[ppxyn1] WEEK 04 solutions

coin-change/ppxyn1.py

@@ -0,0 +1,47 @@
+# idea: DFS/BFS, DP
+
+from collections import deque
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        # Why Greedy is not possiple way?
+        # A greedy is only optimal in specific coin systems (e.g., denominations like 1, 5, 10, 25)
+        # For arbitrary coin denominations, a greedy approach does not always yield the optimal solution.
+        queue = deque([(0,0)]) # (동전갯수, 누적금액)
+        while queue:
+            count, total = queue.popleft()
+            if total == amount:
+                return count
+            for coin in coins:
+                if total + coin <= amount:
+                    queue.append([count+1, total+ coin])
+        return -1
+
+    # # BFS 
+    # def coinChange(self, coins: List[int], amount: int) -> int:
+    #     queue = deque([(0,0)]) # (동전갯수, 누적금액)
+    #     visited = set()
+    #     while queue:
+    #         count, total = queue.popleft()
+    #         if total == amount:
+    #             return count
+    #         if total in visited:
+    #             continue
+    #         visited.add(total)
+    #         for coin in coins:
+    #             if total + coin <= amount:
+    #                 queue.append([count+1, total+ coin])
+    #     return -1
+
+
+    # DP
+    # dp[i] = min(dp[i], dp[i-coin]+1)
+    # from collections import deque
+    # class Solution:
+    #     def coinChange(self, coins: List[int], amount: int) -> int:
+    #        dp=[0]+[amount+1]*amount
+    #        for coin in coins:
+    #             for i in range(coin, amount+1):
+    #                 dp[i] = min(dp[i], dp[i-coin]+1)
+    #         return dp[amount] if dp[amount] < amount else -1
+
+

find-minimum-in-rotated-sorted-array/ppxyn1.py

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+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        # You must write an algorithm that runs in O(log n) time. >> Binary Search 
+        # It's possible because the array is sorted.        
+        left, right = 0, len(nums)-1
+
+        while left < right:
+            mid = (left + right) // 2
+            # If the value at mid is greater than the value at the right end, it means the minimum place to the right of mid.
+            if nums[mid] > nums[right]:
+                left = mid + 1
+            else:
+                # mid can be minimum number
+                # But is it possible to solve it with "right=mid-1"? e.g,[4,5,6,7,0,1,2]
+                right = mid
+        return nums[left]
+
+
+

maximum-depth-of-binary-tree/ppxyn1.py

@@ -0,0 +1,36 @@
+# idea : DFS
+
+# time : O(n)
+# space : O(n)
+
+class Solution:
+    def maxDepth(self, root: Optional[TreeNode]) -> int:
+        # edge case 
+        if not root:
+            return 0 
+        max_depth = 0
+        stack = [(root, 1)]
+        while stack:
+            node, depth = stack.pop()
+            max_depth = max(depth, max_depth)
+            if node.left:
+                stack.append((node.left, depth+1))
+            if node.right:
+                stack.append((node.right, depth+1))
+        return max_depth
+
+
+
+
+# another way : Down-top : recursive
+# class Solution:
+#     def maxDepth(self, root: Optional[TreeNode]) -> int:
+#         if not root:
+#             return 0
+#         return 1 + max(
+#         self.maxDepth(root.left),
+#         self.maxDepth(root.right)
+#         )
+
+
+

merge-two-sorted-lists/ppxyn1.py

@@ -0,0 +1,31 @@
+# idea : Two Pointer
+# It was first time solving a two-pointer problem using nodes instead of arrays, and I realized I'm still not very familiar with nodes yet.
+class Solution:
+    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
+        dummy = ListNode(None)
+        node = dummy
+        while list1 and list2:
+            if list1.val < list2.val:
+                node.next = list1
+                list1 = list1.next
+            else:
+                node.next = list2
+                list2 = list2.next
+            node = node.next
+        node.next = list1 or list2
+        return dummy.next
+
+
+# Another way to solve it
+# class Solution:
+#     def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
+#         if not (list1 and list2):
+#             return list1 or list2
+#         if list1.val < list2.val:
+#             list1.next = self.mergeTwoLists(list1.next, list2)
+#             return list1
+#         else:
+#             list2.next = self.mergeTwoLists(list1, list2.next)
+#             return list2
+
+

word-search/ppxyn1.py

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+# idea : dfs
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        ROWS, COLS = len(board), len(board[0])
+        visited = [[False]*COLS for _ in range(ROWS)]
+
+        def dfs(r, c, index):
+            if r < 0 or r >= ROWS or c < 0 or c >= COLS:
+                return False
+            if visited[r][c] or board[r][c] != word[index]:
+                return False
+
+            if index == len(word) - 1:
+                return True
+
+            visited[r][c] = True
+
+            found = (dfs(r+1, c, index+1) or
+                     dfs(r-1, c, index+1) or
+                     dfs(r, c+1, index+1) or
+                     dfs(r, c-1, index+1))
+
+            # For next backtracking
+            visited[r][c] = False
+
+            return found
+
+        # Run DFS from every cell as a starting point
+        for row in range(ROWS):
+            for col in range(COLS):
+                if dfs(row, col, 0):
+                    return True
+
+        return False
+
+
+