From e150cdb37cf38ccacbe1e6b00ea81e072c0d03bd Mon Sep 17 00:00:00 2001
From: SinnoLn <ok373737@naver.com>
Date: Tue, 6 Oct 2026 22:38:35 +0900
Subject: [PATCH 1/2] 125. Valid Palindrome

---
 .../125. Valid Palindrome.java"               | 36 +++++++++++++++++++
 1 file changed, 36 insertions(+)
 create mode 100644 "leetcode3/\354\235\264\354\247\204\355\235\254/125. Valid Palindrome.java"

diff --git "a/leetcode3/\354\235\264\354\247\204\355\235\254/125. Valid Palindrome.java" "b/leetcode3/\354\235\264\354\247\204\355\235\254/125. Valid Palindrome.java"
new file mode 100644
index 00000000..497531c6
--- /dev/null
+++ "b/leetcode3/\354\235\264\354\247\204\355\235\254/125. Valid Palindrome.java"	
@@ -0,0 +1,36 @@
+/*
+
+1. 아이디어 : 문자열에서 숫자와 알파벳만 별도로 변환 후 펠린드롬 여부 체크
+
+2. 시간복잡도 : O(2N) => O(N)
+
+3. 자료구조/알고리즘 : 완전탐색
+
+ */
+
+class Solution {
+    public boolean isPalindrome(String s) {
+        // 소문자, 숫자만
+        StringBuilder sb = new StringBuilder();
+
+        for(int i=0; i<s.length(); i++) {
+            char c = s.charAt(i);
+
+            if(c>='a' && c<='z') sb.append(c);
+            else if(c>='A' && c<='Z') sb.append((char)(c+32));
+            else if(c>='0' && c<='9') sb.append(c);
+        }
+
+        return isPalindrome(sb);
+    }
+
+    private boolean isPalindrome(StringBuilder sb) {
+        int l = 0;
+        int r = sb.length()-1;
+
+        while(l<r) {
+            if(sb.charAt(l++) != sb.charAt(r--)) return false;
+        }
+        return true;
+    }
+}
\ No newline at end of file

From 03032ec5135bc6f5a8c4de4a01304c2a61609e40 Mon Sep 17 00:00:00 2001
From: SinnoLn <ok373737@naver.com>
Date: Tue, 6 Oct 2026 22:40:43 +0900
Subject: [PATCH 2/2] 447. Number of Boomerangs

---
 .../447. Number of Boomerangs.java"           | 48 +++++++++++++++++++
 1 file changed, 48 insertions(+)
 create mode 100644 "leetcode3/\354\235\264\354\247\204\355\235\254/447. Number of Boomerangs.java"

diff --git "a/leetcode3/\354\235\264\354\247\204\355\235\254/447. Number of Boomerangs.java" "b/leetcode3/\354\235\264\354\247\204\355\235\254/447. Number of Boomerangs.java"
new file mode 100644
index 00000000..6ae904c0
--- /dev/null
+++ "b/leetcode3/\354\235\264\354\247\204\355\235\254/447. Number of Boomerangs.java"	
@@ -0,0 +1,48 @@
+/* 
+
+1. 아이디어 : HashMap 자료구조의 특징 활용
+            한 점을 기준점으로 잡고, 두점 사이의 거리가 같은 개수를 셈 
+            이후 모든 경우의 수를 계산
+
+2. 시간복잡도 : O(N^2)
+
+3. 자료구조/알고리즘 : HashMap 자료구조 특징, 두 점사이의 거리
+
+ */
+
+class Solution {
+    public int numberOfBoomerangs(int[][] points) {
+        // i and j 거리 == i and k 거리
+        // 최대 500개의 고유한 튜플
+        // 22 33 11  22 11 33 (절댓값)
+
+        int ans = 0;
+        Map<Integer, Integer> map = new HashMap<>(); // 매번 객체 생성 불필요
+
+        // (jy - iy)^2 + (jx - ix)^2 = (ky - iy)^2 + (kx - ix)^2
+        
+        for(int i=0; i<points.length; i++) {
+            for(int j=0; j<points.length; j++) {
+                
+                if(i==j) continue;
+
+                int diffY = points[i][1] - points[j][1];
+                int diffX = points[i][0] - points[j][0];
+
+                int distance = diffY*diffY + diffX*diffX;
+
+                if(!map.containsKey(distance)) map.put(distance, 1);
+                else map.put(distance, map.get(distance)+1);
+            }
+
+            for(int key : map.keySet()) {
+                int value = map.get(key);
+                ans += value*(value-1);
+            }
+
+            map.clear();
+        }
+
+        return ans;
+    }
+}
\ No newline at end of file