198. House Robber LeetCode Solution

Last updated: February 15, 2025

In this guide, you will get 198. House Robber LeetCode Solution with the best time and space complexity. The solution to House Robber problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
House Robber solution in C++
House Robber solution in Java
House Robber solution in Python
Additional Resources

198. House Robber LeetCode Solution image

Problem Statement of 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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

198. House Robber LeetCode Solution in C++

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

    // dp[i] := the maximum 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();
  }
};
/* code provided by PROGIEZ */

198. House Robber LeetCode 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] := the maximum 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];
  }
}
// code provided by PROGIEZ

198. House Robber LeetCode Solution in 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]
 # code by PROGIEZ

Additional Resources

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