From 2dc5562e25a404a0af93d21326a374875f93cb24 Mon Sep 17 00:00:00 2001
From: Nakji <yuitfhk@gmail.com>
Date: Tue, 25 Nov 2025 15:21:04 +0900
Subject: [PATCH 1/3] valid palindrome

---
 valid-palindrome/YuuuuuuYu.java | 37 +++++++++++++++++++++++++++++++++
 1 file changed, 37 insertions(+)
 create mode 100644 valid-palindrome/YuuuuuuYu.java

diff --git a/valid-palindrome/YuuuuuuYu.java b/valid-palindrome/YuuuuuuYu.java
new file mode 100644
index 0000000000..3859e55c65
--- /dev/null
+++ b/valid-palindrome/YuuuuuuYu.java
@@ -0,0 +1,37 @@
+/**
+ * Runtime: 2ms
+ * Time Complexity: O(n)
+ *
+ * Memory: 44.29MB
+ * Space Complexity: O(1)
+ *
+ * Approach: 투 포인터
+ * 1) 문자열의 양 끝에서부터 시작하는 두 포인터를 설정
+ * 2) 포인터가 가리키는 문자가 영숫자가 아닌 경우, 해당 포인터를 이동
+ */
+class Solution {
+    public boolean isPalindrome(String s) {
+        int start = 0;
+        int end = s.length()-1;
+
+        while (start < end) {
+            char currLeft = s.charAt(start);
+            char currRight = s.charAt(end);
+
+            if (!Character.isLetterOrDigit(currLeft)) {
+                start++;
+            } else if (!Character.isLetterOrDigit(currRight)) {
+                end--;
+            } else {
+                if (Character.toLowerCase(currLeft) != Character.toLowerCase(currRight)) {
+                    return false;
+                }
+
+                start++;
+                end--;
+            }
+        }
+
+        return true;
+    }
+}

From 5c2fc918339dd0a7aee0196cf92d1ec0904d50b6 Mon Sep 17 00:00:00 2001
From: Nakji <yuitfhk@gmail.com>
Date: Tue, 25 Nov 2025 15:34:40 +0900
Subject: [PATCH 2/3] number of 1 bits

---
 number-of-1-bits/YuuuuuuYu.java | 23 +++++++++++++++++++++++
 1 file changed, 23 insertions(+)
 create mode 100644 number-of-1-bits/YuuuuuuYu.java

diff --git a/number-of-1-bits/YuuuuuuYu.java b/number-of-1-bits/YuuuuuuYu.java
new file mode 100644
index 0000000000..67b8ed9b65
--- /dev/null
+++ b/number-of-1-bits/YuuuuuuYu.java
@@ -0,0 +1,23 @@
+/**
+ * Runtime: 0ms
+ * Time Complexity: O(log n)
+ * - n을 2로 나누는 과정을 반복하므로 log n에 비례
+ *
+ * Memory: 42.32MB
+ * Space Complexity: O(1)
+ *
+ * Approach: 비트 연산 (나눗셈 방식)
+ * 1) n을 2로 나눈 나머지(n%2)를 확인하여 1의 개수를 카운트
+ * 2) 마지막에 1은 무조건 남기 때문에 while문 종료 후 1을 더해줌
+ */
+class Solution {
+    public int hammingWeight(int n) {
+        int count = 0;
+        while (n > 1) {
+            count += n%2;
+            n /= 2;
+        }
+
+        return count+1;
+    }
+}

From 557ca2707581aed249d2a45f8f8313660094d23e Mon Sep 17 00:00:00 2001
From: Nakji <yuitfhk@gmail.com>
Date: Tue, 25 Nov 2025 16:58:35 +0900
Subject: [PATCH 3/3] maximum subarray

---
 maximum-subarray/YuuuuuuYu.java | 24 ++++++++++++++++++++++++
 1 file changed, 24 insertions(+)
 create mode 100644 maximum-subarray/YuuuuuuYu.java

diff --git a/maximum-subarray/YuuuuuuYu.java b/maximum-subarray/YuuuuuuYu.java
new file mode 100644
index 0000000000..92be99d01f
--- /dev/null
+++ b/maximum-subarray/YuuuuuuYu.java
@@ -0,0 +1,24 @@
+/**
+ * Runtime: 1ms
+ * Time Complexity: O(n)
+ *
+ * Memory: 76.88MB
+ * Space Complexity: O(1)
+ *
+ * Approach: 카데인 알고리즘
+ * - 부분 배열의 합이 최대가 되는 값을 찾는 알고리즘
+ * 1) 이전 합보다 현재 숫자가 더 크면 현재 숫자로 sum을 초기화
+ */
+class Solution {
+    public int maxSubArray(int[] nums) {
+        int sum = nums[0];
+        int max = nums[0];
+
+        for (int i=1; i<nums.length; i++) {
+            sum = Math.max(nums[i], sum+nums[i]);
+            max = Math.max(max, sum);
+        }
+
+        return max;
+    }
+}