198. 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.

Dynamic Programming

• We can rob current house or not rob the current house.
• If we rob the current house then amount will be current house amount + i-2th house amount.
• else the amount will be i-1th house amount.
• we can choose whichever is maximum

Code

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

        vector<int> dp(n + 1, 0);
        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[n - 1];
    }
};

Reduced space complexity DP.

• There are only two choices for the robber, either he rob the i the house or he don’t.
• So the maximum he can rob is rob(i) = max(rob(i-1), rob(i-2) + nums[i])
• in the below code we store rob(i-1) in prev1 and rob(i-2) in prev2.

Code

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

        int prev1 = 0, prev2 = 0;
        for (auto& x : nums) {
            int temp = prev1;
            prev1 = max(prev1, x + prev2);
            prev2 = temp;
        }
        return prev1;
    }
};

MUST READ:

• From good to great. How to approach most of DP problems.