Add implementation for leetcode problems 102, 207

leetcode_102.py

@@ -0,0 +1,135 @@
+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Mar 15 00:00:00 2026
+
+@author: rishigoswamy
+
+    LeetCode 102: Binary Tree Level Order Traversal
+    Link: https://leetcode.com/problems/binary-tree-level-order-traversal/
+
+    Problem:
+    Given the root of a binary tree, return the level order traversal of its
+    nodes' values (i.e., from left to right, level by level).
+
+    Approach:
+    DFS with a depth parameter. Maintain a running array where index = level.
+    When visiting a node at depth ht, append a new list if one doesn't exist yet,
+    then append the node's value to array[ht].
+
+    1️⃣ If array has no bucket for depth ht, append a new empty list.
+    2️⃣ Append node.val to array[ht].
+    3️⃣ Recurse left and right with ht+1.
+
+    // Time Complexity : O(n)
+        Every node is visited once.
+    // Space Complexity : O(n)
+        Output array stores all node values; call stack is O(h).
+
+"""
+
+from collections import defaultdict, deque
+from typing import List, Optional
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+
+class Solution:
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        if not root:
+            return []
+        self.array = []
+        self.htMax = 0
+
+        def height(node, ht):
+            self.htMax = max(self.htMax, ht)
+            if not node:
+                return
+            if len(self.array) <= ht:
+                self.array.append([])
+            self.array[ht].append(node.val)
+
+            height(node.left, ht + 1)
+            height(node.right, ht + 1)
+
+        height(root, 0)
+        return self.array
+
+    '''
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        if not root:
+            return []
+        hMap = defaultdict(list)
+        self.htMax = 0
+
+        def height(node, ht):
+            self.htMax = max(self.htMax, ht)
+            if not node:
+                return
+            hMap[ht].append(node.val)
+            height(node.left, ht + 1)
+            height(node.right, ht + 1)
+
+        height(root, 0)
+        ans = []
+        for keys in range(0, self.htMax):
+            ans.append(hMap[keys])
+
+        return ans
+    '''
+
+    '''
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        # Approach uses size of queue for every level
+        if not root:
+            return []
+        queue = deque()
+        queue.append(root)
+        ans = []
+        while queue:
+            level = []
+            size = len(queue)
+            for i in range(size):
+                node = queue.popleft()
+                level.append(node.val)
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+            ans.append(level)
+
+        return ans
+    '''
+
+    '''
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        # Approach uses a None delimiter for every level
+        if not root:
+            return []
+
+        queue = deque()
+        queue.append(root)
+        queue.append(None)
+
+        level = []
+        ans = []
+        while len(queue) > 1:
+            node = queue.popleft()
+
+            if not node:
+                ans.append(level)
+                level = []
+                queue.append(None)
+            else:
+                level.append(node.val)
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+        ans.append(level)
+        return ans
+    '''

leetcode_207.py

@@ -0,0 +1,72 @@
+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Mar 15 00:00:00 2026
+
+@author: rishigoswamy
+
+    LeetCode 207: Course Schedule
+    Link: https://leetcode.com/problems/course-schedule/
+
+    Problem:
+    There are numCourses courses labeled 0 to numCourses-1. You are given an array
+    prerequisites where prerequisites[i] = [a, b] means you must take course b before a.
+    Return true if you can finish all courses, false if a cycle exists.
+
+    Approach:
+    Topological sort (Kahn's algorithm / BFS). Build an adjacency list and indegree
+    array from prerequisites. Seed the queue with all zero-indegree nodes, then
+    repeatedly remove a node and decrement the indegree of its neighbors.
+    If all indegrees reach 0, no cycle exists.
+
+    1️⃣ Build adjMap and indegreeCount from prerequisites.
+    2️⃣ Seed queue with all nodes whose indegree == 0.
+    3️⃣ Early exit: if no zero-indegree nodes → cycle guaranteed → False.
+    4️⃣ Early exit: if all nodes have indegree 0 → no prerequisites → True.
+    5️⃣ BFS: for each dequeued course, decrement neighbors' indegrees; enqueue if 0.
+    6️⃣ After BFS, if any indegree remains > 0 → cycle exists → False.
+
+    // Time Complexity : O(V + E)
+        V = numCourses, E = number of prerequisites.
+    // Space Complexity : O(V + E)
+        Adjacency list and indegree array.
+
+"""
+
+from collections import defaultdict, deque
+from typing import List
+
+
+class Solution:
+    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
+        indegreeCount = [0] * numCourses
+        adjMap = defaultdict(list)
+
+        for preRequisite in prerequisites:
+            adjMap[preRequisite[1]].append(preRequisite[0])
+            indegreeCount[preRequisite[0]] += 1
+
+        queue = deque()
+        for i in range(len(indegreeCount)):
+            if indegreeCount[i] == 0:
+                queue.append(i)
+
+        # Further optimizations:
+        if len(queue) == 0:
+            return False
+
+        if len(queue) == numCourses:
+            return True
+
+        while queue:
+            currCourse = queue.popleft()
+            for childCourse in adjMap[currCourse]:
+                indegreeCount[childCourse] -= 1
+                if indegreeCount[childCourse] == 0:
+                    queue.append(childCourse)
+
+        for i in range(len(indegreeCount)):
+            if indegreeCount[i] != 0:
+                return False
+
+        return True