[alphaorderly] WEEK 04 Solutions (#2737)

* week 1

* [alphaorderly] WEEK 02 Solutions

* fix: description에 맞게 코드 수정

* fix: 좀 더 간결하게 수정

* [alphaorderly] WEEK 03 Solutions

* fix: 불필요한 코드 삭제

* fix: 줄바꿈 린트 문제 수정

* [alphaorderly] WEEK 03 Solutions

* fix: 린트 오류 수정

* fix: 파이썬 내장함수 사용 코드 추가

* fix: valid-palindrome 문제에 투포인터 구현 답안 추가

* fix: 로직 가독성 수정

* [alphaorderly] WEEK 04 Solutions - draft

* fix: 코드 가독성 향상

Author: alphaorderly

coin-change/alphaorderly.py

@@ -0,0 +1,72 @@
+"""
+Time Complexity: O(n * amount)
+Space Complexity: O(n * amount)
+
+- DP / Top-Down approach
+- Use a cache to store the results of the subproblems
+"""
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        LARGE = 10 ** 5
+
+        @cache
+        def dp(coin: int, left: int) -> int:
+            if left == 0:
+                return 0
+
+            if left < 0 or coin >= len(coins):
+                return LARGE
+
+            a = dp(coin, left - coins[coin]) + 1
+            b = dp(coin + 1, left)
+
+            return min(a, b)
+
+        ans = dp(0, amount)
+
+        return ans if ans != LARGE else -1
+
+"""
+Time Complexity: O(n * amount)
+Space Complexity: O(n * amount)
+
+- DP / Bottom-Up approach
+- Use a 2D array to store the results of the subproblems
+"""
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        N = len(coins)
+        dp = [[float('inf')] * (amount + 1) for _ in range(N)]
+
+        for c in range(N):
+            dp[c][0] = 0
+
+            if c > 0:
+                for a in range(amount + 1):
+                    dp[c][a] = dp[c - 1][a]
+
+            for a in range(coins[c], amount + 1):
+                dp[c][a] = min(dp[c][a], dp[c][a - coins[c]] + 1)
+
+        ans = dp[-1][-1]
+
+        return ans if ans != float('inf') else -1
+
+"""
+Time Complexity: O(n * amount)
+Space Complexity: O(amount)
+
+- DP / Bottom-Up approach [Space Optimized]
+- Use a 1D array to store the results of the subproblems
+"""
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        N = len(coins)
+        dp = [float("inf")] * (amount + 1)
+        dp[0] = 0
+
+        for c in range(N):
+            for a in range(coins[c], amount + 1):
+                dp[a] = min(dp[a], dp[a - coins[c]] + 1)
+
+        return dp[-1] if dp[-1] != float("inf") else -1

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

@@ -0,0 +1,40 @@
+"""
+Time Complexity: O(log N)
+Space Complexity: O(1)
+
+### Binary Search Solution ###
+
+1. Initialize two pointers, left and right, to the start and end of the array
+2. While left is less than right, calculate the middle index
+3. If nums[mid] > nums[right], search the right half (left = mid + 1)
+  - Why? mid is in the left sorted portion (larger values), so the rotation point (minimum) must be to the right of mid
+4. Otherwise, search the left half including mid (right = mid)
+  - nums[mid] <= nums[right] means mid is in the right sorted portion or the array is not rotated, so the minimum is at mid or to its left
+5. When left == right, return nums[right]
+"""
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        N = len(nums)
+        left, right = 0, N - 1
+
+        while left < right:
+            mid = (left + right) // 2
+
+            if nums[mid] > nums[right]:
+                left = mid + 1
+            else:
+                right = mid
+
+        return nums[right]
+
+"""
+Time Complexity: O(N)
+Space Complexity: O(1)
+
+### Linear Search Solution ### ( NOT RECOMMENDED )
+
+1. Return the minimum element in the array
+"""
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        return min(nums)

