### The Problem

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.

Check LeetCode link for more details and example input & output.

### A Dynamic Programming Solution

Use the bottom up approach.

`dp[i] = max(nums[i] + dp[i-2], dp[i-1])`

`dp[i]` is the answer of the sub problem “i”, whose input array is a sub-array of the orginal input `nums`. The sub-array starts at index `0` and ends at index `i`.

Base cases:

• `dp[0]` is `nums[0]`.
• `dp[-1]` is 0. (It’s used to calculate `dp[1]`.)

#### Java Code

``````  public int rob(int[] nums) {
int dp1 = nums[0]; // dp[i-1]
int dp2 = 0; // dp[i-2]
for (int i = 1; i < nums.length; i++) {
int tmp = dp1;
dp1 = Math.max(nums[i] + dp2, dp1);
dp2 = tmp;
}
return dp1;
}
``````

The time complexity is `O(n)`.