diff --git a/Arrays & Strings/Duplicates checking cpp b/Arrays & Strings/Duplicates checking cpp
new file mode 100644
index 0000000..bce13b2
--- /dev/null
+++ b/Arrays & Strings/Duplicates checking cpp	
@@ -0,0 +1,13 @@
+class Solution {
+public:
+    bool containsDuplicate(vector<int>& nums) {
+        sort(nums.begin(), nums.end());  // O(n log n)
+        for (int i = 1; i < nums.size(); ++i) {
+            if (nums[i] == nums[i - 1]) {
+                return true;
+            }
+        }
+        return false;
+    }
+};
+
diff --git a/Problems/#14 - Longest Common Prefix - Easy/explanation.md b/Problems/#14 - Longest Common Prefix - Easy/explanation.md
new file mode 100644
index 0000000..cd3cf33
--- /dev/null
+++ b/Problems/#14 - Longest Common Prefix - Easy/explanation.md	
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+# Leetcode Problem #14: Longest Common Prefix (Easy)
+
+## Problem 
+
+Given an array of strings `strs`, return the **longest common prefix** string amongst them.  
+If there is no common prefix, return an **empty string**.
+
+---
+
+## Approach
+
+- Start with the first string as the **initial prefix**.
+- Compare it with the next string:
+  - If the current string doesn't start with the prefix, **trim the prefix** by one character from the end.
+  - Repeat until it matches or becomes an empty string.
+- Continue this for all strings in the list.
+
+---
+
+
+
+
diff --git a/Problems/#14 - Longest Common Prefix - Easy/solution.cpp b/Problems/#14 - Longest Common Prefix - Easy/solution.cpp
new file mode 100644
index 0000000..a6cff60
--- /dev/null
+++ b/Problems/#14 - Longest Common Prefix - Easy/solution.cpp	
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+// LeetCode #14 - Longest Common Prefix
+// Language: C++
+
+#include <string>
+#include <vector>
+
+using namespace std;
+
+class Solution {
+public:
+    string longestCommonPrefix(vector<string>& strs) {
+        if (strs.empty()) return "";
+
+        string prefix = strs[0];
+
+        for (int i = 1; i < strs.size(); ++i) {
+            int j = 0;
+            while (j < prefix.size() && j < strs[i].size() && prefix[j] == strs[i][j])
+                ++j;
+            prefix = prefix.substr(0, j);
+            if (prefix.empty()) return "";
+        }
+
+        return prefix;
+    }
+};
+