Problem
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.
Solution (Java)
class Solution {
public int rob(int[] num) {
int rob = 0; //max monney can get if rob current house
int notrob = 0; //max money can get if not rob current house
for(int i=0; i<num.length; i++) {
int currob = notrob + num[i]; //if rob current value, previous house must not be robbed
notrob = Math.max(notrob, rob); //if not rob ith house, take the max value of robbed (i-1)th house and not rob (i-1)th house
rob = currob;
}
return Math.max(rob, notrob);
}
}
Solution 1
/**
* @param {number[]} nums
* @return {number}
*/
var rob = function (nums) {
return helper([], 0, nums);
};
var helper = function (dp, i, nums) {
if (i >= nums.length) return 0;
if (dp[i] === undefined) {
dp[i] = Math.max(
helper(dp, i + 1, nums),
nums[i] + helper(dp, i + 2, nums)
);
}
return dp[i];
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).
Solution 2
/**
* @param {number[]} nums
* @return {number}
*/
var rob = function (nums) {
var len = nums.length;
var dp = Array(len);
for (var i = len - 1; i >= 0; i--) {
dp[i] = Math.max(
dp[i + 1] || 0,
nums[i] + (dp[i + 2] || 0)
);
}
return dp[0] || 0;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).
Solution 3
/**
* @param {number[]} nums
* @return {number}
*/
var rob = function (nums) {
var len = nums.length;
var dp = [0, 0];
for (var i = len - 1; i >= 0; i--) {
dp = [Math.max(dp[0], nums[i] + dp[1]), dp[0]];
}
return dp[0];
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(1).