[KwonNayeon] Week 2

3sum/KwonNayeon.py

@@ -0,0 +1,42 @@
+"""
+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)
+"""
+
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        nums.sort()
+        result = []
+
+        for i in range(len(nums) - 2):
+            if i > 0 and nums[i] == nums[i-1]:
+                continue
+
+            left, right = i+1, len(nums)-1
+
+            while left < right:
+                sum = nums[i] + nums[left] + nums[right]
+
+                if sum == 0:
+                    result.append([nums[i], nums[left], nums[right]])
+
+                    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 sum < 0:
+                    left += 1
+                else:
+                    right -= 1
+
+        return result

climbing-stairs/KwonNayeon.py

@@ -0,0 +1,23 @@
+"""
+Constraints:
+    - 1 <= n <= 45
+
+Time Complexity:
+    - O(n)
+Space Complexity:
+    - O(n)
+"""
+
+class Solution:
+    def climbStairs(self, n: int) -> int:
+        if n == 1:
+            return 1
+        if n == 2:
+            return 2
+
+        dp = [0] * (n+1)
+        dp[1], dp[2] = 1, 2
+
+        for i in range(3, n+1):
+            dp[i] = dp[i-1] + dp[i-2]
+        return dp[n]

construct-binary-tree-from-preorder-and-inorder-traversal/KwonNayeon.py

@@ -0,0 +1,28 @@
+"""
+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.
+"""
+
+class Solution:
+    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
+        if not preorder or not inorder:
+            return None
+
+        root = TreeNode(preorder[0])
+        mid = inorder.index(preorder[0])
+
+        root.left = self.buildTree(preorder[1:mid+1], inorder[:mid])
+        root.right = self.buildTree(preorder[mid+1:], inorder[mid+1:])
+
+        return root

valid-anagram/KwonNayeon.py

@@ -1 +1,21 @@
+"""
+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)
+"""
+
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+
+        s = s.replace(' ', '').lower()
+        t = t.replace(' ', '').lower()
+
+        if sorted(s) == sorted(t):
+            return True
+        else:
+            return False