[jylee2033] WEEK 04 solutions

find-minimum-in-rotated-sorted-array/jylee2033.py

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+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        # Binary Search: compare mid with right
+        # nums[mid] > nums[right] -> min is in right half
+        # nums[mid] < nums[right] -> min is in left half (including mid)
+
+        left = 0
+        right = len(nums) - 1
+
+        while left < right:
+            mid = (left + right) // 2
+
+            if nums[mid] > nums[right]:
+                left = mid + 1
+            elif nums[mid] < nums[right]:
+                right = mid
+
+        return nums[left]
+
+# Time Complexity: O(log n)
+# Space Complexity: O(1)

maximum-depth-of-binary-tree/jylee2033.py

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+# 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 maxDepth(self, root: Optional[TreeNode]) -> int:
+        # Recursive approach
+
+        if not root:
+            return 0
+
+        left_depth = self.maxDepth(root.left)
+        right_depth = self.maxDepth(root.right)
+
+        return max(left_depth, right_depth) + 1
+
+# Time Complexity: O(n)
+# Space Complexity: O(n)

merge-two-sorted-lists/jylee2033.py

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+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
+        # Use a while loop to compare values and build the merged list
+
+        # Create a new list
+        output = ListNode()
+        cur = output
+
+        while list1 and list2:
+            if list1.val < list2.val:
+                cur.next = list1
+                list1 = list1.next
+            else:
+                cur.next = list2
+                list2 = list2.next
+
+            cur = cur.next
+
+        # Attach the remaining nodes
+        cur.next = list1 or list2
+
+        return output.next
+
+# Time Complexity : O(n + m), n - lenght of list1, m - length of list2
+# Space Complexity : O(1)