diff --git a/longest-repeating-character-replacement/hi-rachel.py b/longest-repeating-character-replacement/hi-rachel.py
index 6f0b220ae..aeeb690ba 100644
--- a/longest-repeating-character-replacement/hi-rachel.py
+++ b/longest-repeating-character-replacement/hi-rachel.py
@@ -44,3 +44,25 @@ def characterReplacement(self, s: str, k: int) -> int:
             max_len = max(max_len, right - left + 1)
 
         return max_len
+
+
+from collections import Counter
+
+class Solution:
+    def characterReplacement(self, s: str, k: int) -> int:
+        count = Counter()
+        l = 0
+        max_freq = 0
+        max_length = 0
+
+        for r, ch in enumerate(s):
+            count[ch] += 1
+            max_freq = max(max_freq, count[ch])
+
+            if (r - l + 1) - max_freq > k:
+                count[s[l]] -= 1
+                l += 1
+            
+            max_length = max(max_length, r - l + 1)
+
+        return max_length
diff --git a/palindromic-substrings/hi-rachel.py b/palindromic-substrings/hi-rachel.py
index b397fa3b2..0fd0ebec2 100644
--- a/palindromic-substrings/hi-rachel.py
+++ b/palindromic-substrings/hi-rachel.py
@@ -55,3 +55,37 @@ def countSubstrings(self, s: str) -> int:
                     cnt += 1
 
         return cnt
+
+
+# 25/09/08 복습
+
+"""
+“모든 회문은 중심에서 좌우 대칭이다”
+→ 따라서 모든 중심점에서 좌우로 확장하면서 회문인지 확인하면
+→ 모든 회문 substring을 탐색 가능!
+
+- 중심점의 개수는 총 2n - 1개:
+- 홀수 길이: center = 0 ~ n-1 (expand(i, i))
+- 짝수 길이: center = 0 ~ n-1 (expand(i, i+1))
+
+TC: O(N^2)
+SC: O(1)
+"""
+
+class Solution:
+    def countSubstrings(self, s: str) -> int:
+        count = 0
+        for center in range(len(s)):
+            count += self.expand(s, center, center)
+
+            count += self.expand(s, center, center + 1)
+        return count
+
+    def expand(self, s, left, right):
+        count = 0
+        # 범위를 벗어나지 않고, palindrome이면 확장
+        while left >= 0 and right < len(s) and s[left] == s[right]:
+            count += 1
+            left -= 1
+            right += 1
+        return count