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# House Robber LeetCode Solution

## Problem – 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.``````

Constraints:

• `1 <= nums.length <= 100`
• `0 <= nums[i] <= 400`

### House Robber LeetCode Solution in C++

``````class Solution {
public:
int rob(vector<int>& A, int i = 0) {
return i < size(A) ? max(rob(A, i+1), A[i] + rob(A, i+2)) : 0;
}
};
``````

### House Robber LeetCode Solution in Python

``````class Solution:
def rob(self, A, i = 0):
return max(self.rob(A, i+1), A[i] + self.rob(A, i+2)) if i < len(A) else 0
``````

### House Robber LeetCode Solution in Java

``````public int rob(int[] num) {
int[][] dp = new int[num.length + 1];
for (int i = 1; i <= num.length; i++) {
dp[i] = Math.max(dp[i - 1], dp[i - 1]);
dp[i] = num[i - 1] + dp[i - 1];
}
return Math.max(dp[num.length], dp[num.length]);
}
``````
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This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

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