Baozi Training Leetcode solution 198. House Robber

Problem Statement 

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 system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine 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:

  • 0 <= nums.length <= 100
  • 0 <= nums[i] <= 400

 Problem link

Video Tutorial

You can find the detailed video tutorial here

Thought Process

Easy problem and Dynamic Programming(DP) should jump into mind given it's only asking for max values (Just think about different combo we have to do without using DP, a little less than 2^N)
 
The mathematical induction formula is below, for any current max money at index i, you either choose to use the i-1 or i-2 + current house's money to not trigger police.
max[i] = max(max[i - 2] + a[i], max[i-1])

Solutions

DP

 1 public int rob(int[] num) {
 2     if (num == null || num.length == 0) {
 3         return 0;
 4     }
 5 
 6     int n = num.length;
 7     int[] lookup = new int[n + 1]; // DP array size normally larger than 1
 8     lookup[0] = 0;
 9     lookup[1] = num[0];
10 
11     for (int i = 2; i <= n; i++) {
12         lookup[i] = Math.max(lookup[i - 1], lookup[i - 2] + num[i - 1]);
13     }
14 
15     return lookup[n];
16 }

 

Time Complexity: O(N) N is the array size
Space Complexity: O(N) since we used an extra array

References

posted @ 2020-06-15 02:04  包子模拟面试  阅读(129)  评论(0编辑  收藏  举报