198. House Robber

Hint

class Solution {
public:
    int rob(vector<int>& nums)
    {
        // If there's only one house, return the value in that house

        // Create a DP array to store the maximum amount of money that can be robbed up to each house

        // Initialize the DP array
        // The max money that can be robbed from the first house is the value of the first house
        // The max money from the first two houses is the max value between the two

        // Fill the DP array using the recurrence relation
        for ()
        {
            // The max money that can be robbed up to the i-th house is either:
            // - the max money robbed up to the (i-1)-th house (i.e., skipping the i-th house), or
            // - the max money robbed up to the (i-2)-th house plus the value in the i-th house
        }
        // The last element in the DP array contains the maximum amount of money that can be robbed from all houses
    }
};

• DP

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

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

• T: $O(N)$

• S: $O(N)$

• DP Improved

class Solution {
public:
    int rob(vector<int>& nums)
    {
        int n = nums.size();

        if (n == 1) return nums[0];

        int first = nums[0];
        int second = max(nums[0], nums[1]);
        int curMax = INT_MIN;

        for (int i = 2; i < n; ++i)
        {
            curMax = max(second, first + nums[i]);
            first = second;
            second = curMax;
        }

        return second;
    }
};

• T: $O(N)$
• S: $O(1)$