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
import java.util.*; import java.lang.*; public class HouseRobber { public static int rob(int[] nums) { int n=nums.length; if(n==0) return 0; if(n==1) return nums[0]; int a[]=new int[n]; a[0]=nums[0]; a[1]=Math.max(nums[0],nums[1]); for(int i=2;i<n;i++) a[i]=Math.max(nums[i]+a[i2], a[i1]); return a[n1]; } public static void main(String args[]) { int arr[]={1,2,3,1}; System.out.println(rob(arr)); } }
Complexity Analysis for House Robber Solution
Time Complexity
O(n) : we are iterating from 1st house to nth house in a single loop, where n is the number of total houses.
Space Complexity
O(n) : we have used an array of size n to store the intermediate result.