diff --git a/solution/2000-2099/2014.Longest Subsequence Repeated k Times/Solution.java b/solution/2000-2099/2014.Longest Subsequence Repeated k Times/Solution.java
new file mode 100644
index 0000000000000..a468db531301b
--- /dev/null
+++ b/solution/2000-2099/2014.Longest Subsequence Repeated k Times/Solution.java	
@@ -0,0 +1,88 @@
+import java.util.*;
+
+public class Solution {
+    public String longestSubsequenceRepeatedK(String s, int k) {
+        // Step 1: Frequency counter for each character in the string
+        Map<Character, Integer> freq = new HashMap<>();
+        for (char c : s.toCharArray()) {
+            freq.put(c, freq.getOrDefault(c, 0) + 1);
+        }
+
+        // Step 2: Collect all characters with at least k frequency
+        List<Character> candidates = new ArrayList<>();
+        for (Map.Entry<Character, Integer> entry : freq.entrySet()) {
+            if (entry.getValue() >= k) {
+                candidates.add(entry.getKey());
+            }
+        }
+
+        // Step 4: Breadth-First Search (BFS) to generate subsequences
+        Queue<String> queue = new LinkedList<>();
+        queue.offer(""); // Start with an empty sequence
+        String longest = "";
+
+        while (!queue.isEmpty()) {
+            String curr = queue.poll();
+
+            // Try appending each candidate character to the current subsequence
+            for (char ch : candidates) {
+                String newSubseq = curr + ch;
+
+                // Check if the new subsequence is repeatable k times
+                if (canBeRepeatedKTimes(newSubseq, s, k)) {
+                    queue.offer(newSubseq);
+
+                    // Update longest if newSubseq is longer or lexicographically larger
+                    if (newSubseq.length() > longest.length() || 
+                        (newSubseq.length() == longest.length() && newSubseq.compareTo(longest) > 0)) {
+                        longest = newSubseq;
+                    }
+                }
+            }
+        }
+
+        return longest;
+    }
+
+    // Helper function to check if a subsequence can be repeated k times
+    private boolean canBeRepeatedKTimes(String subseq, String s, int k) {
+        int i = 0, count = 0;
+        for (char ch : s.toCharArray()) {
+            if (ch == subseq.charAt(i)) {
+                i++;
+                if (i == subseq.length()) {
+                    i = 0;
+                    count++;
+                    if (count == k) {
+                        return true;
+                    }
+                }
+            }
+        }
+        return false;
+    }
+
+    public static void main(String[] args) {
+        Solution sol = new Solution();
+
+        // Test case 1
+        String s1 = "letsleetcode";
+        int k1 = 2;
+        System.out.println(sol.longestSubsequenceRepeatedK(s1, k1)); // Output: "let"
+
+        // Test case 2
+        String s2 = "bb";
+        int k2 = 2;
+        System.out.println(sol.longestSubsequenceRepeatedK(s2, k2)); // Output: "b"
+
+        // Test case 3
+        String s3 = "ab";
+        int k3 = 2;
+        System.out.println(sol.longestSubsequenceRepeatedK(s3, k3)); // Output: ""
+
+        // Test case 4
+        String s4 = "bbabbabbbbabaababab";
+        int k4 = 3;
+        System.out.println(sol.longestSubsequenceRepeatedK(s4, k4)); // Output: "bbbb"
+    }
+}