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[krokerdile] WEEK 02 solutions #1225

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Apr 14, 2025
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58 changes: 58 additions & 0 deletions 3sum/krokerdile.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,58 @@
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
res = []
n = len(nums)

for i in range(n):
if i > 0 and nums[i] == nums[i-1]:
continue

target = -nums[i]
seen = set()
j = i + 1

while j < n:
complement = target - nums[j]
if complement in seen:
res.append([nums[i], complement, nums[j]])
while j + 1 < n and nums[j] == nums[j+1]:
j += 1
seen.add(nums[j])
j += 1
return list(set(tuple(x) for x in res))


class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
result = []

for i in range(len(nums)):
# 중복된 첫 번째 수는 skip
if i > 0 and nums[i] == nums[i - 1]:
continue

left, right = i + 1, len(nums) - 1

while left < right:
total = nums[i] + nums[left] + nums[right]

if total == 0:
result.append([nums[i], nums[left], nums[right]])

# 중복된 두 번째, 세 번째 수 건너뛰기
while left < right and nums[left] == nums[left + 1]:
left += 1
while left < right and nums[right] == nums[right - 1]:
right -= 1

left += 1
right -= 1

elif total < 0:
left += 1 # 합이 작으면 왼쪽을 오른쪽으로
else:
right -= 1 # 합이 크면 오른쪽을 왼쪽으로

return result
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포인터를 사용해서 하셨군요!
여러가지 방법을 배웁니다
저도 이 방법으로 다시 풀어봐야 겠어요
혹시 2가지 방법 중 어떤 것이 속도가 더 빨랐나요?

10 changes: 10 additions & 0 deletions climbing-stairs/krokerdile.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,10 @@
class Solution:
def climbStairs(self, n: int) -> int:
one, two = 1,1

for i in range(n-1):
temp = one;
one = one + two;
two = temp;

return one;
18 changes: 18 additions & 0 deletions product-of-array-except-self/krokerdile.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,18 @@
class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
n = len(nums)
answer = [1] * n

# 1. 왼쪽 곱 저장
left_product = 1
for i in range(n):
answer[i] = left_product
left_product *= nums[i]

# 2. 오른쪽 곱을 곱해주기
right_product = 1
for i in reversed(range(n)):
answer[i] *= right_product
right_product *= nums[i]

return answer
13 changes: 13 additions & 0 deletions valid-anagram/krokerdile.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,13 @@
class Solution:
def isAnagram(self, s: str, t: str) -> bool:
a = {}
b = {}

if len(s) != len(t):
return False

for i in range(len(s)):
a[s[i]] = a.get(s[i], 0) + 1
b[t[i]] = b.get(t[i], 0) + 1
Comment on lines +9 to +11
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제가 파이썬을 하지 않아서 이 부분이 잘 이해가 가지 않는데 어떤 순서로 리스트에 넣는 건가요?


return a == b
15 changes: 15 additions & 0 deletions validate-binary-search-tree/krokerdile.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,15 @@
class Solution:
def isValidBST(self, root: Optional[TreeNode]) -> bool:
def validate(node, low, high):
if not node:
return True

# 현재 노드의 값이 범위를 벗어나면 False
if not (low < node.val < high):
return False

# 왼쪽 서브트리는 최대값을 현재 노드 값으로 제한
# 오른쪽 서브트리는 최소값을 현재 노드 값으로 제한
return validate(node.left, low, node.val) and validate(node.right, node.val, high)

return validate(root, float('-inf'), float('inf'))
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저는 자바로 해서 long을 사용했는데 파이썬에서는 float로 하군요!
혹시 파이썬에도 int 보다는 크고 float 보다는 작은 범위가 있나요?
그것이 있다면 그것으로 해도 될 듯 하네요