Merge pull request #303 from HaJunYoo/main

[HaJunYoo][WEEK 01] 리트코드 문제 풀이

contains-duplicate/hajunyoo.py

@@ -0,0 +1,7 @@
+class Solution:
+ # Time complexity: O(n)
+ def containsDuplicate(self, nums: List[int]) -> bool:
+ string_len = len(nums)
+ set_len = len(set(nums))
+
+ return string_len != set_len

kth-smallest-element-in-a-bst/hajunyoo.py

@@ -0,0 +1,30 @@
+# 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
+from collections import defaultdict
+
+
+class Solution:
+ # Time complexity: O(n)
+ def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
+ self.count = 0
+ self.result = 0
+
+ def dfs(node):
+ if not node:
+ return
+
+ dfs(node.left)
+
+ self.count += 1
+ if self.count == k:
+ self.result = node.val
+ return
+
+ dfs(node.right)
+
+ dfs(root)
+ return self.result

number-of-1-bits/hajunyoo.py

@@ -0,0 +1,9 @@
+class Solution:
+ # Time complexity: O(log n)
+ def hammingWeight(self, n: int) -> int:
+ cnt = 0
+ while n > 0:
+ if n % 2 == 1:
+ cnt += 1
+ n = n // 2
+ return cnt

palindromic-substrings/hajunyoo.py

@@ -0,0 +1,19 @@
+class Solution:
+ # Time complexity: O(n^2) = O(n) * O(n)
+ def countSubstrings(self, s: str) -> int:
+ self.count = 0
+ n = len(s)
+
+ def two_pointer_expand(left, right):
+ while left >= 0 and right < n and s[left] == s[right]:
+ self.count += 1
+ left -= 1
+ right += 1
+
+ for i in range(0, n):
+ # 1, 3, 5 ...
+ two_pointer_expand(i, i)
+ # 2, 4, 6 ...
+ two_pointer_expand(i, i + 1)
+
+ return self.count

top-k-frequent-elements/hajunyoo.py

@@ -0,0 +1,20 @@
+from collections import defaultdict
+from typing import List
+
+
+class Solution:
+ # Time complexity: O(nlogn) -> O(n) + O(nlogn) + O(k)
+ def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+ counter_dict = defaultdict(int)
+
+ for n in nums:
+ counter_dict[n] += 1
+
+ count_list = []
+ for key, val in counter_dict.items():
+ count_list.append((key, val))
+
+ count_list.sort(key=lambda x: x[1], reverse=True)
+ answer = [a for a, b in count_list[:k]]
+
+ return answer