From 763a2950c672c3198f505b8aedebacaefe27ccb9 Mon Sep 17 00:00:00 2001
From: SinnoLn <ok373737@naver.com>
Date: Wed, 26 Aug 2026 12:22:27 +0900
Subject: [PATCH] 3940. Limit Occurrences in Sorted Array

---
 ...40_Limit Occurrences in Sorted Array.java" | 36 ++++++++++++++++++-
 1 file changed, 35 insertions(+), 1 deletion(-)

diff --git "a/leetcode3/\354\235\264\354\247\204\355\235\254/3940_Limit Occurrences in Sorted Array.java" "b/leetcode3/\354\235\264\354\247\204\355\235\254/3940_Limit Occurrences in Sorted Array.java"
index 42ed95b6..f49e72a3 100644
--- "a/leetcode3/\354\235\264\354\247\204\355\235\254/3940_Limit Occurrences in Sorted Array.java"	
+++ "b/leetcode3/\354\235\264\354\247\204\355\235\254/3940_Limit Occurrences in Sorted Array.java"	
@@ -1,4 +1,38 @@
-/*
+/* 2차 풀이
+
+1. 아이디어 : 정렬된 배열을 순회하며, 현재 값 개수가 k 이상이면 continue, 이하면 붙인다 (이전 풀이와 동일)
+
+2. 시간복잡도 : O(N)
+
+3. 자료구조/알고리즘 : 완전탐색
+
+ */
+
+class Solution {
+    public int[] limitOccurrences(int[] nums, int k) {
+        int cnt = 1;
+        int prev = nums[0];
+        int len = 0;
+        int[] ans = new int[nums.length];
+        
+        ans[len++] = nums[0];
+
+        for(int i=1; i<nums.length; i++) {
+            if(prev != nums[i]) {
+                cnt = 1;
+                prev = nums[i];
+            }
+            else cnt++;
+
+            if(cnt>k) continue;
+            ans[len++] = nums[i];
+        }
+
+        return Arrays.copyOfRange(ans, 0, len);
+    }
+}
+
+/* 1차 풀이
 
 1. 아이디어 : 배열을 순회하며, 현재 값 개수가 k 이상이면 continue, 이하면 붙인다