diff --git a/dsa-solutions/lc-solutions/0500-0599/0522-Longest-Uncommon-Subsequence-II.md b/dsa-solutions/lc-solutions/0500-0599/0522-Longest-Uncommon-Subsequence-II.md
new file mode 100644
index 000000000..70930814d
--- /dev/null
+++ b/dsa-solutions/lc-solutions/0500-0599/0522-Longest-Uncommon-Subsequence-II.md
@@ -0,0 +1,163 @@
+---
+id: longest-uncommon-subsequence-ii
+title: Longest Uncommon Subsequence II
+sidebar_label: 0522-Longest-Uncommon-Subsequence-II
+tags:
+- String
+- Sorting
+description: "Given an array of strings strs, return the length of the longest uncommon subsequence between them. If the longest uncommon subsequence does not exist, return -1."
+---
+
+## Problem
+
+Given an array of strings `strs`, return the length of the longest uncommon subsequence between them. An uncommon subsequence is a subsequence that is not common to any other string in the array.
+
+A **subsequence** of a string `s` is a string that can be obtained after deleting any number of characters from `s`.
+
+### Examples
+
+**Example 1:**
+
+**Input:** `strs = ["aba","cdc","eae"]`  
+**Output:** `3`
+
+**Example 2:**
+
+**Input:** `strs = ["aaa","aaa","aa"]`  
+**Output:** `-1`
+
+### Constraints
+
+- `2 <= strs.length <= 50`
+- `1 <= strs[i].length <= 10`
+- `strs[i]` consists of lowercase English letters.
+
+---
+
+## Approach
+
+To find the longest uncommon subsequence, we need to consider the following:
+
+1. If a string is unique in the list (i.e., it does not appear more than once), we need to check if it is not a subsequence of any other string. 
+2. If a string is not unique (i.e., it appears more than once), it cannot be an uncommon subsequence.
+3. We can start by checking the strings in descending order of their lengths. This way, we can return the first string that meets the criteria as soon as we find it.
+
+### Steps:
+
+1. Sort the strings by length in descending order.
+2. Check each string:
+   - If the string appears only once in the list.
+   - If it is not a subsequence of any other string longer than it.
+3. If such a string is found, return its length.
+4. If no such string is found, return `-1`.
+
+### Helper Function
+
+A helper function `is_subsequence` can be used to determine if one string is a subsequence of another.
+
+### Solution
+
+#### Java Solution
+
+```java
+import java.util.*;
+
+class Solution {
+    public int findLUSlength(String[] strs) {
+        Arrays.sort(strs, (a, b) -> b.length() - a.length());
+        for (int i = 0; i < strs.length; i++) {
+            boolean isSubsequence = false;
+            for (int j = 0; j < strs.length; j++) {
+                if (i != j && isSubsequence(strs[i], strs[j])) {
+                    isSubsequence = true;
+                    break;
+                }
+            }
+            if (!isSubsequence) {
+                return strs[i].length();
+            }
+        }
+        return -1;
+    }
+
+    private boolean isSubsequence(String a, String b) {
+        int i = 0, j = 0;
+        while (i < a.length() && j < b.length()) {
+            if (a.charAt(i) == b.charAt(j)) {
+                i++;
+            }
+            j++;
+        }
+        return i == a.length();
+    }
+}
+```
+#### C++ Solution
+
+```cpp
+#include <vector>
+#include <string>
+#include <algorithm>
+using namespace std;
+
+class Solution {
+public:
+    int findLUSlength(vector<string>& strs) {
+        sort(strs.begin(), strs.end(), [](const string &a, const string &b) {
+            return b.size() < a.size();
+        });
+
+        for (int i = 0; i < strs.size(); i++) {
+            bool isSubsequence = false;
+            for (int j = 0; j < strs.size(); j++) {
+                if (i != j && isSubsequence(strs[i], strs[j])) {
+                    isSubsequence = true;
+                    break;
+                }
+            }
+            if (!isSubsequence) {
+                return strs[i].size();
+            }
+        }
+        return -1;
+    }
+
+private:
+    bool isSubsequence(const string &a, const string &b) {
+        int i = 0, j = 0;
+        while (i < a.size() && j < b.size()) {
+            if (a[i] == b[j]) {
+                i++;
+            }
+            j++;
+        }
+        return i == a.size();
+    }
+};
+```
+#### Python Solution
+
+```python
+class Solution:
+    def findLUSlength(self, strs: List[str]) -> int:
+        strs.sort(key=len, reverse=True)
+        
+        def is_subsequence(a, b):
+            it = iter(b)
+            return all(c in it for c in a)
+        
+        for i, s in enumerate(strs):
+            if all(not is_subsequence(s, strs[j]) for j in range(len(strs)) if i != j):
+                return len(s)
+        
+        return -1
+```
+### Complexity Analysis
+**Time Complexity:** O(n^2 * l)
+>Reason: Sorting the list takes O(n log n). Checking if a string is a subsequence of another string takes O(l) time, and this is done for each pair of strings, leading to O(n^2 * l) in total.
+
+**Space Complexity:** O(1)
+>Reason: The space complexity is constant as we are not using any extra space proportional to the input size, other than the space required for sorting.
+
+### References
+LeetCode Problem: Longest Uncommon Subsequence II
\ No newline at end of file