House Robber LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the House Robber in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – House Robber– LeetCode Problem

House Robber– LeetCode Problem

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.

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 Solutions

House Robber  Solution in C++:

class Solution {
 public:
  int rob(vector<int>& nums) {
    int prev1 = 0;  // dp[i - 1]
    int prev2 = 0;  // dp[i - 2]

    for (const int num : nums) {
      const int dp = max(prev1, prev2 + num);
      prev2 = prev1;
      prev1 = dp;
    }

    return prev1;
  }
};

House Robber Solution in Java:

class Solution {
  public int rob(int[] nums) {
    final int n = nums.length;
    if (n == 0)
      return 0;
    if (n == 1)
      return nums[0];

    // dp[i] := max money of robbing nums[0..i]
    int[] dp = new int[n];
    dp[0] = nums[0];
    dp[1] = Math.max(nums[0], nums[1]);

    for (int i = 2; i < n; ++i)
      dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);

    return dp[n - 1];
  }
}

House Robber Solution in Python:

class Solution:
  def rob(self, nums: List[int]) -> int:
    prev1 = 0  # dp[i - 1]
    prev2 = 0  # dp[i - 2]

    for num in nums:
      dp = max(prev1, prev2 + num)
      prev2 = prev1
      prev1 = dp

    return prev1