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32 changes: 32 additions & 0 deletions 3sum/sangbeenmoon.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,32 @@
# time : O(n^2)
# space : O(n)
class Solution:
def threeSum(self, nums: list[int]) -> list[list[int]]:
nums.sort()
answer = []

for i, num in enumerate(nums):
# 중복 제거
if i > 0 and nums[i] == nums[i-1]:
continue

target = num * -1

visited = dict() # 중복 제거
left = i + 1
right = len(nums) - 1

while left < right:
sum_ = nums[left] + nums[right]
if sum_ < target:
left = left + 1
elif sum_ > target:
right = right - 1
else:
if nums[left] not in visited: # 중복 제거
answer.append([num, nums[left], nums[right]])
visited[nums[left]] = True

left = left + 1
right = right - 1
return answer
15 changes: 15 additions & 0 deletions climbing-stairs/sangbeenmoon.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,15 @@
# dp[i] = dp[i-2] + dp[i-1]
# time : O(n)
# complexity : O(n)

class Solution:
def climbStairs(self, n: int) -> int:
dp = [0] * n
dp[0] = 1
if n == 1:
return dp[0]
dp[1] = 2
for i in range(2, n):
dp[i] = dp[i-2] + dp[i-1]

return dp[n-1]
26 changes: 26 additions & 0 deletions product-of-array-except-self/sangbeenmoon.py
Original file line number Diff line number Diff line change
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# idea
# 1. left[i] = left[0] * left[1] * ... * left[i-1]
# 2. right[i] = right[i+1] * right[i+2] * ... * right[n-1]
# 3. answer[i] = left[i-1] * right[i+1]

# time : O(n)
# space : O(n)

class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
left = [1] * len(nums)
left[0] = nums[0]
for i in range(1, len(nums)):
left[i] = left[i-1] * nums[i]

right = [1] * len(nums)
right[len(nums) - 1] = nums[len(nums) - 1]
for i in range(len(nums) - 2, 0, -1):
right[i] = right[i+1] * nums[i]

answer = []
answer.append(right[1])
for i in range(1, len(nums) - 1):
answer.append(left[i-1] * right[i+1])
answer.append(left[len(nums) - 2])
return answer
6 changes: 6 additions & 0 deletions valid-anagram/sangbeenmoon.py
Original file line number Diff line number Diff line change
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# time : O(nlogn)
# space : O(n)

class Solution:
def isAnagram(self, s: str, t: str) -> bool:
return sorted(s) == sorted(t)
40 changes: 40 additions & 0 deletions validate-binary-search-tree/sangbeenmoon.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,40 @@
# time: O(n)
# space: O(h) : 추가 자료구조는 없지만 재귀 스택으로 인해 트리 높이만큼 공간 사용.

# 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:
answer = True
def isValidBST(self, root: Optional[TreeNode]) -> bool:

if root:
if root.left:
self.go_left(root.left, -2**31 - 1, root.val )
if root.right:
self.go_right(root.right, root.val, 2**31)

return self.answer

def go_left(self, cur:TreeNode, mm:int, MM:int):
if mm >= cur.val or cur.val >= MM :
self.answer = False
return
if cur.left:
self.go_left(cur.left, mm, cur.val)
if cur.right:
self.go_right(cur.right, cur.val, MM)

def go_right(self, cur:TreeNode, mm:int, MM:int):
if mm >= cur.val or cur.val >= MM :
self.answer = False
return
if cur.left:
self.go_left(cur.left, mm, cur.val)
if cur.right:
self.go_right(cur.right, cur.val, MM)

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