Merge pull request #586 from lymchgmk/week13

[EGON] Week13 Solutions

find-median-from-data-stream/EGON.py

@@ -0,0 +1,67 @@
+from heapq import heappush, heappop
+from unittest import TestCase, main
+
+
+class MedianFinder:
+    """
+    Runtime: 137 ms (Beats 56.16%)
+    Time Complexity:
+        1) addNum
+            - 최악의 경우 heappush + (heappop + heappush) + (heappop + heapppush) 가 발생할 수 있으므로 O(5 * log n)
+            > O(5 * log n) ~= O(log n)
+        2) findMedian
+            > heap의 루트 값을 가지고 사칙연산만 하므로 O(1)
+
+    Memory: 39.94 MB (Beats 5.85%)
+    Space Complexity: O(n)
+        - max_heap은 최대 n // 2 개 혹은 n // 2 + 1개의 원소를 가지므로 O(n / 2 + 1), upper bound
+        - min_heap은 최대 n // 2개의 원소를 가지므로 O(n / 2)
+        > O(n / 2 + 1) + O(n / 2) ~= O(n) + O(n) ~= O(n)
+    """
+
+    def __init__(self):
+        self.max_heap = []
+        self.min_heap = []
+
+    def addNum(self, num):
+        heappush(self.max_heap, -num)
+        if self.min_heap and (-self.max_heap[0] > self.min_heap[0]):
+            heappush(self.min_heap, -heappop(self.max_heap))
+
+        if len(self.max_heap) > len(self.min_heap) + 1:
+            heappush(self.min_heap, -heappop(self.max_heap))
+        elif len(self.max_heap) < len(self.min_heap):
+            heappush(self.max_heap, -heappop(self.min_heap))
+
+    def findMedian(self):
+        if self.max_heap and self.min_heap:
+            if len(self.max_heap) == len(self.min_heap):
+                return ((-self.max_heap[0]) + self.min_heap[0]) / 2
+            else:
+                return -self.max_heap[0]
+        elif self.min_heap and not self.max_heap:
+            return self.min_heap[0]
+        elif not self.min_heap and self.max_heap:
+            return -self.max_heap[0]
+        else:
+            return 0.0
+
+
+class _LeetCodeTestCases(TestCase):
+
+    def test_1(self):
+        medianFinder = MedianFinder()
+        medianFinder.addNum(-1)
+        self.assertEqual(medianFinder.findMedian(), -1.0)
+        medianFinder.addNum(-2)
+        self.assertEqual(medianFinder.findMedian(), -1.5)
+        medianFinder.addNum(-3)
+        self.assertEqual(medianFinder.findMedian(), -2.0)
+        medianFinder.addNum(-4)
+        self.assertEqual(medianFinder.findMedian(), -2.5)
+        medianFinder.addNum(-5)
+        self.assertEqual(medianFinder.findMedian(), -3.0)
+
+
+if __name__ == '__main__':
+    main()

house-robber/EGON.py

@@ -0,0 +1,42 @@
+from typing import List
+from unittest import TestCase, main
+
+
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        return self.solve_dp(nums)
+
+    """
+    Runtime: 0 ms (Beats 100.00%)
+    Time Complexity: O(n)
+        - nums 배열을 조회하며 dp 배열을 갱신하므로 O(n)
+            - 2항에 대한 max 연산을 사용하므로 * O(2)
+        > O(2 * n) ~= O(n)
+
+    Memory: 16.62 MB (Beats 24.05%)
+    Space Complexity: O(n)
+        > 길이가 n인 dp 배열을 사용하므로 O(n)
+    """
+
+    def solve_dp(self, nums: List[int]) -> int:
+        if len(nums) <= 2:
+            return max(nums)
+
+        dp = [0] * len(nums)
+        dp[0] = nums[0]
+        dp[1] = max(nums[0], nums[1])
+        for i in range(2, len(nums)):
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+
+        return dp[-1]
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        nums = [2,1,1,2]
+        output = 4
+        self.assertEqual(Solution().rob(nums), output)
+
+
+if __name__ == '__main__':
+    main()

