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 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.
Solution:
Time Complexity : O(n)
Space Complexity: O(n)
var rob = function(nums) {
if (nums.length === 0) return 0;
if (nums.length === 1) return nums[0]
// Keep track of the max money we can make with x amount of houses available
// dp[0] = max amount if we only have the first house to rob
// dp[1] = max amount if we only have the first 2 houses to rob
let dp = [nums[0], Math.max(nums[0], nums[1])];
for (let i = 2; i < nums.length; i++) {
// Compare current max with the previous max
// Check if the money from the current house + max of 2 houses away is greater than the current max
dp[i] = Math.max(dp[i-2] + nums[i], dp[i-1]);
}
return dp[nums.length - 1];
};
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