feat: Add Solution.py

Author: CodersAcademy006

Diff for: solution/2000-2099/2014.Longest Subsequence Repeated k Times/Solution.py

@@ -0,0 +1,52 @@
+from collections import Counter, deque
+
+class Solution:
+    def longestSubsequenceRepeatedK(self, s: str, k: int) -> str:
+        # Step 1: Frequency counter for each character in the string
+        freq = Counter(s)
+        
+        # Step 2: Collect all characters with at least k frequency
+        candidates = [ch for ch, count in freq.items() if count >= k]
+        
+        # Step 3: Helper function to check if a given sequence can be repeated k times
+        def canBeRepeatedKTimes(subseq):
+            i = 0
+            count = 0
+            for ch in s:
+                if ch == subseq[i]:
+                    i += 1
+                    if i == len(subseq):
+                        i = 0
+                        count += 1
+                        if count == k:
+                            return True
+            return False
+
+        # Step 4: Breadth-First Search (BFS) to generate subsequences
+        queue = deque([""])  # Start with an empty sequence
+        longest = ""
+
+        while queue:
+            curr = queue.popleft()
+
+            # Try appending each candidate character to the current subsequence
+            for ch in candidates:
+                new_subseq = curr + ch
+                
+                # Check if the new subsequence is repeatable k times
+                if canBeRepeatedKTimes(new_subseq):
+                    queue.append(new_subseq)
+                    
+                    # Update longest if new_subseq is longer or lexicographically larger
+                    if len(new_subseq) > len(longest) or (len(new_subseq) == len(longest) and new_subseq > longest):
+                        longest = new_subseq
+
+        return longest
+
+
+# Example usage:
+sol = Solution()
+print(sol.longestSubsequenceRepeatedK("letsleetcode", 2))  # Output: "let"
+print(sol.longestSubsequenceRepeatedK("bb", 2))  # Output: "b"
+print(sol.longestSubsequenceRepeatedK("ab", 2))  # Output: ""
+print(sol.longestSubsequenceRepeatedK("bbabbabbbbabaababab", 3))  # Output: "bbbb"