docs(docs): rename variable to window and add Longest Repeating Character Replacement page

apps/docs/content/(hash-table)/3-longest-substring-without-repeating-characters.mdx

@@ -35,12 +35,12 @@ class Solution:
     def lengthOfLongestSubstring(self, s: str) -> int:
         left = 0
         ans = 0
-        seen = set()
+        window = set()
         for right, c in enumerate(s):
-            while c in seen:
+            while c in window:
-                seen.remove(s[left])
+                window.remove(s[left])
                 left += 1
-            seen.add(c)
+            window.add(c)
             ans = max(ans, right - left + 1)
 
         return ans

apps/docs/content/(hash-table)/424-longest-repeating-character-replacement.mdx

@@ -0,0 +1,47 @@
+---
+title: '424. Longest Repeating Character Replacement'
+description: You are given a string s and an integer k. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most k times
+sidebar:
+  label: 'Longest Repeating Character Replacement'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Sliding Window</Badge>
+
+### Example 1:
+- Input: `s = "ABAB", k = 2`
+- Output: `4`
+- Explanation: Replace the two 'A's with two 'B's or vice versa.
+
+### Example 2:
+- Input: `s = "AABABBA", k = 1`
+- Output: `4`
+- Explanation: Replace the one 'A' in the middle with 'B' and form "AABBBBA". The substring "BBBB" has the longest repeating letters, which is `4`. There may exists other ways to achieve this answer too.
+
+### Constraints:
+
+- `1 <= s.length <= 10^5`
+- `s` consists of only uppercase English letters.
+- `0 <= k <= s.length`
+
+## Solution
+
+```py
+class Solution:
+    def characterReplacement(self, s: str, k: int) -> int:
+        count = {}
+        res = 0
+
+        l = 0
+        maxF = 0
+        for r, c in enumerate(s):
+            count[c] = 1 + count.get(c, 0)
+            maxF = max(maxF, count[c])
+
+            while (r - l + 1) - maxF > k:
+                count[s[l]] -= 1
+                l += 1
+
+            res = max(res, r - l + 1)
+        return res
+```