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

@@ -0,0 +1,122 @@
+# 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
+"""
+Time Complexity: O(N)
+Space Complexity: O(N) - Recursive stack space
+
+### Recursive Solution ( DFS ) ###
+
+1. If the root is None, return 0
+2. Return the maximum of the depth of the left and right subtrees + 1
+"""
+class Solution:
+    def maxDepth(self, root: Optional[TreeNode]) -> int:
+        if not root:
+            return 0
+
+        return max(self.maxDepth(root.left), self.maxDepth(root.right)) + 1
+
+"""
+Time Complexity: O(N)
+Space Complexity: O(N) - Stack space
+
+### Iterative Solution ( DFS ) ###
+
+1. If the root is None, return 0
+2. Use a stack to store the nodes and the level of the nodes
+3. Pop the nodes from the stack and update the maximum depth
+4. If the node has a left child, add the left child and the level + 1 to the stack
+5. If the node has a right child, add the right child and the level + 1 to the stack
+6. Return the maximum depth
+"""
+class Solution:
+    def maxDepth(self, root: Optional[TreeNode]) -> int:
+        if not root:
+            return 0
+
+        s = [(root, 1)]
+        ans = 1
+
+        while s:
+            node, level = s.pop()
+            ans = max(ans, level)
+
+            if node.left:
+                s.append((node.left, level + 1))
+
+            if node.right:
+                s.append((node.right, level + 1))
+
+        return ans
+
+"""
+Time Complexity: O(N)
+Space Complexity: O(N) - Queue space
+
+### Iterative Solution ( BFS ) ###
+
+1. If the root is None, return 0
+2. Use a queue to store the nodes and the level of the nodes
+3. Pop the nodes from the queue and update the maximum depth
+4. If the node has a left child, add the left child and the level + 1 to the queue
+5. If the node has a right child, add the right child and the level + 1 to the queue
+6. Return the maximum depth
+"""
+class Solution:
+    def maxDepth(self, root: Optional[TreeNode]) -> int:
+        if not root:
+            return 0
+
+        q = deque([(root, 1)])
+        ans = 1
+
+        while q:
+            node, level = q.popleft()
+            ans = max(ans, level)
+
+            if node.left:
+                q.append((node.left, level + 1))
+
+            if node.right:
+                q.append((node.right, level + 1))
+
+        return ans
+
+"""
+Time Complexity: O(N)
+Space Complexity: O(N) - Queue space
+
+### Iterative Solution ( Level Order Traversal ) ###
+
+1. If the root is None, return 0
+2. Use a queue to store the nodes
+3. Pop the nodes from the queue and update the maximum depth
+4. If the node has a left child, add the left child to the queue
+5. If the node has a right child, add the right child to the queue
+6. Return the maximum depth
+"""
+class Solution:
+    def maxDepth(self, root: Optional[TreeNode]) -> int:
+        if not root:
+            return 0
+
+        q = deque([root])
+        level = 0
+
+        while q:
+            level += 1
+            N = len(q)
+            for _ in range(N):
+                node = q.popleft()
+
+                if node.left:
+                    q.append(node.left)
+
+                if node.right:
+                    q.append(node.right)
+        
+        return level

merge-two-sorted-lists/alphaorderly.py

@@ -0,0 +1,57 @@
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+"""
+Time Complexity: O(N)
+Space Complexity: O(1)
+
+1. Create a dummy node to store the head of the merged list ( return_node )
+2. Traverse the two lists until one of the lists is empty
+3. Compare the values of the two nodes and add the smaller node to the dummy node
+4. Move the pointer of the list with the smaller node to the next node
+5. Add the remaining nodes of the non-empty list to the dummy node
+6. Return the next node of the dummy node
+"""
+class Solution:
+    def mergeTwoLists(
+        self, list1: Optional[ListNode], list2: Optional[ListNode]
+    ) -> Optional[ListNode]:
+        head = ListNode()
+        node = head
+
+        while list1 and list2:
+            if list1.val > list2.val:
+                head.next = list2
+                list2 = list2.next
+            else:
+                head.next = list1
+                list1 = list1.next
+
+            head = head.next
+
+        head.next = list1 or list2
+
+        return node.next
+
+"""
+Time Complexity: O(N)
+Space Complexity: O(N) - Recursive stack space
+
+1. If one of the lists is empty, return the other list
+2. Compare the values of the two nodes and add the smaller node to the merged list
+3. Move the pointer of the list with the smaller node to the next node
+4. Return the next node of the merged list
+"""
+class Solution:
+    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
+        if not list1 or not list2:
+            return list1 or list2
+
+        if list1.val > list2.val:
+            list2.next = self.mergeTwoLists(list1, list2.next)
+            return list2
+        else:
+            list1.next = self.mergeTwoLists(list1.next, list2)
+            return list1