Completed BFS-1

[viveksharma180]

No description provided.

[super30admin]

Strengths:

• The solution correctly implements the level order traversal using BFS with a queue.
• The code is well-commented, making it easy to understand the logic.
• The time and space complexity are optimal.

Areas for Improvement:

• Avoid using instance variables (self.queue and self.result) for storing temporary data in the method. These should be local variables to prevent issues if the same Solution instance is used multiple times. For example, if you call levelOrder twice on the same instance, the second call will have the queue and result already populated from the first call, leading to incorrect results.
• Remove the print statement (print(curr.val)) as it is not needed for the solution and may affect performance for large trees.
• Consider using a more standard approach without instance variables. For example, you can initialize the queue and result inside the method.

Here's a revised version of your code without instance variables:

from collections import deque
from typing import Optional, List

class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        queue = deque([root])
        result = []
        while queue:
            size = len(queue)
            level = []
            for _ in range(size):
                curr = queue.popleft()
                level.append(curr.val)
                if curr.left:
                    queue.append(curr.left)
                if curr.right:
                    queue.append(curr.right)
            result.append(level)
        return result

This version is cleaner and avoids potential state issues.

[viveksharma180]

@super30admin What about the Course Scheduler?

[super30admin]

Level Order Traversal in Binary tree (BinaryTreeLevelOrderTraversal.py)

Evaluation completed, but no feedback text was generated.

VERDICT: NEEDS_IMPROVEMENT

---

Scheduling Courses (CourseSchedule.py)

Strengths:

• The solution correctly implements Kahn's algorithm for topological sorting.
• Comments are provided, making the code easy to understand.
• The time and space complexity are optimal.

Areas for Improvement:

• There is a typo in the variable name "adjajency" – it should be "adjacency" for clarity.
• The variable "dep" and "indep" in the loop might be confusing; consider using more descriptive names like "dependent" and "independent" or "course" and "prerequisite".
• The code uses q.pop() which from a deque without specifying (by default pops from the right) is efficient, but note that for BFS, typically we use popleft() to make it a true queue. However, in this algorithm, the order of processing nodes with zero in-degree does not matter, so using pop() (which behaves like a stack) is also correct. But for consistency with BFS, you might want to use popleft().
• The condition if curr not in adjajency can be avoided by using adjacency.get(curr, []) to get the list of dependents, which would return an empty list if the key is not present. This would make the code slightly cleaner.

Suggested changes:

from collections import deque

class Solution:
    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        q = deque()
        indegrees = [0] * numCourses
        adjacency = {}  # Corrected spelling
        count = 0

        for pair in prerequisites:
            course, prereq = pair[0], pair[1]  # More descriptive names
            indegrees[course] += 1
            if prereq not in adjacency:
                adjacency[prereq] = []
            adjacency[prereq].append(course)
        
        for idx, num in enumerate(indegrees):
            if num == 0:
                q.append(idx)
                count += 1
        
        while q:
            curr = q.popleft()  # Use popleft for BFS
            for neighbor in adjacency.get(curr, []):  # Avoids the need to check if key exists
                indegrees[neighbor] -= 1
                if indegrees[neighbor] == 0:
                    q.append(neighbor)
                    count += 1
        
        return count == numCourses

Overall, the solution is correct and efficient. The minor issues do not affect the functionality.

VERDICT: PASS