LeetCode – House Robber (Java)

 

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.

Java Solution 1 - Dynamic Programming

The key is to find the relation dp[i] = Math.max(dp[i-1], dp[i-2]+nums[i]).

public int rob(int[] nums) {
    if(nums==null||nums.length==0)
        return 0;
 
    if(nums.length==1)
        return nums[0];
 
    int[] dp = new int[nums.length];
    dp[0]=nums[0];
    dp[1]=Math.max(nums[0], nums[1]);
 
    for(int i=2; i<nums.length; i++){
        dp[i] = Math.max(dp[i-2]+nums[i], dp[i-1]);
    }
 
    return dp[nums.length-1];
}

Java Solution 2

We can use two variables, even and odd, to track the maximum value so far as iterating the array. You can use the following example to walk through the code. 

50 1 1 50

house-robber

public int rob(int[] num) {
	if(num==null || num.length == 0)
		return 0;
 
	int even = 0;
	int odd = 0;
 
	for (int i = 0; i < num.length; i++) {
		if (i % 2 == 0) {
			even += num[i];
			even = even > odd ? even : odd;
		} else {
			odd += num[i];
			odd = even > odd ? even : odd;
		}
	}
 
	return even > odd ? even : odd;
}

Java Solution 3 - Dynamic Programming with Memorization

public int rob(int[] nums) {
    if(nums.length==0){
        return 0;
    }
 
    int[] mem = new int[nums.length+1]; 
    Arrays.fill(mem, -1);
 
    mem[0] = 0;
 
    return helper(nums.length, mem, nums);
}
 
private int helper(int size, int[] mem, int[] nums){
    if(size <1){
        return 0;
    }
 
    if(mem[size]!=-1){
        return mem[size];
    }
 
    //two cases
    int firstSelected = helper(size-2, mem, nums) + nums[nums.length -size];
    int firstUnselected = helper(size-1, mem, nums);
 
    return mem[size] = Math.max(firstSelected, firstUnselected);
}

Related posts:

  1. LeetCode – House Robber II (Java)
  2. LeetCode – House Robber III (Java)
  3. LeetCode – Paint House (Java)
  4. LeetCode – Paint House II (Java)
Category >> Algorithms >> Interview  
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  • Ajay V

    store the index of which ever is maximum out of dp[i-2]+num[i] , dp[i-1] in an array houses. Now backtrack from houses[n].

  • Neha Rao

    how to print indexes of robbed houses??

  • Internet Hero

    In the last there will be a comparison between 51 and 150 and 150 will be picked. Looks like it works.

  • OJ

    It will work

  • rafaelolg

    You are right!
    When the robber is in the ith house he has to decide if he is going to steal ith or i+1th house:
    So, if N is the number of houses and v_i the value in the ith house we can define recursively the maximum value as:

    maxvalor(i) = { 0 if i >= N;

    max(v_i + max(i+2) , max(i+1))
    }
    For each i there is two calls of max(i) in this definition, so a plain recursive implementation will give you an exponential solution. So you need to make a memoization of results.

  • Mario

    counter example showing that your odd-even solution may not work: [50, 1, 1, 100]