Introduction

In this blog post, we will explore how to solve the LeetCode problem "213" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. Example 1: 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. Example 2: 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. Example 3: Input: nums = [1,2,3] Output: 3 Constraints: 1 <= nums.length <= 100 0 <= nums[i] <= 1000

Explanation

Here's the optimal solution to the circular house robber problem, focusing on efficiency and clarity:

• Decomposition: Break the circular problem into two linear subproblems. One subproblem considers robbing houses from index 0 to n-2 (inclusive), effectively excluding the last house. The second subproblem considers robbing houses from index 1 to n-1 (inclusive), excluding the first house.

• Dynamic Programming: Use dynamic programming (specifically, the iterative approach) to solve each linear subproblem efficiently. The core idea is to maintain two variables that represent the maximum loot possible up to a certain house, either including or excluding that house.

• Maximum Selection: The final result is the maximum of the loot obtained from the two linear subproblems. This ensures we cover all possible scenarios without robbing adjacent houses in the circular arrangement.

• Complexity: O(N) runtime, O(1) storage

Code

    def rob(nums):
    """
    Calculates the maximum amount of money that can be robbed from a circular arrangement of houses.

    Args:
        nums: A list of integers representing the amount of money in each house.

    Returns:
        The maximum amount of money that can be robbed without alerting the police.
    """
    n = len(nums)
    if n == 0:
        return 0
    if n == 1:
        return nums[0]

    def rob_linear(arr):
        """Calculates the maximum amount that can be robbed in a linear arrangement."""
        rob1, rob2 = 0, 0

        for num in arr:
            temp = max(num + rob1, rob2)
            rob1 = rob2
            rob2 = temp
        return rob2

    # Rob houses 0 to n-2 (exclude the last house)
    max_rob1 = rob_linear(nums[:-1])

    # Rob houses 1 to n-1 (exclude the first house)
    max_rob2 = rob_linear(nums[1:])

    return max(max_rob1, max_rob2)