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[Lyla] Week 14 #1095
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[Lyla] Week 14 #1095
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''' | ||
시간 복잡도: O(n) | ||
공간 복잡도: O(n) | ||
''' | ||
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from collections import deque | ||
from typing import List, Optional | ||
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class TreeNode: | ||
def __init__(self, val=0, left=None, right=None): | ||
self.val = val | ||
self.left = left | ||
self.right = right | ||
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class Solution: | ||
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: | ||
queue = deque([root]) if root else [] | ||
result = [] | ||
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while queue: | ||
same_level_nodes = [] | ||
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for _ in range(len(queue)): | ||
node = queue.popleft() | ||
same_level_nodes.append(node.val) | ||
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if node.left: | ||
queue.append(node.left) | ||
if node.right: | ||
queue.append(node.right) | ||
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result.append(same_level_nodes) | ||
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return result |
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''' | ||
시간 복잡도: O(n) | ||
공간 복잡도: O(n) | ||
''' | ||
from typing import List | ||
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class Solution: | ||
def countBits(self, n: int) -> List[int]: | ||
dp = [0] * (n + 1) | ||
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for i in range(1, n + 1): | ||
dp[i] = dp[i >> 1] + (i & 1) | ||
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return dp |
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''' | ||
시간 복잡도: O(n) | ||
공간 복잡도: O(n) | ||
''' | ||
from typing import List | ||
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class Solution: | ||
def rob(self, nums: List[int]) -> int: | ||
n = len(nums) | ||
if n == 1: | ||
return nums[0] | ||
if n == 2: | ||
return max(nums[0], nums[1]) | ||
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dp_first = [0] * n | ||
dp_second = [0] * n | ||
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dp_first[0], dp_first[1] = nums[0], max(nums[0], nums[1]) | ||
dp_second[1], dp_second[2] = nums[1], max(nums[1], nums[2]) | ||
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for i in range(2, n): | ||
dp_first[i] = max(dp_first[i - 1], dp_first[i - 2] + nums[i]) | ||
dp_second[i] = max(dp_second[i - 1], dp_second[i - 2] + nums[i]) | ||
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return max(dp_first[-2], dp_second[-1]) |
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''' | ||
시간 복잡도: O(n log n) | ||
공간 복잡도: O(n) | ||
''' | ||
from typing import List | ||
from heapq import heappush, heappop | ||
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class Interval(object): | ||
def __init__(self, start, end): | ||
self.start = start | ||
self.end = end | ||
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class Solution: | ||
def min_meeting_rooms(self, intervals: List[Interval]) -> int: | ||
intervals.sort(key=lambda x:x.start) | ||
ends = [] | ||
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for interval in intervals: | ||
if ends and ends[0] <= interval.start: | ||
heappop(ends) | ||
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heappush(ends, interval.end) | ||
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return len(ends) | ||
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def run_tests(): | ||
solution = Solution() | ||
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test_cases = [ | ||
# Test Case 1: 문제의 Example 1 | ||
{ | ||
"input": [Interval(0, 30), Interval(5, 10), Interval(15, 20)], | ||
"expected": 2, | ||
"description": "Example 1: [(0,30), (5,10), (15,20)] - 2 rooms needed" | ||
}, | ||
# Test Case 2: 문제의 Example 2 | ||
{ | ||
"input": [Interval(2, 7)], | ||
"expected": 1, | ||
"description": "Example 2: [(2,7)] - 1 room needed" | ||
}, | ||
# Test Case 3: 겹치지 않는 회의들 | ||
{ | ||
"input": [Interval(0, 8), Interval(8, 10), Interval(10, 12)], | ||
"expected": 1, | ||
"description": "Non-overlapping meetings: [(0,8), (8,10), (10,12)]" | ||
}, | ||
# Test Case 4: 모든 회의가 겹치는 경우 | ||
{ | ||
"input": [Interval(1, 5), Interval(2, 6), Interval(3, 7)], | ||
"expected": 3, | ||
"description": "All overlapping: [(1,5), (2,6), (3,7)]" | ||
}, | ||
# Test Case 5: 빈 입력 | ||
{ | ||
"input": [], | ||
"expected": 0, | ||
"description": "Empty input: []" | ||
}, | ||
# Test Case 6: 복잡한 경우 | ||
{ | ||
"input": [Interval(1, 10), Interval(2, 7), Interval(3, 19), Interval(8, 12)], | ||
"expected": 3, | ||
"description": "Complex case: [(1,10), (2,7), (3,19), (8,12)]" | ||
}, | ||
{ | ||
"input": [Interval(1, 4), Interval(2, 5), Interval(3, 6)], | ||
"expected": 3, | ||
"description": "Multiple overlaps: [(1,4), (2,5), (3,6)]" | ||
}, | ||
{ | ||
"input": [Interval(1, 10), Interval(2, 3), Interval(4, 5), Interval(6, 7)], | ||
"expected": 2, | ||
"description": "Short and long meetings: [(1,10), (2,3), (4,5), (6,7)]" | ||
}, | ||
{ | ||
"input": [Interval(1, 5), Interval(5, 10), Interval(10, 15), Interval(2, 7)], | ||
"expected": 2, | ||
"description": "Mixed overlaps: [(1,5), (5,10), (10,15), (2,7)]" | ||
} | ||
] | ||
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# 테스트 실행 | ||
for i, test in enumerate(test_cases, 1): | ||
intervals = test["input"] | ||
expected = test["expected"] | ||
result = solution.min_meeting_rooms(intervals) | ||
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print(f"Test Case {i}: {test['description']}") | ||
print(f"Input: {[(interval.start, interval.end) for interval in intervals]}") | ||
print(f"Expected Output: {expected}") | ||
print(f"Your Output: {result}") | ||
print(f"Result: {'✅ PASS' if result == expected else '❌ FAIL'}") | ||
print("-" * 50) | ||
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if __name__ == "__main__": | ||
run_tests() |
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''' | ||
시간 복잡도: O(m * n * 4^s) s = word 최대 길이 | ||
공간 복잡도: O(w) w = 모든 단어에 포함된 문자 수의 합 | ||
''' | ||
from typing import List | ||
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class Solution: | ||
def findWords(self, board: List[List[str]], words: List[str]) -> List[str]: | ||
root = {} | ||
for word in words: | ||
node = root | ||
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for char in word: | ||
if char not in node: | ||
node[char] = {} | ||
node = node[char] | ||
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node['$'] = word | ||
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def dfs(i, j, node, visited): | ||
if '$' in node: | ||
result.append(node['$']) | ||
del node['$'] | ||
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if not (0 <= i < m and 0 <= j < n) or (i, j) in visited: | ||
return | ||
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char = board[i][j] | ||
if char not in node: | ||
return | ||
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visited.add((i, j)) | ||
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for di, dj in [(0, 1), (1, 0), (0, -1), (-1, 0)]: | ||
dfs(i + di, j + dj, node[char], visited) | ||
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visited.remove((i, j)) | ||
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m, n = len(board), len(board[0]) | ||
result = [] | ||
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for i in range(m): | ||
for j in range(n): | ||
dfs(i, j, root, set()) | ||
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return result |
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Queue 를 활용한 BFS 방식으로 문제를 풀어주셨네요!
DFS를 사용하여 풀어보시는것도 재미있을것 같습니다 :)