From c9df170a9bc76eee24aabee8c36bf4b96155b050 Mon Sep 17 00:00:00 2001
From: DongHun <hune0329@gmail.com>
Date: Mon, 16 Mar 2026 17:01:16 +0900
Subject: [PATCH 1/3] Solve valid-palindrome in two-way

---
 valid-palindrome/OstenHun.cpp | 84 +++++++++++++++++++++++++++++++++++
 1 file changed, 84 insertions(+)
 create mode 100644 valid-palindrome/OstenHun.cpp

diff --git a/valid-palindrome/OstenHun.cpp b/valid-palindrome/OstenHun.cpp
new file mode 100644
index 000000000..1267d0da8
--- /dev/null
+++ b/valid-palindrome/OstenHun.cpp
@@ -0,0 +1,84 @@
+/*
+    A phrase is a palindrome if,
+    after converting all uppercase letters into lowercase letters and
+    removing all non-alphanumeric characters, it reads the same forward and backward.
+    Alphanumeric characters include letters and numbers.
+
+    Given a string s, return true if it is a palindrome, or false otherwise.
+
+    Example 1:
+
+    Input: s = "A man, a plan, a canal: Panama"
+    Output: true
+    Explanation: "amanaplanacanalpanama" is a palindrome.
+
+    Constraints:
+
+    1 <= s.length <= 2 * 105
+    s consists only of printable ASCII characters.
+*/
+
+#include <cctype>
+#include <string>
+#include <vector>
+using namespace std;
+
+#pragma region ExtraSpaceIdea
+// 새로운 배열을 만들어서 문자열을 정리한 후에 팰린드롬 판별
+// 시간 복잡도 : O(n)
+// 공간 복잡도 : O(n)
+// n은 문자열의 길이일 것이다.
+namespace extra_space_idea {
+
+class Solution {
+public:
+    bool isPalindrome(string s) {
+        vector<char> str;
+        str.reserve(s.size());
+
+        size_t len = s.length();
+        for (size_t i = 0; i < len; i++) {
+            if (isalnum(s[i])) {
+                str.push_back(tolower(s[i]));
+            }
+        }
+
+        size_t vec_len = str.size();
+        for (size_t i = 0; i < (vec_len + 1) / 2; i++) {
+            if (str[i] != str[vec_len - 1 - i]) return false;
+        }
+
+        return true;
+    }
+};
+
+}  // namespace extra_space_idea
+#pragma endregion
+
+#pragma region FinalSolution
+// 투 포인터를 이용해 추가 배열 없이 구현
+// 시간 복잡도 : O(n)
+// 공간 복잡도 : O(1)
+class Solution {
+public:
+    bool isPalindrome(string s) {
+        int left = 0;
+        int right = s.length() - 1;
+
+        while(left < right) {
+            if (!isalnum(s[left]))
+                left++;
+            else if (!isalnum(s[right]))
+                right--;
+            else {
+                if (tolower(s[left]) != tolower(s[right]))
+                    return false;
+                left++;
+                right--;
+            }
+        }
+
+        return true;
+    }
+};
+#pragma endregion

From cd02770a7efd3bfe65744da1bf24bbcb1e4edd70 Mon Sep 17 00:00:00 2001
From: DongHun <hune0329@gmail.com>
Date: Wed, 18 Mar 2026 00:00:00 +0900
Subject: [PATCH 2/3] Solve number-of-1-bits

---
 number-of-1-bits/OstenHun.cpp | 42 +++++++++++++++++++++++++++++++++++
 1 file changed, 42 insertions(+)
 create mode 100644 number-of-1-bits/OstenHun.cpp

diff --git a/number-of-1-bits/OstenHun.cpp b/number-of-1-bits/OstenHun.cpp
new file mode 100644
index 000000000..6381f1892
--- /dev/null
+++ b/number-of-1-bits/OstenHun.cpp
@@ -0,0 +1,42 @@
+/*
+    191. Number of 1 Bits
+
+    Given a positive integer n, write a function 
+    that returns the number of set bits in its binary representation 
+    (also known as the Hamming weight).
+
+    Example 1:
+        Input: n = 11
+        Output: 3
+
+    Explanation:
+        The input binary string 1011 has a total of three set bits.     
+    
+    Constraints:
+        1 <= n <= 2^31 - 1
+*/
+
+// 비트 연산을 이용하여 풀기
+#include <iostream>
+using namespace std;
+
+class Solution {
+public:
+    int hammingweight(int n) {
+        unsigned int answer = 0;
+        while(n>0) {
+            if (n & 1)
+                answer++;
+            n = n >> 1;
+        }
+        return answer;
+    }
+};
+
+int main() {
+    int n;
+    cin >> n;
+    Solution s;
+
+    cout << s.hammingweight(n);
+}

From c055eff98e02f225eef3b46b03f8a99a25079597 Mon Sep 17 00:00:00 2001
From: DongHun <hune0329@gmail.com>
Date: Wed, 18 Mar 2026 01:46:06 +0900
Subject: [PATCH 3/3] Refine number-of-1-bits solution

---
 number-of-1-bits/OstenHun.cpp | 25 +++++++++++++++++--------
 1 file changed, 17 insertions(+), 8 deletions(-)

diff --git a/number-of-1-bits/OstenHun.cpp b/number-of-1-bits/OstenHun.cpp
index 6381f1892..7e280e0e7 100644
--- a/number-of-1-bits/OstenHun.cpp
+++ b/number-of-1-bits/OstenHun.cpp
@@ -20,6 +20,9 @@
 #include <iostream>
 using namespace std;
 
+// 시간 복잡도 : O(logn) -> n >> 1 연산을 하기 때문에 절반씩 줄어듬
+// 문제의 조건은 32bit 내의 범위이기 때문에 32번의 반복으로 항상 끝나기에 O(1)이라고 할 수 있다.
+// 공간 복잡도 : O(1)
 class Solution {
 public:
     int hammingweight(int n) {
@@ -29,14 +32,20 @@ class Solution {
                 answer++;
             n = n >> 1;
         }
+
+        // 처음 풀었던 풀이.
+        // unsigned int answer = 0;
+        // for (int i = 0; i < 32; i++) {
+        //     if ((n >> i) & 1) answer++;
+        // }
+        
+        // 생각 못 한 풀이
+        // -> n & (n-1) 을 하면 가장 오른쪽 1비트를 지운다.
+        // while (n > 0) {
+        //     n &= (n - 1);
+        //     answer++;
+        // }
+
         return answer;
     }
 };
-
-int main() {
-    int n;
-    cin >> n;
-    Solution s;
-
-    cout << s.hammingweight(n);
-}