From 09741bdb06f61e79e2813999f3e67f405978789a Mon Sep 17 00:00:00 2001
From: aa601 <vmop899@gmail.com>
Date: Sat, 1 Feb 2025 04:36:04 +0900
Subject: [PATCH] FEAT : upload #232 #244 #274 #284

---
 longest-common-subsequence/aa601.py           | 16 ++++++++++
 .../aa601.py                                  | 30 +++++++++++++++++++
 number-of-1-bits/aa601.py                     | 13 ++++++++
 sum-of-two-integers/aa601.py                  | 17 +++++++++++
 4 files changed, 76 insertions(+)
 create mode 100644 longest-common-subsequence/aa601.py
 create mode 100644 longest-repeating-character-replacement/aa601.py
 create mode 100644 number-of-1-bits/aa601.py
 create mode 100644 sum-of-two-integers/aa601.py

diff --git a/longest-common-subsequence/aa601.py b/longest-common-subsequence/aa601.py
new file mode 100644
index 000000000..90c76f031
--- /dev/null
+++ b/longest-common-subsequence/aa601.py
@@ -0,0 +1,16 @@
+'''
+TC : O(nm)
+SC : O(nm)
+'''
+class Solution:
+    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
+        #dp 2차원 배열 생성
+        dp = [[0 for _ in range(len(text2) + 1)] for _ in range(len(text1) + 1)]
+        #i와 j는 dp기준
+        for i in range(1, len(text1) + 1):
+            for j in range(1, len(text2) + 1):
+                if text1[i - 1] == text2[j - 1]:
+                    dp[i][j] = dp[i - 1][j - 1] + 1
+                else:
+                    dp[i][j] = max(dp[i][j - 1], dp[i - 1][j])
+        return dp[i][j]
diff --git a/longest-repeating-character-replacement/aa601.py b/longest-repeating-character-replacement/aa601.py
new file mode 100644
index 000000000..c352b00ba
--- /dev/null
+++ b/longest-repeating-character-replacement/aa601.py
@@ -0,0 +1,30 @@
+'''
+TC : O(n) 
+	문자열 s를 총 한번 순회한다
+SC : O(1)
+	최대 26개의 글자를 딕셔너리에 저장한다
+'''
+class Solution:
+    def characterReplacement(self, s: str, k: int) -> int:
+        start = 0
+        max_len = 0
+        max_cnt = 0
+        substr_len = 0
+        # 문자열 s에 대한 해시테이블을 만들고 초기 값을 0으로 설정
+        dic = {char : 0 for char in s}
+
+        # 기본적으로 end를 오른쪽으로 이동시키는 방향
+        for end in range(len(s)):
+            dic[s[end]] += 1
+
+            # 현재까지의 최다 빈도 글자수 계산
+            max_cnt = max(max_cnt, dic[s[end]])
+
+            # 현재 부분 문자열 안에서(즉 end가 고정된 상태에서)
+            # 바꿀 수 있는 글자의 개수가 k에 못미친다면
+            # start를 오른쪽으로 한칸 당긴다
+            if end - start + 1 - max_cnt > k:
+                dic[s[start]] -= 1
+                start += 1
+            max_len = max(max_len, end - start + 1)
+        return max_len
diff --git a/number-of-1-bits/aa601.py b/number-of-1-bits/aa601.py
new file mode 100644
index 000000000..62e17172e
--- /dev/null
+++ b/number-of-1-bits/aa601.py
@@ -0,0 +1,13 @@
+'''
+TC : O(logn)
+	비트연산이 n을 2로 나누는 것과 같으므로 로그 시간복잡도를 갖는다
+SC : O(1)
+'''
+class Solution:
+    def hammingWeight(self, n: int) -> int:
+        cnt = 0
+        while n != 0:
+            if n & 1 == 1:
+                cnt += 1
+            n >>= 1
+        return cnt
diff --git a/sum-of-two-integers/aa601.py b/sum-of-two-integers/aa601.py
new file mode 100644
index 000000000..1fda88258
--- /dev/null
+++ b/sum-of-two-integers/aa601.py
@@ -0,0 +1,17 @@
+'''
+TC : O(1)
+    최대 32번 반복
+SC : O(1)
+'''
+class Solution:
+    def getSum(self, a: int, b: int) -> int:
+        # 32비트 정수 연산을 위한 마스크 비트 설정
+        mask = 0xFFFFFFFF
+        while b & mask :
+            tmp = a
+            a = a ^ b
+            b = (tmp & b) << 1
+        if b > 0: # 연산결과 32비트를 벗어난 경우 a의 32비트까지 반환
+            return a & mask
+        else: # 32비트 안에서 연산이 끝났다면 마스크할 필요 없음
+            return a