */
}

What's the maximum amount of money we can have if we have only one house to rob?

That's also easy:

function rob(nums: number[]): number {
  /* ... */

  if (nums.length === 1) {
    return nums[0];
  }
}

Since this is a dynamic programming problem, we know that it contains subproblems, and if we can solve them, we'll eventually reach a solution for the problem itself.

At each house that we can possibly be in, we'll have a maximum amount that we can have so far.
So, we can first create a maxPossibles array to hold the values that represent the maximum amount of money we can have up until the $i^\text{th}$ house, and initialize all values as 0 for now:

let maxPossibles = Array.from({ length: nums.length }, () => 0);

Now we can initialize the first item.
The maximum amount of money we can have up until the first house is just that amount in the house itself:

maxPossibles[0] = nums[0];

Up until the second house, the maximum amount is either the value in there or the value in the first house:

maxPossibles[1] = Math.max(nums[0], nums[1]);

Now that we're done with the two base cases, we can continue, starting with the third house.

Whatever house we're in, we have two options for the possible maximum we can have there:

• We can take the maximum amount of money that we can have up until the previous house (maxPossibles[i - 1])
• We can rob the current house, taking the money there and adding it to the maximum that we can have up until the house that is two before us (nums[i] + maxPossibles[i - 2])

So, the maximum amount of money we can have up until the house that we're currently in is the maximum of those two choices:

for (let i = 2; i < nums.length; i++) {
  maxPossibles[i] = Math.max(maxPossibles[i - 1], nums[i] + maxPossibles[i - 2]);
}

Now, finally, we can return the maximum of all those maximum amounts:

function rob(nums: number[]): number {
  /* ... */
  return Math.max(...maxPossibles);
}

The final solution looks like this in TypeScript:

function rob(nums: number[]): number {
  if (nums.length === 0) {
    return 0;
  }

  if (nums.length === 1) {
    return nums[0];
  }

  let maxPossibles = Array.from({ length: nums.length }, () => 0);
  maxPossibles[0] = nums[0];
  maxPossibles[1] = Math.max(nums[0], nums[1]);

  for (let i = 2; i < nums.length; i++) {
    maxPossibles[i] = Math.max(maxPossibles[i - 1], nums[i] + maxPossibles[i - 2]);
  }

  return Math.max(...maxPossibles);
}

Time and space complexity

The time and space complexities are both $O(n)$ since we're iterating over each house doing a constant operation, and keeping an array whose size will grow as $n$ increases.

---

We don't even have to keep an array to hold all the maximum values for each house. Since we're using a bottom-up approach, we can keep the bottom two values.
We can initialize two variables to do just that:

let twoPrevMax = 0; 
let prevMax = 0; 

twoPrevMark will keep the maximum amount of money we can have up until two houses prior. prevMax is the maximum until the previous house.

Then, for each house, we can get the maximum so far, like this:

for (const n of nums) {
  let maxUpToN = Math.max(prevMax, n + twoPrevMax);
  twoPrevMax = prevMax;
  prevMax = maxUpToN;
}

At the end, we can return prevMax as it'll hold the last value, the maximum that we can have of all the houses.

This is how this version looks like:

function rob(nums: number[]): number {
  let twoPrevMax = 0;
  let prevMax = 0;

  for (const n of nums) {
    let maxUpToN = Math.max(prevMax, n + twoPrevMax);
    twoPrevMax = prevMax;
    prevMax = maxUpToN;
  }

  return prevMax;
}

Time and space complexity

The time complexity is still $O(n)$ as we iterate over the houses, but the space complexity is $O(1)$ as we don't need additional storage whose size will grow as our input increases.

---

This was, in my opinion, a quite challenging problem, so it's now time to take a deep breath. Next up, we'll take a look at House Robber II. Until then, happy coding.

Comments

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