From 1d2dd1e6c860be1193473c0dc84cac64e5a92bea Mon Sep 17 00:00:00 2001
From: SinnoLn <ok373737@naver.com>
Date: Wed, 12 Aug 2026 13:21:37 +0900
Subject: [PATCH] 914. X of a Kind in a Deck of Cards

---
 .../914. X of a Kind in a Deck of Cards.java" | 35 +++++++++++++++++++
 1 file changed, 35 insertions(+)
 create mode 100644 "leetcode3/\354\235\264\354\247\204\355\235\254/914. X of a Kind in a Deck of Cards.java"

diff --git "a/leetcode3/\354\235\264\354\247\204\355\235\254/914. X of a Kind in a Deck of Cards.java" "b/leetcode3/\354\235\264\354\247\204\355\235\254/914. X of a Kind in a Deck of Cards.java"
new file mode 100644
index 00000000..a5995f08
--- /dev/null
+++ "b/leetcode3/\354\235\264\354\247\204\355\235\254/914. X of a Kind in a Deck of Cards.java"	
@@ -0,0 +1,35 @@
+/*
+1. 아이디어: 크기가 1보다 큰 파티션으로 배열 나누기
+             이때 각 파티션은 같은 숫자로만 이루어져야 한다.
+             
+             최대공약수 - 유클리드 호제법 사용. 
+             먼저 숫자의 빈도수를 세고, 첫 빈도수 기준 최대 공약수를 각각 계산한다.
+
+2. 시간복잡도: O(10000*logN)
+
+3. 자료구조/알고리즘: 최대공약수, 카운팅
+
+*/
+
+class Solution {
+    public boolean hasGroupsSizeX(int[] deck) {
+        int[] cnt = new int[10001];
+
+        for(int i=0; i<deck.length; i++) cnt[deck[i]]++;
+
+        int g = -1;
+        for(int i=0; i<10001; i++) {
+            if(cnt[i] == 0) continue;
+            
+            if(g == -1) g = cnt[i]; 
+            else g = gcd(g, cnt[i]);
+        }
+        return g > 1;
+
+    }
+
+    private int gcd(int a, int b) {
+        if(b==0) return a;
+        return gcd(b, a%b);
+    }
+}
\ No newline at end of file