Course Scheduling - Amazon Top Interview Questions


Problem Statement :


You are given a two-dimensional integer list courses representing an adjacency list where courses[i] is the list of prerequisite courses needed to take course i.

Return whether it's possible to take all courses.

Constraints

n, m ≤ 250 where n and m are the number of rows and columns in matrix.

Example 1

Input

courses = [
    [1],
    [],
    [0]
]

Output

True

Explanation

We can take course 1, then course 0, and then course 2


Solution :



title-img 




                        Solution in C++ :

bool solve(vector<vector<int>>& courses) {
    int n = courses.size();
    vector<int> indeg(n + 1, 0);

    // populate indegree
    for (int u = 0; u < n; u++) {
        for (int v : courses[u]) {
            indeg[v]++;
        }
    }
    // push all nodes with zero indegree
    queue<int> q;
    for (int i = 0; i < n; i++) {
        if (!indeg[i]) q.push(i);
    }

    // edge case
    if (q.size() == 0) return false;

    // topo sort
    vector<int> order;
    while (!q.empty()) {
        int u = q.front();
        order.push_back(u);
        q.pop();
        for (int v : courses[u]) {
            if (--indeg[v] == 0) {
                q.push(v);
            }
        }
    }
    // if there's no cycle topo order will have n nodes in it
    return order.size() == n;
}
                    




                        Solution in Python : 
                            
class Solution:
    def solve(self, courses):
        # Initialize list of indegrees for each vertex
        indegrees = [0] * len(courses)

        # Populate the indegrees by going through the adjacency list
        for adj in courses:
            for dest in adj:
                indegrees[dest] += 1

        # Similarly to a classical BFS, we take advantage of a double-ended queue
        # The queue contains all elements which have indegree == 0
        queue = deque([i for i in range(len(indegrees)) if indegrees[i] == 0])

        while queue:
            for _ in range(len(queue)):
                source = queue.popleft()

                # Go through neighbours and remove that edge, insert in queue if
                # queue condition is verified -> indegree == 0
                for elem in courses[source]:
                    indegrees[elem] -= 1
                    if indegrees[elem] == 0:
                        queue.append(elem)

        # Ensure that all the indegrees have been correctly removed
        return all(indegree == 0 for indegree in indegrees)


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