Merge pull request #554 from lymchgmk/feat/week11

[EGON] Week11 Solutions

Author: EgonD3V

binary-tree-maximum-path-sum/EGON.py

@@ -0,0 +1,67 @@
+from typing import Optional
+from unittest import TestCase, main
+
+
+# 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 maxPathSum(self, root: Optional[TreeNode]) -> int:
+        return self.solve_dfs(root)
+
+    """
+    Runtime: 11 ms (Beats 98.62%)
+    Time Complexity: O(n)
+    > dfs를 통해 모든 node를 방문하므로 O(n)
+
+    Memory: 22.10 MB (Beats 10.70%)
+    Space Complexity: O(n)
+        - dfs 재귀 호출 스택의 깊이는 이진트리가 최악으로 편향된 경우 O(n), upper bound
+        - 나머지 변수는 O(1)
+        > O(n), upper bound
+    """
+    def solve_dfs(self, root: Optional[TreeNode]) -> int:
+        max_path_sum = float('-inf')
+
+        def dfs(node: Optional[TreeNode]) -> int:
+            nonlocal max_path_sum
+
+            if not node:
+                return 0
+
+            max_left = max(dfs(node.left), 0)
+            max_right = max(dfs(node.right), 0)
+            max_path_sum = max(max_path_sum, node.val + max_left + max_right)
+
+            return node.val + max(max_left, max_right)
+
+        dfs(root)
+
+        return max_path_sum
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        root = TreeNode(-10)
+        node_1 = TreeNode(9)
+        node_2 = TreeNode(20)
+        node_3 = TreeNode(15)
+        node_4 = TreeNode(7)
+        node_2.left = node_3
+        node_2.right = node_4
+        root.left = node_1
+        root.right = node_2
+
+        # root = [-10, 9, 20, None, None, 15, 7]
+        output = 42
+
+        self.assertEqual(Solution().maxPathSum(root), output)
+
+
+if __name__ == '__main__':
+    main()

graph-valid-tree/EGON.py

@@ -0,0 +1,83 @@
+from typing import List
+from unittest import TestCase, main
+
+
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+class Solution:
+    def valid_tree(self, n: int, edges: List[List[int]]) -> bool:
+        return self.solve_union_find(n, edges)
+
+    """
+    LintCode 로그인이 안되어서 https://neetcode.io/problems/valid-tree에서 실행시키고 통과만 확인했습니다.
+    
+    Runtime: ? ms (Beats ?%)
+    Time Complexity: O(max(m, n))
+        - UnionFind의 parent 생성에 O(n)
+        - edges 조회에 O(m)
+            - Union-find 알고리즘의 union을 매 조회마다 사용하므로, * O(α(n)) (α는 아커만 함수의 역함수)
+        - UnionFind의 모든 노드가 같은 부모, 즉 모두 연결되어 있는지 확인하기 위해, n번 find에 O(n * α(n))
+        > O(n) + O(m * α(n)) + O(n * α(n)) ~= O(max(m, n) *  α(n)) ~= O(max(m, n)) (∵ α(n) ~= C)
+
+    Memory: ? MB (Beats ?%)
+    Space Complexity: O(n)
+        - UnionFind의 parent와 rank가 크기가 n인 리스트이므로, O(n) + O(n)
+        > O(n) + O(n) ~= O(n)
+    """
+    def solve_union_find(self, n: int, edges: List[List[int]]) -> bool:
+
+        class UnionFind:
+            def __init__(self, size: int):
+                self.parent = [i for i in range(size)]
+                self.rank = [1] * size
+
+            def union(self, first: int, second: int) -> bool:
+                first_parent, second_parent = self.find(first), self.find(second)
+                if first_parent == second_parent:
+                    return False
+
+                if self.rank[first_parent] > self.rank[second_parent]:
+                    self.parent[second_parent] = first_parent
+                elif self.rank[first_parent] < self.rank[second_parent]:
+                    self.parent[first_parent] = second_parent
+                else:
+                    self.parent[second_parent] = first_parent
+                    self.rank[first_parent] += 1
+
+                return True
+
+            def find(self, node: int):
+                if self.parent[node] != node:
+                    self.parent[node] = self.find(self.parent[node])
+
+                return self.parent[node]
+
+        unionFind = UnionFind(size=n)
+        for first, second in edges:
+            is_cycle = unionFind.union(first, second) is False
+            if is_cycle:
+                return False
+
+        root = unionFind.find(0)
+        for i in range(1, n):
+            if unionFind.find(i) != root:
+                return False
+
+        return True
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        n = 5
+        edges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]]
+        output = False
+
+        self.assertEqual(Solution().valid_tree(n, edges), output)
+
+
+if __name__ == '__main__':
+    main()

