House Robber LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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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 Solutions

Time: O(n)
Space: O(n)

C++

class Solution {
 public:
  int rob(vector<int>& nums) {
    if (nums.empty())
      return 0;
    if (nums.size() == 1)
      return nums[0];

    // dp[i] := max money of robbing nums[0..i]
    vector<int> dp(nums.size());
    dp[0] = nums[0];
    dp[1] = max(nums[0], nums[1]);

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

    return dp.back();
  }
};

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];
  }
}

Python

 class Solution:
  def rob(self, nums: List[int]) -> int:
    if not nums:
      return 0
    if len(nums) == 1:
      return nums[0]

    # dp[i]: = max money of robbing nums[0..i]
    dp = [0] * len(nums)
    dp[0] = nums[0]
    dp[1] = max(nums[0], nums[1])

    for i in range(2, len(nums)):
      dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])

    return dp[-1]

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