Merge pull request #634 from pmjuu/main

[Lyla] Week 1

Author: lylaminju

contains-duplicate/pmjuu.py

@@ -0,0 +1,3 @@
+class Solution:
+    def containsDuplicate(self, nums: List[int]) -> bool:
+        return len(nums) != len(set(nums))

house-robber/pmjuu.py

@@ -0,0 +1,16 @@
+from typing import List
+
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        if len(nums) == 1:
+            return nums[0]
+        
+        # prev1: 이전 집까지의 최대 이익
+        # prev2: 전전 집까지의 최대 이익
+        prev1, prev2 = 0, 0
+        for num in nums:
+            temp = prev1
+            prev1 = max(prev2 + num, prev1) # 현재 집을 털었을 때와 안 털었을 때 중 더 큰 이익 선택
+            prev2 = temp
+        
+        return prev1

longest-consecutive-sequence/pmjuu.py

@@ -0,0 +1,22 @@
+from typing import List
+
+class Solution:
+    def longestConsecutive(self, nums: List[int]) -> int:
+        # Convert to set for O(1) lookups
+        num_set = set(nums)
+        longest_length = 0
+
+        for num in num_set:
+            # Only start counting if num is the start of a sequence
+            if num - 1 not in num_set:
+                current_num = num
+                current_length = 1
+
+                # Count the length of the sequence
+                while current_num + 1 in num_set:
+                    current_num += 1
+                    current_length += 1
+
+                longest_length = max(longest_length, current_length)
+        
+        return longest_length

top-k-frequent-elements/pmjuu.py

@@ -0,0 +1,24 @@
+from collections import Counter
+from typing import List
+
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+        # 빈도 계산
+        count = Counter(nums)
+        n = len(nums)
+        
+        # 빈도수를 기준으로 버킷 생성 (0에서 n까지)
+        buckets = [[] for _ in range(n + 1)]
+        
+        # 각 숫자를 해당 빈도수의 버킷에 추가
+        for num, freq in count.items():
+            buckets[freq].append(num)
+        
+        # 빈도가 높은 순서대로 k개의 숫자를 추출
+        result = []
+        for freq in range(n, 0, -1):
+            if buckets[freq]:
+                result.extend(buckets[freq])
+                
+                if len(result) == k:
+                    return result

valid-palindrome/pmjuu.py

@@ -0,0 +1,21 @@
+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+        # two pointer
+        left, right = 0, len(s) - 1
+
+        while left < right:
+            # compare only alphanumeric characters
+            while left < right and not s[left].isalnum():
+                left += 1
+            while left < right and not s[right].isalnum():
+                right -= 1
+            
+            # compare with lowercase
+            if s[left].lower() != s[right].lower():
+                return False
+            
+            # move pointers
+            left += 1
+            right -= 1
+        
+        return True