LeetCode 198. House Robber

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 andit 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 tonightwithout alerting the police.

At the first sight, this is very simple..... why not just add the even indexes and odd indexes, then compare.....

After a second thought, I found this is really stupid if I was the thief. For example: [4, 2, 3, 13, 8] --> even indexes will sum to 4+3+8 = 15. Odd indexes will sum to 2+13 = 15. But if I steal 4 + 13 = 17, I can get the maximum.

Thus, in order to steal the max, I need to consider whether the max sum of (i - 1, i -2) if sum(i-2)  + num[i] > sum(i-1), I can steal i, otherwise, I will drop i to keep the value of sum(i-1).

This is DP then.... to remember the max value I passed by before.

    int rob(vector<int>& nums) {
        if(nums.size() == 0) return 0;
        if(nums.size() == 1) return nums[0];
        if(nums.size() == 2) return max(nums[0], nums[1]);
        int n = nums.size();
        vector<int> dp(n, 0);
        dp[0] = nums[0];
        dp[1] = max(nums[0], nums[1]);
        for(int i = 2; i < n; ++i) {
            dp[i] = max(dp[i-2] + nums[i], dp[i-1]);
        }
        return dp[n-1];
    }

To further save space: use rolling array.

int robII(vector<int>& nums) {
  if(nums.size() == 0) return 0;
  if(nums.size() == 1) return nums[0];
  if(nums.size() == 2) return max(nums[0], nums[1]);
  int prev = nums[0];
  int current = max(nums[0], nums[1]);
  for(int i = 3; i < nums.size(); ++i) {
    int tmp = max(current, prev + nums[i]);
    prev = current;
    current = tmp;
  }
  return current;
}

Follow up: out the robbed house number to make the largest robbing value.

#include "header.h"
using namespace std;

// return the max robbing value and the robbed houses.
vector<bool> maxProfit(vector<int>& houses) {
  int n = houses.size();
  vector<bool> houseNumber(n, false);  // house being robbed.
  vector<int> dp(n, 0);
  dp[0] = houses[0]; houseNumber[0] = true;
  dp[1] = max(houses[0], houses[1]);
  if(houses[0] < houses[1]) {houseNumber[0] = false, houseNumber[1] = true;}
  for(int i = 2; i < n; ++i) {
    dp[i] = max(dp[i-1], dp[i-2] + houses[i]);
    if(dp[i-1] < (dp[i-2] + houses[i])) {  if we would like to rob house i, we will have to drop house i - 1.
      houseNumber[i-1] = false;
      houseNumber[i] = true;
    }
  }
  return houseNumber;
}

int main(void) {
  vector<int> houses{500, 3, 300, 13, 8};
  vector<bool> res = maxProfit(houses);
  for(int i = 0; i < res.size(); ++i) {
    if(res[i] == true) cout << i << endl;
  }
}