Merge pull request #1019 from yolophg/main

[�Helena] Week 10

Author: SamTheKorean

course-schedule/yolophg.py

@@ -0,0 +1,43 @@
+# Time Complexity: O(V + E) - visit each course once and process all edges
+# Space Complexity: O(V + E) - storing the graph + recursion stack (worst case)
+
+class Solution:
+    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
+
+        # Build the adjacency list (prereq map)
+        preMap = {i: [] for i in range(numCourses)}
+        for crs, pre in prerequisites:
+            preMap[crs].append(pre)
+        
+        # tracks courses currently being checked (for cycle detection)
+        visited = set()  
+
+        def dfs(crs):
+            # found a cycle
+            if crs in visited:  
+                return False
+            # no more prereqs left
+            if preMap[crs] == []: 
+                return True
+            
+            # mark as visiting
+            visited.add(crs)  
+            
+            # check all prereqs for this course
+            for pre in preMap[crs]:
+                if not dfs(pre):
+                    # cycle detected, return False
+                    return False  
+            
+            # done checking, remove from visited
+            visited.remove(crs)  
+            # mark as completed
+            preMap[crs] = []  
+            return True
+
+        # run dfs on every course
+        for crs in range(numCourses):
+            if not dfs(crs):
+                return False
+        
+        return True  

invert-binary-tree/yolophg.py

@@ -0,0 +1,18 @@
+# Time Complexity: O(N) - visit each node once.
+# Space Complexity: O(N) - skewed tree, recursion goes N levels deep.
+
+class Solution:
+    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
+        if root is None:
+            return root  # If the tree is empty, just return None.
+        
+        # swap left and right child nodes.
+        root.left, root.right = root.right, root.left
+        
+        # recursively invert left and right subtrees.
+        if root.left:
+            self.invertTree(root.left)
+        if root.right:
+            self.invertTree(root.right)
+        
+        return root

jump-game/yolophg.py

@@ -0,0 +1,29 @@
+# Time Complexity: O(n) - go through the array once, checking jump reachability.
+# Space Complexity: O(1) - no extra space used, just a few variables.
+
+class Solution:
+    def canJump(self, nums: List[int]) -> bool:
+
+        if len(nums) == 1:
+            # if there's only one element, we're already at the last index
+            return True  
+
+        # start from second-to-last index
+        backtrack_index = len(nums) - 2 
+        # need to jump at least once to reach the last index 
+        jump = 1  
+
+        while backtrack_index > 0:
+            # If we can jump from this position to the next "safe" spot
+            if nums[backtrack_index] >= jump:
+                # reset jump counter since we found a new reachable point
+                jump = 1  
+            else:
+                # otherwise, increase the jump required
+                jump += 1  
+
+            # move one step left
+            backtrack_index -= 1  
+
+        # finally, check if we can reach the last index from the starting position
+        return jump <= nums[0]

merge-k-sorted-lists/yolophg.py

@@ -0,0 +1,22 @@
+# Time Complexity: O(N log N) - collecting all nodes is O(N), sorting is O(N log N)
+# Space Complexity: O(N) - storing all nodes in an array
+
+class Solution:
+    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
+
+        arr = []
+        
+        # collect all nodes from the linked lists
+        for node in lists:
+            while node:
+                arr.append(node)
+                node = node.next
+
+        # sort all nodes based on their value
+        arr.sort(key=lambda x: x.val)
+
+        # reconnect nodes to form the merged linked list
+        for i in range(len(arr) - 1):
+            arr[i].next = arr[i + 1]
+
+        return arr[0] if arr else None  

search-in-rotated-sorted-array/yolophg.py

@@ -0,0 +1,37 @@
+# Time Complexity: O(log n) - using a modified binary search, 
+# Space Complexity: O(1) - no extra space used, just a few variables.
+
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        # default result, in case the target isn't found
+        res = -1  
+         # set up the binary search range
+        left, right = 0, len(nums) - 1 
+
+        while left <= right:
+            # calculate the middle index
+            mid = left + (right - left) // 2 
+            
+            # found the target, store the index
+            if nums[mid] == target:  
+                res = mid
+            
+            # check if the left half is sorted
+            if nums[left] <= nums[mid]:  
+                if nums[left] <= target < nums[mid]:  
+                    # target in the left half, move right pointer
+                    right = mid - 1  
+                else: 
+                    # otherwise, search in the right half
+                    left = mid + 1 
+            
+            # otherwise, the right half must be sorted
+            else:  
+                if nums[mid] < target <= nums[right]:  
+                    # target in the right half, move left pointer
+                    left = mid + 1 
+                else: 
+                    # otherwise, search in the left half
+                    right = mid - 1 
+
+        return res