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.

Example 1:

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

Solution:

1. Recursive + memoization:

public class Solution {
    public static int rob(int[] houses) {
        int[] memo = new int[houses.length];
        Arrays.fill(memo, -1);
        int res = helper(houses, 0, memo);
        return res;
    }
 
    public static int helper(int[] houses, int src, int[] memo) {
        if (src >= houses.length) {
            return 0;
        }
 
        if (memo[src] != -1) {
            return memo[src];
        }
 
        int pick = houses[src] + helper(houses, src + 2, memo);
        int notpick = 0 + helper(houses, src + 1, memo);
 
        return memo[src] = Math.max(pick, notpick);
    }
}

2. Tabulation:

public class Solution {
    public int rob(int[] houses) {
        int n = houses.length;
        if (n == 1) {
            return houses[0];
        }
        if (n == 2) {
            return Math.max(houses[0], houses[1]);
        }
 
        int[] dp = new int[n + 1];
        dp[1] = houses[0];
        dp[2] = Math.max(houses[0], houses[1]);
 
        for (int i = 3; i <= n; i++) {
	        int pick = houses[i - 1] + dp[i - 2];
	        int notPick = dp[i - 1];
            
            dp[i] = Math.max(pick, notPick);
        }
 
        return dp[n];
    }
}

3. Space optimized

public class Solution {
    public int rob(int[] houses) {
        int n = houses.length;
        if (n == 1) {
            return houses[0];
        }
        if (n == 2) {
            return Math.max(houses[0], houses[1]);
        }
 
        int prev1 = houses[0];
        int prev2 = Math.max(houses[0], houses[1]);
		
        for (int i = 2; i < n; i++) {
            int res = Math.max(houses[i] + prev1, prev2);
            prev1 = prev2;
            prev2 = res;
        }
 
        return prev2;
    }
}