[Sam] Week 2 solutions

invert-binary-tree/SamTheKorean.py

@@ -0,0 +1,17 @@
+# Time complexity : O(n)
+# Space complexity : O(n)
+class Solution:
+    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
+        stack = [root]
+
+        while stack:
+            node = stack.pop()
+
+            # Not accessing child if node is Null
+            if not node:
+                continue
+
+            node.left, node.right = node.right, node.left
+            stack += [node.left, node.right]
+
+        return root

linked-list-cycle/SamTheKorean.py

@@ -0,0 +1,15 @@
+# Time complexity : O(n)
+# Space complexity : O(1)
+class Solution:
+    def hasCycle(self, head: Optional[ListNode]) -> bool:
+        slow = head
+        fast = head
+
+        # check fast and fast.next are not the last node to ensure we can access fast.next.next
+        while fast and fast.next:
+            slow = slow.next
+            fast = fast.next.next
+            if slow == fast:
+                return True
+
+        return False

merge-two-sorted-lists/SamTheKorean.py

@@ -0,0 +1,32 @@
+# Time complexity : O(m*n)
+# Space complexity : O(1)
+class Solution:
+    def mergeTwoLists(
+        self, list1: Optional[ListNode], list2: Optional[ListNode]
+    ) -> Optional[ListNode]:
+        # Create a dummy node to simplify the logic
+        dummy = ListNode(None)
+        # Pointer to the current node in the merged list
+        node = dummy
+
+        # Loop until both lists have nodes
+        while list1 and list2:
+            # Choose the smaller value between list1 and list2
+            if list1.val < list2.val:
+                # Append list1 node to the merged list
+                node.next = list1
+                # Move to the next node in list1
+                list1 = list1.next
+            else:
+                # Append list2 node to the merged list
+                node.next = list2
+                # Move to the next node in list2
+                list2 = list2.next
+            # Move to the next node in the merged list
+            node = node.next
+
+        # Append the remaining nodes from list1 or list2
+        node.next = list1 or list2
+
+        # Return the merged list (skip the dummy node)
+        return dummy.next

reverse-linked-list/SamTheKorean.py

@@ -0,0 +1,27 @@
+# Time complexity : O(n)
+# Space complexity : O(n)
+class Solution:
+    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        # Return None when head is None
+        if not head:
+            return None
+
+        stack = []
+        node = head
+
+        while node:
+            stack.append(node)
+            node = node.next
+
+        head = stack.pop()
+        node = head
+
+        # End the loop when stack is empty
+        while stack:
+            node.next = stack.pop()
+            node = node.next
+
+        # Set the last node's next to None to indicate the end of the reversed list
+        node.next = None
+
+        return head

valid-parentheses/SamTheKorean.py

@@ -0,0 +1,22 @@
+# Time complexity : O(n)
+# Space complexity : O(n)
+class Solution:
+    def isValid(self, s: str) -> bool:
+        parentheses = {"[": "]", "{": "}", "(": ")"}
+        stack = []
+
+        # if ch is opening bracket
+        for ch in s:
+            if ch in parentheses:
+                stack.append(ch)
+            # if ch is closing bracket
+            else:
+                # if closing bracket comes without opening bracket previously
+                if not stack:
+                    return False
+                # if closing bracket doesnt match with the latest type of opening bracket
+                if ch != parentheses[stack.pop()]:
+                    return False
+
+        # True if stack is empty after going through the process above
+        return not stack