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...

Course Schedule

Question:

http://www.lintcode.com/en/problem/course-schedule/
There are a total of n courses you have to take, labeled from 0 to n - 1.
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?

Answer:

class Solution {
public:
    /*
     * @param numCourses: a total of n courses
     * @param prerequisites: a list of prerequisite pairs
     * @return: true if can finish all courses or false
     */
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        // Write your code here
        vector<vector<int>> adj(numCourses);
        vector<int> indegree(numCourses);
        for (int i = 0; i < numCourses; i++)
             indegree[i] = 0;
        for (auto pq : prerequisites) {
            adj[pq.first].push_back(pq.second);
            indegree[pq.second]++;
        }
        stack<int> st;
        for (int i = 0; i < numCourses; i++) {
            if (indegree[i] == 0)
                st.push(i);
        }
        int count = 0;
        while (!st.empty()) {
            int v1 = st.top();
            st.pop();
            count++;
            for (auto v2 : adj[v1]) {
                indegree[v2]--;
                if (indegree[v2] == 0)
                    st.push(v2);
            }
        }
        return count == numCourses;
    }
}

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