House Robber II

Problem

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

Solution

/**
 * @param {number[]} nums
 * @return {number}
 */
var rob = function(nums) {
    if (nums.length === 1) {
        return nums[0];
    }
    const cp1 = [...nums];
    const cp2 = [...nums];
    cp1.shift(); // cp1 has all houses except first
    cp2.pop(); // cp2 has all houses except last
    return Math.max(rob2(cp1), rob2(cp2));
};

var rob2 = function(nums) {
    const dp = [0, nums[0]];
    for (let i = 1; i < nums.length; i++) {
        dp.push(Math.max(dp[i - 1] + nums[i], dp[i]));
    }
    return dp[nums.length];
};

We will implement a reduction solution. Essentially, this problem reduces down to the House Robber problem. Since the only additional constraint is that the houses are arranged in a circle, we just need to test whether excluding the first or last house gives a greater amount of money.

Problem​

Solution​

Problem

Solution