[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 +1,31 @@
+# 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 +1,10 @@
+class Solution:
+    # Time complexity: O(log n)
+    def hammingWeight(self, n: int) -> int:
+        n = int(n)
+        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