Description
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: [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: [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.
Algorithm
This is a dynamic programming problem.
Let dp[i] represent the maximum amount of money you can rob from houses 0 to i inclusive.
dp[i] =
0 if i < 0
nums[0] if i == 0
Math.max(nums[0], nums[1]) if i == 1
Math.max(dp[i - 1], dp[i - 2] + nums[i]) otherwise
which can be more concisely represented as:
dp[i] =
0 if i < 0
Math.max(dp[i - 1], dp[i - 2] + nums[i]) otherwise
Solutions
Solution 1 - Dynamic Programming: Recursive
class Solution {
public int rob(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int cache[] = new int[nums.length];
Arrays.fill(cache, -1); // we fill -1 since a house can have 0 money.
return rob(nums, cache, nums.length - 1);
}
private int rob(int[] nums, int[] cache, int n) {
if (n < 0) { // this can happen due to our recursive calls
return 0;
} else if (cache[n] >= 0) {
return cache[n];
}
cache[n] = Math.max(rob(nums, cache, n - 1), rob(nums, cache, n - 2) + nums[n]);
return cache[n];
}
}
Solution 2 - Dynamic Programming: Iterative
class Solution {
public int rob(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
} else 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 - 1], dp[i - 2] + nums[i]);
}
return dp[nums.length - 1];
}
}
Solution 3 - Iterative, no array
Notice that dp only depends on dp[i-1] and dp[i-2]. So instead of storing all the results in the dp array, we can just save the previous 2 values.
class Solution {
public int rob(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
} else if (nums.length == 1) {
return nums[0];
} else if (nums.length == 2) { // new base case needed for 'no array' solution
return Math.max(nums[0], nums[1]);
}
int prev2 = nums[0];
int prev1 = Math.max(nums[0], nums[1]);
int curr = 0;
for (int i = 2; i < nums.length; i++) {
curr = Math.max(prev1, prev2 + nums[i]);
prev2 = prev1;
prev1 = curr;
}
return curr;
}
}