From 5b436c6b25ba5410d1140c722c3cba63f275f2de Mon Sep 17 00:00:00 2001
From: SinnoLn <ok373737@naver.com>
Date: Mon, 27 Jul 2026 15:33:53 +0900
Subject: [PATCH 1/2] =?UTF-8?q?[=EC=9D=B4=EC=A7=84=ED=9D=AC]=20Day28?=
MIME-Version: 1.0
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---
 .../3663. Find The Least Frequent Digit.java" | 35 +++++++++++++++++++
 1 file changed, 35 insertions(+)
 create mode 100644 "leetcode3/\354\235\264\354\247\204\355\235\254/3663. Find The Least Frequent Digit.java"

diff --git "a/leetcode3/\354\235\264\354\247\204\355\235\254/3663. Find The Least Frequent Digit.java" "b/leetcode3/\354\235\264\354\247\204\355\235\254/3663. Find The Least Frequent Digit.java"
new file mode 100644
index 00000000..686f9b7f
--- /dev/null
+++ "b/leetcode3/\354\235\264\354\247\204\355\235\254/3663. Find The Least Frequent Digit.java"	
@@ -0,0 +1,35 @@
+/*
+
+1. 아이디어 : 숫자의 빈도수 계산 후, 가장 빈도수가 적은 숫자 반환
+              정답이 여러개일 경우 더 작은 숫자 반환
+              각 숫자를 카운팅하여 계산 후 해결
+
+2. 시간복잡도 : O(31 + 10) => O(1)
+
+3. 자료구조/알고리즘 : 배열
+
+ */
+
+class Solution {
+    public int getLeastFrequentDigit(int n) {
+        int[] nums = new int[10];
+
+        while(n > 0) {
+            int num = n%10;
+            nums[num]++;
+
+            n/=10;
+        }   
+
+        int minNum = 0;
+        int minCnt = 100;
+        for(int i=0; i<10; i++) {
+            if(nums[i] == 0 || minCnt<=nums[i]) continue;
+
+            minNum = i;
+            minCnt = nums[i];
+        }
+
+        return minNum;
+    }
+}
\ No newline at end of file

From d23e1530c8ff4eb25a3adacf1878d000f83b56ca Mon Sep 17 00:00:00 2001
From: SinnoLn <ok373737@naver.com>
Date: Mon, 27 Jul 2026 16:41:02 +0900
Subject: [PATCH 2/2] =?UTF-8?q?[=EC=9D=B4=EC=A7=84=ED=9D=AC]=20Day26?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 .../809. Expressive Words.java"               | 69 +++++++++++++++++++
 1 file changed, 69 insertions(+)
 create mode 100644 "leetcode3/\354\235\264\354\247\204\355\235\254/809. Expressive Words.java"

diff --git "a/leetcode3/\354\235\264\354\247\204\355\235\254/809. Expressive Words.java" "b/leetcode3/\354\235\264\354\247\204\355\235\254/809. Expressive Words.java"
new file mode 100644
index 00000000..a0696746
--- /dev/null
+++ "b/leetcode3/\354\235\264\354\247\204\355\235\254/809. Expressive Words.java"	
@@ -0,0 +1,69 @@
+/*
+
+1. 아이디어 : 조건에 맞춰 구현
+              리스트를 사용하여 각 문자열을 압축, 각각의 개수 구함
+              words 배열을 완전탐색하여 똑같이 계산 후, 늘릴수 있는지, 조건에 맞춰 하나하나 비교
+
+2. 시간복잡도 : O(S+N*M) (N = words.length, M = words 배열의 각 단어의 길이)
+
+3. 자료구조/알고리즘 : 구현
+
+ */
+
+class Solution {
+    private List<int[]> ori;
+
+    public int expressiveWords(String s, String[] words) {
+        ori = new ArrayList<>();
+        char memo = s.charAt(0);
+        int cnt = 1;
+        int ans = 0;
+
+        for(int i=1; i<s.length(); i++) {
+            char c = s.charAt(i);
+            if(c != memo) {
+                ori.add(new int[]{memo, cnt});
+                memo = c;
+                cnt = 1;
+            }
+            else cnt++;
+        }
+        ori.add(new int[]{memo, cnt});
+
+        for(String word : words) {
+            if(checkExpressiveWords(word)) ans++;
+        }
+
+        return ans;
+    }
+
+    private boolean checkExpressiveWords(String s) {
+        List<int[]> tmp = new ArrayList<>();
+        char memo = s.charAt(0);
+        int cnt = 1;
+        int ans = 0;
+
+        for(int i=1; i<s.length(); i++) {
+            char c = s.charAt(i);
+            if(c != memo) {
+                tmp.add(new int[]{memo, cnt});
+                memo = c;
+                cnt = 1;
+            }
+            else cnt++;
+        }
+        tmp.add(new int[]{memo, cnt});
+
+        if(tmp.size() != ori.size()) return false;
+        for(int i=0; i<tmp.size(); i++) {
+            int[] o = ori.get(i);
+            int[] t = tmp.get(i);
+
+            if(o[0] != t[0]) return false;
+            if(o[1] < t[1]) return false;
+            if(o[1] != t[1] && o[1] < 3) return false;
+        }
+
+        return true;
+    }
+}
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