insert-interval/EGON.py

@@ -0,0 +1,66 @@
+from bisect import bisect_left
+from typing import List
+from unittest import TestCase, main
+
+
+class Solution:
+    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
+        return self.solve(intervals, newInterval)
+
+    """
+    Runtime: 1 ms (Beats 95.76%)
+    Time Complexity: O(n)
+    > intervals의 전체를 선형적으로 조회하므로 O(n), 그 외의 append등의 연산들은 O(1)이므로 무시
+    
+    Memory: 18.70 MB (Beats 99.60%)
+    Space Complexity: O(n)
+    > result의 크기는 intervals와 newInterval이 하나도 겹치지 않는 경우, 최대 n + 1이므로, O(n + 1) ~= O(n)
+    """
+    def solve(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
+        if not intervals:
+            return [newInterval]
+
+        result = []
+        new_s, new_e = newInterval
+        for interval in intervals:
+            s, e = interval
+            if e < new_s:
+                result.append(interval)
+            elif new_e < s:
+                if new_s != -1 and new_e != -1:
+                    result.append([new_s, new_e])
+                    new_s = new_e = -1
+
+                result.append(interval)
+            else:
+                new_s = min(new_s, s)
+                new_e = max(new_e, e)
+
+        if new_s != -1 and new_e != -1:
+            result.append([new_s, new_e])
+
+        return result
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]]
+        newInterval = [4,8]
+        output = [[1,2],[3,10],[12,16]]
+        self.assertEqual(Solution().insert(intervals, newInterval), output)
+
+    def test_2(self):
+        intervals = [[1,2]]
+        newInterval = [3,4]
+        output = [[1,2], [3,4]]
+        self.assertEqual(Solution().insert(intervals, newInterval), output)
+
+    def test_3(self):
+        intervals = [[1,3], [5,6]]
+        newInterval = [4,5]
+        output = [[1,3], [4,6]]
+        self.assertEqual(Solution().insert(intervals, newInterval), output)
+
+
+if __name__ == '__main__':
+    main()

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

@@ -0,0 +1,80 @@
+from typing import Optional
+from unittest import TestCase, main
+
+
+# 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:
+        return self.solve_dfs_iterable(root)
+
+    """
+    Runtime: 0 ms (Beats 100.00%)
+    Time Complexity: O(n)
+    > 트리의 모든 노드의 갯수를 n개라고 하면, 트리의 모든 노드를 stack에 넣어 조회하므로 O(n)
+
+    Memory: 17.75 MB (Beats 21.97%)
+    Space Complexity: O(n)
+    > 최악의 경우 트리의 최대 길이가 n인 경우이므로, stack의 최대 크기가 n에 비례하므로 O(n), upper bound
+    """
+    def solve_dfs_iterable(self, root: Optional[TreeNode]) -> int:
+        max_depth = 0
+        stack = [(root, 0)]
+        while stack:
+            curr_node, curr_depth = stack.pop()
+            if curr_node is None:
+                continue
+
+            if curr_node.left is None and curr_node.right is None:
+                max_depth = max(max_depth, curr_depth + 1)
+                continue
+
+            if curr_node.left:
+                stack.append((curr_node.left, curr_depth + 1))
+            if curr_node.right:
+                stack.append((curr_node.right, curr_depth + 1))
+
+        return max_depth
+
+
+    """
+    Runtime: 0 ms (Beats 100.00%)
+    Time Complexity: O(n)
+    
+    Memory: 17.90 MB (Beats 9.05%)
+    Space Complexity: O(n)
+    """
+    def solve_dfs_recursive(self, root: Optional[TreeNode]) -> int:
+        max_depth = 0
+
+        def dfs(node: Optional[TreeNode], depth: int):
+            nonlocal max_depth
+
+            if not node:
+                return max_depth
+
+            if node.left is None and node.right is None:
+                max_depth = max(max_depth, depth + 1)
+                return
+
+            dfs(node.left, depth + 1)
+            dfs(node.right, depth + 1)
+
+        dfs(root, 0)
+
+        return max_depth
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        self.assertEqual(True, True)
+
+
+if __name__ == '__main__':
+    main()

reorder-list/EGON.py

@@ -0,0 +1,71 @@
+from heapq import heappush, heappop
+from typing import List, Optional
+from unittest import TestCase, main
+
+
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+class Solution:
+    def reorderList(self, head: Optional[ListNode]) -> None:
+        return self.solve(head)
+
+    """
+    Runtime: 15 ms (Beats 88.30%)
+    Time Complexity: O(n)
+        - 역방향 링크드 리스트인 backward를 생성하는데, 원본 링크드 리스트의 모든 노드를 조회하는데 O(n)
+        - reorder하는데 원본 링크드 리스트의 모든 노드의 길이만큼 backward와 forward의 노드들을 조회하는데 O(n)
+        > O(n) + O(n) = 2 * O(n) ~= O(n)
+
+    Memory: 23.20 MB (Beats 88.27%)
+    Space Complexity: O(n)
+    > 역방향 링크드 리스트인 backward를 생성하는데, backward의 길이는 원본 링크드 리스트의 길이와 같으므로 O(n)
+    """
+    def solve(self, head: Optional[ListNode]) -> None:
+        backward = ListNode(head.val)
+        backward_node = head.next
+        length = 1
+        while backward_node:
+            length += 1
+            temp_node = ListNode(backward_node.val)
+            temp_node.next = backward
+            backward = temp_node
+            backward_node = backward_node.next
+
+        node = head
+        forward = head.next
+        for i in range(length):
+            if i == length - 1:
+                node.next = None
+                return
+
+            if i % 2 == 0:
+                node.next = backward
+                backward = backward.next
+                node = node.next
+            else:
+                node.next = forward
+                forward = forward.next
+                node = node.next
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        node_1 = ListNode(1)
+        node_2 = ListNode(2)
+        node_3 = ListNode(3)
+        node_4 = ListNode(4)
+        node_5 = ListNode(5)
+        node_1.next = node_2
+        node_2.next = node_3
+        node_3.next = node_4
+        node_4.next = node_5
+
+        self.assertEqual(Solution().reorderList(node_1), True)
+
+
+if __name__ == '__main__':
+    main()