feat(work): add longest‑repeating‑character‑replacement solution and rename variable in longest‑substring solution

work/Medium/Hash Table/3.longest-substring-without-repeating-characters.py

@@ -45,12 +45,12 @@ 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:
-                seen.remove(s[left])
+            while c in window:
+                window.remove(s[left])
                 left += 1
-            seen.add(c)
+            window.add(c)
             ans = max(ans, right - left + 1)
 
         return ans

work/Medium/Hash Table/424.longest-repeating-character-replacement.py

@@ -0,0 +1,60 @@
+"""
+424. Longest Repeating Character Replacement
+Difficulty: Medium
+https://leetcode.com/problems/longest-repeating-character-replacement/
+
+──────────────────────────────────────────────────
+
+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.
+
+Return the length of the longest substring containing the same letter
+you can get after performing the above operations.
+
+
+
+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
+"""
+
+
+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