Merge pull request #2114 from devyulbae/main

[devyulbae] WEEK 03 solutions

Author: devyulbae

decode-ways/devyulbae.py

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+class Solution:
+    def numDecodings(self, s: str) -> int:
+        

longest-substring-without-repeating-characters/devyulbae.py

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+"""
+Blind75 - length of longest substring without repeating characters
+https://leetcode.com/problems/longest-substring-without-repeating-characters/
+시간복잡도 : O(n)
+공간복잡도 : O(min(m, n)) (문자 집합 char_index_map의 크기, 최대는 n = len(s))
+풀이 : 슬라이딩 윈도우 기법을 사용한 문자열 순회
+"""
+
+
+class Solution:
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        max_count = 0
+        start = 0
+        char_index_map = {}
+
+        for i, char in enumerate(s):
+            if char in char_index_map and char_index_map[char] >= start:
+                start = char_index_map[char] + 1
+                char_index_map[char] = i
+            else:
+                char_index_map[char] = i
+                max_count = max(max_count, i - start + 1)
+
+        return max_count
+    

maximum-subarray/devyulbae.py

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+"""
+Blind75 - 5. Maximum Subarray
+LeetCode Problem Link: https://leetcode.com/problems/maximum-subarray/
+
+통과는 했는데 시간복잡도 O(n^2)라서 좀 아쉽다...투포인터로 풀 수도 있을 거 같은데...
+"""
+from typing import List
+
+class Solution:
+    def maxSubArray(self, nums: List[int]) -> int:
+        max_total = nums[0]
+
+        for s in range(len(nums)):
+            total = 0
+            for e in range(s, len(nums)):
+                total += nums[e]
+                if total > max_total:
+                    max_total = total
+        return max_total
+    

number-of-1-bits/devyulbae.py

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+"""
+Blind75 - Number of 1 Bits
+https://leetcode.com/problems/number-of-1-bits/
+시간복잡도 : O(log n)
+공간복잡도 : O(1)
+
+풀이 :
+    파이썬의 내장함수 bin() -> O(log n)
+    문자열 메서드 count() -> O(log n)
+
+"""
+
+class Solution:
+    def hammingWeight(self, n: int) -> int:
+        return str(bin(n)).count('1')
+    
+

valid-palindrome/devyulbae.py

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+"""
+Blind75 - Valid Palindrome
+https://leetcode.com/problems/valid-palindrome/
+
+시간복잡도 O(n), 공간복잡도 O(n)
+
+isalnum() 함수는 시간복잡도 O(1)이므로 필터링하는 과정도 O(n)이다.
+[::-1] 슬라이싱도 O(n)이다.
+따라서 전체 시간복잡도는 O(n)이다.
+
+
+Runtime   Beats
+7ms       82.67%
+
+Memory    Beats
+23.14 MB  17.23%
+"""
+
+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+        filtered_s = ''.join(char.lower() for char in s if char.isalnum())
+        return filtered_s == filtered_s[::-1]
+