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Merge pull request #716 from KwonNayeon/main
[KwonNayeon] Week 2
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""" | ||
Constraints: | ||
1. 3 <= nums.length <= 3000 | ||
2. -10^5 <= nums[i] <= 10^5 | ||
Time Complexity: | ||
- O(n^2) (정렬은 O(n log n), 이중 반복문은 O(n^2)) | ||
Space Complexity: | ||
- O(n) (결과 리스트) | ||
""" | ||
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class Solution: | ||
def threeSum(self, nums: List[int]) -> List[List[int]]: | ||
nums.sort() | ||
result = [] | ||
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for i in range(len(nums) - 2): | ||
if i > 0 and nums[i] == nums[i-1]: | ||
continue | ||
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left, right = i+1, len(nums)-1 | ||
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while left < right: | ||
sum = nums[i] + nums[left] + nums[right] | ||
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if sum == 0: | ||
result.append([nums[i], nums[left], nums[right]]) | ||
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while left < right and nums[left] == nums[left+1]: | ||
left += 1 | ||
while left < right and nums[right] == nums[right-1]: | ||
right -= 1 | ||
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left += 1 | ||
right -= 1 | ||
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elif sum < 0: | ||
left += 1 | ||
else: | ||
right -= 1 | ||
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return result |
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""" | ||
Constraints: | ||
- 1 <= n <= 45 | ||
Time Complexity: | ||
- O(n) | ||
Space Complexity: | ||
- O(n) | ||
""" | ||
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class Solution: | ||
def climbStairs(self, n: int) -> int: | ||
if n == 1: | ||
return 1 | ||
if n == 2: | ||
return 2 | ||
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dp = [0] * (n+1) | ||
dp[1], dp[2] = 1, 2 | ||
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for i in range(3, n+1): | ||
dp[i] = dp[i-1] + dp[i-2] | ||
return dp[n] |
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construct-binary-tree-from-preorder-and-inorder-traversal/KwonNayeon.py
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""" | ||
Constraints: | ||
1. 1 <= preorder.length <= 3000 | ||
2. inorder.length == preorder.length | ||
3. -3000 <= preorder[i], inorder[i] <= 3000 | ||
4. preorder and inorder consist of unique values | ||
5. Each value of inorder also appears in preorder | ||
6. preorder is guaranteed to be the preorder traversal of the tree | ||
7. inorder is guaranteed to be the inorder traversal of the tree | ||
Time Complexity: | ||
- O(N^2). 각 노드(N)마다 inorder에서 index를 찾는 연산(N)이 필요하고, 각 노드를 한 번씩 방문하여 트리를 구성하기 때문. | ||
Space Complexity: | ||
- O(N). 재귀 호출 스택을 위한 공간이 필요하며, 최악의 경우(한쪽으로 치우친 트리) 재귀 깊이가 N까지 갈 수 있기 때문. | ||
""" | ||
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class Solution: | ||
def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]: | ||
if not preorder or not inorder: | ||
return None | ||
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root = TreeNode(preorder[0]) | ||
mid = inorder.index(preorder[0]) | ||
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root.left = self.buildTree(preorder[1:mid+1], inorder[:mid]) | ||
root.right = self.buildTree(preorder[mid+1:], inorder[mid+1:]) | ||
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return root |
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""" | ||
Constraints: | ||
- 1 <= len(s), len(t) <= 50_000 | ||
- s and t consist of lowercase English letters (a-z) only | ||
Time Complexity: | ||
- O(n log n) | ||
Space Complexity: | ||
- O(n) | ||
""" | ||
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class Solution: | ||
def isAnagram(self, s: str, t: str) -> bool: | ||
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s = s.replace(' ', '').lower() | ||
t = t.replace(' ', '').lower() | ||
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if sorted(s) == sorted(t): | ||
return True | ||
else: | ||
return False |