Add 395

Author: trung-hn

Readme.md

@@ -89,6 +89,7 @@
 | 389 | Find the Difference | Medium | O(N) | O(1) | Hash Table, Bit Manipulation | | |
 | 387 | First Unique Character in a String | Easy | O(N) | O(N) | | | |
 | 394 | Decode String | Medium | O(maxK*N) | O(N) | Stack, DFS | | |
+| 395 | Longest Substring with At Least K Repeating Characters | Medium | O(NlogN) | O(N) | Divide and Conquer, Recursion, Sliding Window | | |
 | 399 | Evaluate Division | Medium | O(N*M) | O(N) | Union Find, Graph | | |
 | 402 | Remove K Digits | Medium | O(N) | O(N) | Greedy | | |
 | 404 | Sum of Left Leaves | Easy | O(N) | O(H) | Tree | | |

src/395.longest-substring-with-at-least-k-repeating-characters.py

@@ -0,0 +1,32 @@
+#
+# @lc app=leetcode id=395 lang=python3
+#
+# [395] Longest Substring with At Least K Repeating Characters
+#
+
+# @lc code=start
+# TAGS: Divide and Conquer, Recursion, Sliding Window
+class Solution:
+    # LTE. Time: O(N^2)
+    def longestSubstring(self, s: str, k: int) -> int:
+        def valid(counter, k):
+            return all(c >= k for c in counter.values())
+        
+        ans = 0
+        for i in range(len(s)):
+            counter = collections.Counter()
+            for j in range(i, len(s)):
+                counter[s[j]] += 1
+                if valid(counter, k):
+                    ans = max(ans, j - i + 1)
+        return ans
+                    
+    # Divide and conquer. 32 ms, 83.66%. 
+    # Worst case is still O(N^2) but O(NlogN) on average
+    def longestSubstring(self, s: str, k: int) -> int:
+        for c in set(s):
+            if s.count(c) < k: # split here
+                return max(self.longestSubstring(sub, k) for sub in s.split(c))
+        return len(s)
+# @lc code=end
+

testcases/395

@@ -0,0 +1,18 @@
+"aaabb"
+3
+"ababbc"
+2
+"abaaaccaabb"
+3
+"a"
+1
+"aabbaaabbccddeeaa"
+3
+"baaaccaaabdd"
+2
+"baaacccaaabdd"
+3
+"aa"
+3
+"baaabcb"
+3