From d5e5a440838efb52d0a10fdb5d8f31b2f330f41e Mon Sep 17 00:00:00 2001
From: socow <socowjh@gmail.com>
Date: Tue, 25 Nov 2025 12:15:25 +0900
Subject: [PATCH 1/3] =?UTF-8?q?valid-palindrome=ED=92=80=EC=9D=B4?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 valid-palindrome/socow.py | 15 +++++++++++++++
 1 file changed, 15 insertions(+)
 create mode 100644 valid-palindrome/socow.py

diff --git a/valid-palindrome/socow.py b/valid-palindrome/socow.py
new file mode 100644
index 0000000000..d7f33c768e
--- /dev/null
+++ b/valid-palindrome/socow.py
@@ -0,0 +1,15 @@
+"""
+        문제 요약
+        - 영숫자만 고려, 대소문자 무시 → 팰린드롬 여부
+
+        아이디어
+        - 소문자 변환 → 영숫자만 필터 → 역순과 동일한지 비교
+
+        시간복잡도: O(n)
+        공간복잡도: O(n)
+"""
+
+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+        cleaned = [c for c in s.lower() if c.isalnum()]
+        return cleaned == cleaned[::-1]
\ No newline at end of file

From f6ed285c424edbdfae9e5e24b5ca00cf5982d1a2 Mon Sep 17 00:00:00 2001
From: socow <socowjh@gmail.com>
Date: Tue, 25 Nov 2025 12:22:24 +0900
Subject: [PATCH 2/3] =?UTF-8?q?number-of-1-bits=ED=92=80=EC=9D=B4?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 number-of-1-bits/socow.py | 19 +++++++++++++++++++
 valid-palindrome/socow.py |  2 +-
 2 files changed, 20 insertions(+), 1 deletion(-)
 create mode 100644 number-of-1-bits/socow.py

diff --git a/number-of-1-bits/socow.py b/number-of-1-bits/socow.py
new file mode 100644
index 0000000000..b45ceb274b
--- /dev/null
+++ b/number-of-1-bits/socow.py
@@ -0,0 +1,19 @@
+"""
+        문제 요약
+        - 32비트 부호 없는 정수 n의 이진 표현에서 '1'의 개수(= set bits)를 구하라.
+
+        아이디어 A: 브라이언 커니핸(Brian Kernighan) 알고리즘
+        - 핵심 관찰: n & (n - 1)은 n의 최하위 1비트를 '0'으로 만든다.
+        예) n = 0b101100 → n-1 = 0b101011 → n & (n-1) = 0b101000 (최하위 1 하나 삭제)
+        - 따라서 '1'의 개수만큼 루프가 돈다.
+        - 시간복잡도: O(k)  (k = n의 1비트 개수)  ← 매우 빠름
+        - 공간복잡도: O(1)
+"""
+
+class Solution:
+    def hammingWeight(self, n: int) -> int:
+        cnt = 0
+        while n:
+            n &= n - 1   # 최하위 1비트 제거
+            cnt += 1
+        return cnt
diff --git a/valid-palindrome/socow.py b/valid-palindrome/socow.py
index d7f33c768e..fc885c9576 100644
--- a/valid-palindrome/socow.py
+++ b/valid-palindrome/socow.py
@@ -12,4 +12,4 @@
 class Solution:
     def isPalindrome(self, s: str) -> bool:
         cleaned = [c for c in s.lower() if c.isalnum()]
-        return cleaned == cleaned[::-1]
\ No newline at end of file
+        return cleaned == cleaned[::-1]

From faf6d9826007db5f6f7d4e1571a9aee086362f08 Mon Sep 17 00:00:00 2001
From: socow <socowjh@gmail.com>
Date: Wed, 26 Nov 2025 14:39:11 +0900
Subject: [PATCH 3/3] combination-sum

---
 combination-sum/socow.py | 34 ++++++++++++++++++++++++++++++++++
 1 file changed, 34 insertions(+)
 create mode 100644 combination-sum/socow.py

diff --git a/combination-sum/socow.py b/combination-sum/socow.py
new file mode 100644
index 0000000000..d82fb59d86
--- /dev/null
+++ b/combination-sum/socow.py
@@ -0,0 +1,34 @@
+"""
+        문제 요약
+        - candidates 배열에서 숫자를 무제한 사용하여 합이 target이 되는 모든 조합 찾기
+
+        아이디어
+        - 백트래킹: i번째 인덱스부터 탐색하여 중복 조합 방지
+        - 같은 숫자 재사용 가능 → 재귀 시 인덱스 i 유지
+        - 정렬 후 target 초과 시 break로 가지치기
+
+        시간복잡도: O(N^(T/M)) - N: candidates 길이, T: target, M: 최소값
+        공간복잡도: O(T/M) - 재귀 깊이
+"""
+
+class Solution:
+    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
+        result = []
+        sol = []
+        candidates.sort()
+        n = len(candidates)
+
+        def backtrack(start, cur_sum):
+            if cur_sum == target:
+                result.append(sol.copy())
+                return
+
+            for i in range(start, n):
+                if cur_sum + candidates[i] > target:
+                    break
+                sol.append(candidates[i])
+                backtrack(i, cur_sum + candidates[i])
+                sol.pop()
+
+        backtrack(0, 0)
+        return result