A New Collection of Thoughtful Learning Apps — Now Available on iOS & Android

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I’m excited to share a set of mobile apps I’ve recently completed and published on both the Google Play Store and the Apple App Store. These apps are designed with a simple goal in mind: to make meaningful, structured content more accessible, whether you’re studying theology or improving your English vocabulary. 📱 Now Available on Both Platforms All apps are live and available for download: Google Play Developer Page: https://play.google.com/store/apps/dev?id=5835943159853189043 Apple App Store Developer Page: https://apps.apple.com/ca/developer/q-z-l-corp/id1888794100 📖 Theology & Confession Study Apps For those interested in Reformed theology and classical Christian teachings, I’ve developed a series of apps that present foundational texts in a clean, focused reading format: The Belgic Confession Canons of Dort Heidelberg Catechism Westminster Shorter Catechism Each app is designed to provide a distraction-free experience, making it easier to read, reflect, and revisit these im...

House Robber

Question:
http://www.lintcode.com/en/problem/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 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.

Answer:
time:O(N)
~~~c++
class Solution {
public:
    /*
     * @param A: An array of non-negative integers
     * @return: The maximum amount of money you can rob tonight
     */
    long long houseRobber(vector<int> &A) {
        size_t as = A.size();
        if (as == 0)
          return 0;
        if (as == 1)
          return A[0];
        if (as == 2)
          return max(A[0], A[1]);
        vector<long long> dp(as);
        dp[0] = A[0];
        dp[1] = max(A[0], A[1]);
        for (size_t i = 2; i < as; i++)
          dp[i] = max(dp[i-1], dp[i-2] + A[i]);
        return dp[as-1];
    }
};
~~~
  
time:O(N)
memory:O(1)
~~~c++
class Solution {
public:
    /*
     * @param A: An array of non-negative integers
     * @return: The maximum amount of money you can rob tonight
     */
    long long houseRobber(vector<int> &A) {
        size_t as = A.size();
        long long odd, even;
        odd = even = 0;
        for (size_t i = 0; i < as; i++)
          if (i % 2)
            odd = max(odd+A[i], even);
          else
            even = max(even + A[i], odd);
        return max(odd, even);
    }
};
~~~
                                 
                                 

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