lowest-common-ancestor-of-a-binary-search-tree/EGON.py

@@ -0,0 +1,106 @@
+from collections import deque
+from unittest import TestCase, main
+
+
+# Definition of TreeNode:
+class TreeNode:
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+
+class Solution:
+    def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:
+        return self.solve_bfs(root, p, q)
+
+    """
+    Runtime: 50 ms (Beats 81.68%)
+    Time Complexity:`
+        - bfs를 이용해 ancetor와 nodes를 초기화 하는데 트리의 모든 p와 q를 포함할 때까지 모든 node를 조회하는데, O(n)
+        - p로 부터 부모를 따라 올라가는데, 트리의 모든 node가 편향적으로 연결된 경우 최대 O(n), upper bound
+            - q로 부터 부모를 따라 올라가는데, 단, 위 p의 추적 path와 겹치지 않는 곳만 올라가므로, 총합이 O(n)
+        > O(n) + (O(P) + O(Q)) = O(n) + O(n) ~= O(n) 
+
+    Memory: 21.20 MB (Beats 14.40%)
+    Space Complexity: O(n)
+        - p, q가 트리의 리프노드인 경우 dq의 크기는 O(n), upper bound
+        - p_ancestors와 q_anscestor의 크기는 합쳐서 O(n)
+        > O(n) + O(n) ~= O(n)
+    """
+    def solve_bfs(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:
+        dq = deque([root])
+        ancestor = {root.val: root.val}
+        nodes = {root.val: root}
+        while dq:
+            if p.val in ancestor and q.val in ancestor:
+                break
+
+            curr_node = dq.popleft()
+            if curr_node.left:
+                ancestor[curr_node.left.val] = curr_node.val
+                dq.append(curr_node.left)
+                nodes[curr_node.left.val] = curr_node.left
+            if curr_node.right:
+                ancestor[curr_node.right.val] = curr_node.val
+                dq.append(curr_node.right)
+                nodes[curr_node.right.val] = curr_node.right
+
+        p_val = p.val
+        p_ancestors = set()
+        p_ancestors.add(root.val)
+        while p_val in ancestor and p_val != root.val:
+            p_ancestors.add(p_val)
+            p_ancestors.add(ancestor[p_val])
+            p_val = ancestor[p_val]
+
+        q_val = q.val
+        q_ancestors = set()
+        q_ancestors.add(q_val)
+        while q_val in ancestor and q_val not in p_ancestors:
+            q_ancestors.add(q_val)
+            q_ancestors.add(ancestor[q_val])
+            q_val = ancestor[q_val]
+
+        common_ancestor = p_ancestors & q_ancestors
+        if common_ancestor:
+            return nodes[common_ancestor.pop()]
+        else:
+            return root
+
+
+class _LeetCodeTestCases(TestCase):
+
+    def test_1(self):
+        root = TreeNode(5)
+        node1 = TreeNode(3)
+        node2 = TreeNode(6)
+        node3 = TreeNode(2)
+        node4 = TreeNode(4)
+        node5 = TreeNode(1)
+
+        root.left = node1
+        root.right = node2
+        node1.left = node3
+        node1.right = node4
+        node3.left = node5
+
+        p = node5
+        q = node1
+        output = root
+        self.assertEqual(Solution().lowestCommonAncestor(root, p, q), output)
+
+    def test_2(self):
+        root = TreeNode(2)
+        node1 = TreeNode(1)
+
+        root.left = node1
+
+        p = TreeNode(2)
+        q = TreeNode(1)
+        output = root
+        self.assertEqual(Solution().lowestCommonAncestor(root, p, q), output)
+
+
+if __name__ == '__main__':
+    main()

meeting-rooms/EGON.py

@@ -0,0 +1,66 @@
+from typing import List
+from unittest import TestCase, main
+
+
+# Definition of Interval:
+class Interval(object):
+    def __init__(self, start, end):
+        self.start = start
+        self.end = end
+
+
+class Solution:
+    def canAttendMeetings(self, intervals: List[Interval]) -> bool:
+        return self.solve_sort_stack(intervals)
+
+    """
+    LintCode 로그인이 안되어서 hhttps://neetcode.io/problems/meeting-schedule 에서 실행시키고 통과만 확인했습니다.
+
+    Runtime: ? ms (Beats ?%)
+    Time Complexity: O(n log n)
+        - intervals 정렬에 O(n log n)
+        - intervals 조회하며 stack의 마지막 값의 end와 현재 interval의 start를 비교하는데 O(n)
+        > O(n log n) + O(n) ~= O(n log n)
+
+    Memory: ? MB (Beats ?%)
+    Space Complexity: O(n)
+        - intervals 메모리를 그대로 사용하면서 정렬했으므로 O(1)
+        - stack의 크기는 최대 intervals와 같아질 수 있으므로 O(n)
+    """
+
+    def solve_sort_stack(self, intervals: List[Interval]) -> bool:
+        intervals.sort(key=lambda x: x.start)
+
+        stack: List[Interval] = []
+        for interval in intervals:
+            if not stack:
+                stack.append(interval)
+                continue
+
+            if stack[-1].end > interval.start:
+                return False
+            else:
+                stack.append(interval)
+
+        return True
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        intervals = [Interval(0,30), Interval(5,10), Interval(15,20)]
+        output = False
+        self.assertEqual(Solution().canAttendMeetings(intervals), output)
+
+    def test_2(self):
+        intervals = [Interval(5, 8), Interval(9,15)]
+        output = False
+        self.assertEqual(Solution().canAttendMeetings(intervals), output)
+
+    def test_3(self):
+        intervals = [Interval(0, 1), Interval(1, 2)]
+        output = False
+        self.assertEqual(Solution().canAttendMeetings(intervals), output)
+
+
+if __name__ == '__main__':
+    main()

non-overlapping-intervals/EGON.py

@@ -0,0 +1,45 @@
+from typing import List
+from unittest import TestCase, main
+
+
+class Solution:
+    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
+        return self.solve_stack(intervals)
+
+    """
+    LintCode 로그인이 안되어서 hhttps://neetcode.io/problems/meeting-schedule 에서 실행시키고 통과만 확인했습니다.
+
+    Runtime: 66 ms (Beats 85.10%)
+    Time Complexity: O(n log n)
+        - intervals를 정렬하는데 O(n log n)
+        - intervals를 조회하며 result stack을 갱신하는데 O(n)
+            - 2항에 대한 or 연산 및 append 메서드만 쓰므로 * O(1)
+        > O(n log n) + O(n) ~= O(n log n)
+    Memory: 50.98 MB (Beats 74.30%)
+    Space Complexity: O(n)
+        - intervals 정렬도 기존 intervals를 사용하므로 O(1)
+        - result의 크기가 최대 interval과 같아질 수 있으므로 O(n)
+        > O(1) + O(n) ~= O(n)
+    """
+    def solve_stack(self, intervals: List[List[int]]) -> int:
+        if not intervals:
+            return 0
+
+        intervals.sort(key=lambda x: x[1])
+        result = []
+        for interval in intervals:
+            if not result or interval[0] >= result[-1][1]:
+                result.append(interval)
+
+        return len(intervals) - len(result)
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        intervals = [[1,2],[2,3],[3,4],[1,3]]
+        output = 1
+        self.assertEqual(Solution().eraseOverlapIntervals(intervals), output)
+
+
+if __name__ == '__main__':
+    main()