House Robber - Array - Easy - LeetCode
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House Robber - Array - Easy - LeetCode

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  • 1The problem involves maximizing money robbed from houses without alerting police by avoiding adjacent house robberies.
  • 2Examples demonstrate that optimal strategies yield maximum amounts of 4 and 12 from given house values.
  • 3The solution employs a dynamic programming approach with O(n) time complexity and O(1) space complexity.

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"The problem involves maximizing money robbed from houses without alerting police by avoiding adjacent house robberies."

House Robber - Array - Easy - LeetCode

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: 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:

0 <= nums.length <= 100 0 <= nums[i] <= 400

public class Solution {
 public int Rob(int[] nums) {
 if(nums.Length==0){
 return 0;
 }
 if(nums.Length == 1){
 return nums[0];
 }
 int skipSum = 0;
 int currSum = nums[0];
 
 for(int i=1;i<nums.Length;i++){
 int max = Math.Max(nums[i]+skipSum, currSum);
 skipSum = currSum;
 currSum = max;
 }
 
 return Math.Max(skipSum, currSum);
 }
}

Time Complexity: O(n)

Space Complexity: O(1)

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sumitc91

Published on 3 October 2020 · 1 min read · 223 words

Part of AskGif Blog · coding

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