LeetCode Meditations: House Robber II

The description for House Robber II is:

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system 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 = [2, 3, 2]
Output: 3

Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.

Or:

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 = [1, 2, 3]
Output: 3

We've seen the previous problem that was very similar, except that the first and last houses weren't counted as adjacent.

In fact, for the previous problem, our final solution looked like this:

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;
}

We're using a bottom-up approach where we keep the "bottom" two values: twoPrevMark keeps the maximum amount of money we can have up until two houses prior. prevMax is the maximum until the previous house.

In the for loop, we calculate the maximum value for each house. Then, we return prevMax as it'll hold the last value, the maximum that we can have of all the houses.

We can reuse this solution, but we can't consider the first and last houses at the same time. In order to do that, we can pass the two versions of nums to this function as arguments: one will start from the first house and end at the house previous to the last one:

nums.slice(0, nums.length - 1)

The other will start from the second house and end at the last house:

nums.slice(1)

Then, we can get the maximum of those two values, which will be our return value.

First, we can start by renaming the above function to robHelper.

Our base cases will stay the same as the previous problem:

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

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

Then, we can consider those slices separately and get the maximum value from either of them:

function rob(nums: number[]): number {
  /* ... */
  return Math.max(
    robHelper(nums.slice(0, nums.length - 1)),
    robHelper(nums.slice(1)),
  );
}

And, that's pretty much it. The final version looks like this:

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

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

  return Math.max(
    robHelper(nums.slice(0, nums.length - 1)),
    robHelper(nums.slice(1)),
  );
}

function robHelper(houses: number[]): number {
  let twoPrevMax = 0;
  let prevMax = 0;

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

  return prevMax;
}

Time and space complexity

The time complexity is $O(n)$ as we're iterating over each house twice. The space complexity is $O(1)$ because we don't require additional storage whose size will grow as the input size grows.

Next up is Longest Palindromic Substring. Until then, happy coding.