[Chaedie] Week 10

course-schedule/Chaedie.py

@@ -0,0 +1,39 @@
+"""
+(해설을 보면서 따라 풀었습니다.)
+
+Solution: graph 에서 circular 가 생기면 false, 아니면 true
+
+n: numCourses
+p: number of preRequisites
+Time: O(n + p) 
+Space: O(n + p)
+"""
+
+
+class Solution:
+    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
+        graph = {i: [] for i in range(numCourses)}
+        for crs, pre in prerequisites:
+            graph[crs].append(pre)
+
+        traversing = set()
+        finished = set()
+
+        def dfs(crs):
+            if crs in traversing:
+                return False
+            if crs in finished:
+                return True
+
+            traversing.add(crs)
+            for pre in graph[crs]:
+                if not dfs(pre):
+                    return False
+            traversing.remove(crs)
+            finished.add(crs)
+            return True
+
+        for crs in graph:
+            if not dfs(crs):
+                return False
+        return True

invert-binary-tree/Chaedie.py

@@ -0,0 +1,16 @@
+"""
+Time: O(n)
+Space: O(n)
+"""
+
+
+class Solution:
+    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
+        if not root:
+            return None
+
+        temp = root.right
+        root.right = self.invertTree(root.left)
+        root.left = self.invertTree(temp)
+
+        return root

jump-game/Chaedie.py

@@ -0,0 +1,37 @@
+class Solution:
+    """
+    Solution 1:
+        뒤에서 부터 푸는 방법이 있을거라 생각했습니다.
+        풀이는 성공했지만 성능은 느린 5% 라는 느린 성능을 보입니다.
+
+    Time: O(n * 10^5) - 10^5 는 nums 원소의 최대 값
+    Space: O(n)
+    """
+
+    def canJump(self, nums: List[int]) -> bool:
+        dp = [False for i in range(len(nums))]
+
+        dp[len(nums) - 1] = True
+
+        for i in range(len(nums) - 1 - 1, -1, -1):
+            for j in range(1, nums[i] + 1):
+                if i + j < len(nums) and dp[i + j]:
+                    dp[i] = True
+                    break
+
+        return dp[0]
+
+    """
+    Solution 2:
+        Greedy - 솔루션을 참고했습니다.
+
+    Time: O(n)
+    Space: O(1)
+    """
+
+    def canJump(self, nums: List[int]) -> bool:
+        reach = 0
+        for idx in range(len(nums)):
+            if idx <= reach:
+                reach = max(reach, idx + nums[idx])
+        return reach >= len(nums) - 1

merge-k-sorted-lists/Chaedie.py

@@ -0,0 +1,27 @@
+"""
+Solution:
+    1) 모든 linked list 의 value 를 arr에 담는다.
+    2) arr 를 정렬한다.
+    3) arr 로 linked list 를 만든다.
+
+Time: O(n log(n)) = O(n) arr 만들기 + O(n log(n)) 정렬하기 + O(n) linked list 만들기
+Space: O(n)
+"""
+
+
+class Solution:
+    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
+        arr = []
+        for linked_list in lists:
+            while linked_list:
+                arr.append(linked_list.val)
+                linked_list = linked_list.next
+
+        dummy = ListNode()
+        node = dummy
+
+        arr.sort()
+        for num in arr:
+            node.next = ListNode(num)
+            node = node.next
+        return dummy.next

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

@@ -0,0 +1,77 @@
+"""
+1) binary search 로 pivot 찾아 2번 binary search 하기
+Time: O(log(n))
+Space: O(1)
+"""
+
+
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+
+        def find_pivot():
+            low, high = 0, len(nums) - 1
+            while low <= high:
+                mid = (low + high) // 2
+                if mid > 0 and nums[mid - 1] >= nums[mid]:
+                    return mid
+                if nums[0] <= nums[mid]:
+                    low = mid + 1
+                else:
+                    high = mid - 1
+            return 0
+
+        def binary_search(low, high):
+            while low <= high:
+                mid = (low + high) // 2
+                if nums[mid] == target:
+                    return mid
+                if nums[mid] < target:
+                    low = mid + 1
+                else:
+                    high = mid - 1
+            return -1
+
+        pivot = find_pivot()
+
+        idx = binary_search(0, pivot - 1)
+        return idx if idx > -1 else binary_search(pivot, len(nums) - 1)
+
+
+"""
+2) binary search
+    left sorted case
+    4 5 6 7 0 1 2 3
+        m
+        left target case
+        right target case
+
+    right sorted case
+    6 7 0 1 2 3 4 5
+            m
+        left target case
+        right target case
+Time: O(log(n))
+Space: O(1)
+"""
+
+
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+
+        low, high = 0, len(nums) - 1
+        while low <= high:
+            mid = (low + high) // 2
+            if target == nums[mid]:
+                return mid
+
+            if nums[low] <= nums[mid]:
+                if nums[low] <= target <= nums[mid]:
+                    high = mid - 1
+                else:
+                    low = mid + 1
+            else:
+                if nums[mid] <= target <= nums[high]:
+                    low = mid + 1
+                else:
+                    high = mid - 1
+        return -1