Completed BFS-1

BinaryTreeLevelOrderTraversal.py

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+# Time Complexity : O(n), where n is the number of nodes in the binary tree
+# Space Complexity : O(w), where w is the maximum width of the binary tree
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Your code here along with comments explaining your approach
+
+from collections import deque
+from typing import Optional, List
+
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+class Solution:
+    def __init__(self):
+        self.queue = deque() # Intialize a queue to keep track of the nodes at each level
+        self.result = [] # Intialize a result list to store the values of nodes at each level
+
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        if not root: # If the root is None, return an empty list
+            return self.result
+        self.queue.append(root) # Add the root node to the queue to start the level order traversal
+
+        while len(self.queue) > 0: # While there are nodes in the queue, continue the traversal
+            size = len(self.queue) # Get the number of nodes at the current level (size of the queue)
+            res=[] # Initialize a list to store the values of nodes at the current level
+            for i in range(size): # Iterate through the nodes at the current level
+                curr = self.queue.popleft() # Remove the current node from the queue
+                print(curr.val)
+                if curr.left is not None: # If the left child of the current node is not None, add it to the queue
+                    self.queue.append(curr.left)
+                if curr.right is not None: # If the right child of the current node is not None, add it to the queue
+                    self.queue.append(curr.right)
+                res.append(curr.val) # Add the value of the current node to the res list for the current level
+            self.result.append(res) # After processing all nodes at the current level, add the res list to the result list
+
+        return self.result # Return the final result list containing the values of nodes at each level
+
+if __name__ == "__main__":
+    root = TreeNode(3)
+    root.left = TreeNode(9)
+    root.right = TreeNode(20)
+    root.right.left = TreeNode(15)
+    root.right.right = TreeNode(7)
+
+    sol = Solution()
+    print(sol.levelOrder(root))

CourseSchedule.py

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+# Time Complexity : O(V+E), where V is the number of courses/Vertices and E is the number of dependencies/Edges
+# Space Complexity : O(V+E), where V is the number of courses/Vertices and E is the number of dependencies/Edges
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+
+# Your code here along with comments explaining your approach
+
+from collections import deque
+from typing import List
+
+class Solution:
+
+    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
+        q = deque() # Initializing queue
+        indegrees = [0] * numCourses # Intiailizing  indegrees array to store count of dependent on particular index/course
+        adjajency = {} # Adjajency hashmap to store Indexpendent course as key and list of dependent course as value to access faster
+        count = 0 # initializing count which would be incremented when a course has been satisfied
+
+        for pair in prerequisites: # Iterating the list of tuples
+            dep = pair[0] # first course is dependent
+            indep = pair[1] # second course is independent
+
+            indegrees[dep] += 1 # increasing the count of dependencies on each course
+
+            if indep not in adjajency: 
+                adjajency[indep] = []
+
+            adjajency[indep].append(dep) # Adding the dependent course to its independent course
+
+        for idx, num in enumerate(indegrees): # storing courses with 0 dependencies in q and increasing count
+            if num == 0:
+                q.append(idx)
+                count += 1
+
+        while len(q): # If any course with 0 dependencies present, otherwise skip
+            curr = q.pop() 
+            if curr not in adjajency: # Handling Null if no dependency on any course
+                continue
+            dep_list = adjajency[curr]
+            for num in dep_list:
+                indegrees[num] -= 1 # Reducing the dependency of that course OR completing pre req
+                if indegrees[num] == 0:
+                    q.append(num)
+                    count += 1
+
+        if count == numCourses: # At the end if count is equal to nuCourses, it means we can complete all the courses 
+            return True
+
+        return False