# 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 :

```                        ```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
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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