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[sejineer] Week 02 solutions #1209

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Apr 14, 2025
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24 changes: 24 additions & 0 deletions 3sum/sejineer.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,24 @@
"""
시간 복잡도: O(N^2)
공간 복잡도: O(N)
"""
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()

result = set()

for i in range(len(nums) - 2):
start, end = i + 1, len(nums) - 1
while start < end:
temp = nums[i] + nums[start] + nums[end]
if temp == 0:
result.add((nums[i], nums[start], nums[end]))
start += 1
end -= 1
elif temp < 0:
start += 1
else:
end -= 1

return list(result)
13 changes: 13 additions & 0 deletions climbing-stairs/sejineer.py
Original file line number Diff line number Diff line change
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"""
시간 복잡도: O(N)
공간 복잡도: O(N)
"""
class Solution:
def climbStairs(self, n: int) -> int:
dp = [0] * 46
dp[1], dp[2] = 1, 2

for i in range(3, n + 1):
dp[i] = dp[i - 1] + dp[i - 2]

return dp[n]
19 changes: 19 additions & 0 deletions product-of-array-except-self/sejineer.py
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"""
시간 복잡도: O(N)
공간 복잡도: O(N)
"""
class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
prefix = [1] * len(nums)
for i in range(len(nums) - 1):
prefix[i + 1] = prefix[i] * nums[i]

suffix = [1] * len(nums)
for i in range(len(nums) - 1, 0, -1):
suffix[i - 1] = suffix[i] * nums[i]

result = []
for i, j in zip(prefix, suffix):
result.append(i * j)

return result
26 changes: 26 additions & 0 deletions valid-anagram/sejineer.py
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"""
시간 복잡도: O(N)
공간 복잡도: O(N)

코드 가독성 개선 코드:
from collections import Counter

class Solution:
def isAnagram(self, s: str, t: str) -> bool:
return Counter(s) == Counter(t)
"""
from collections import Counter

class Solution:
def isAnagram(self, s: str, t: str) -> bool:
if len(s) != len(t):
return False

s_counter = Counter(s)
t_counter = Counter(t)

for num, freq in t_counter.items():
if s_counter[num] != freq:
return False

return True
24 changes: 24 additions & 0 deletions validate-binary-search-tree/sejineer.py
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DFS 사용하시면서 공간 복잡도는 스택의 높이 만큼 차지해서 O(h), h 는 스택의 높이 라고 생각하는데 의견 여쭙고 싶습니다!

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DFS 사용하시면서 공간 복잡도는 스택의 높이 만큼 차지해서 O(h), h 는 스택의 높이 라고 생각하는데 의견 여쭙고 싶습니다!

전 num_list를 기준으로 O(N)으로 생각했는데 그렇게도 생각할 수 있을 것 같습니다 ㅎㅎ 좋은 피드백 감사합니다.

Original file line number Diff line number Diff line change
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"""
시간 복잡도: O(N)
공간 복잡도: O(N)
"""
class Solution:
def isValidBST(self, root: Optional[TreeNode]) -> bool:
num_list = []

def dfs(node: TreeNode):
if not node:
return

dfs(node.left)
num_list.append(node.val)
dfs(node.right)

dfs(root)

mem = num_list[0]
for i in range(1, len(num_list)):
if mem >= num_list[i]:
return False
mem = num_list[i]
return True