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+# Two Sum
+
+## Description
+
+Given an array of integers `nums` and an integer `target`, return the indices of two numbers in the array such that they add up to the `target`.
+
+You may assume that each input would have exactly one solution, and you may not use the same element twice.
+
+You can return the answer in any order.
+
+Link to problem: [Two Sum](https://leetcode.com/problems/two-sum/)
+
+## Example
+
+Example 1:
+Input: `nums = [2,7,11,15]`, `target = 9`
+Output: `[0,1]`
+Explanation: Because `nums[0] + nums[1]` equals 9, we return [0, 1].
+
+Example 2:
+Input: `nums = [3,2,4]`, `target = 6`
+Output: `[1,2]`
+
+Example 3:
+Input: `nums = [3,3]`, `target = 6`
+Output: `[0,1]`
+
+## Approach
+
+The approach for solving this problem can be as follows:
+1. For each number `x` in the array, calculate the complementary number `y` such that `y = target - x`.
+2. Check if `y` exists in the hash table. If it does, return the indices of `x` and `y`.
+3. If `y` is not found in the hash table, store `x` in the hash table with its index as the value.
+4. If no two numbers are found that add up to the target, return `[0,0]`.
+
+## Complexity
+
+- Time complexity: `O(n)`
+- Space complexity: `O(n)`
+
+## Code
+
+```cpp
+class Solution {
+public:
+    vector<int> twoSum(vector<int>& nums, int target) {
+        unordered_map<int, int> mp;
+        for(int i = 0; i < nums.size(); i++) {
+            if(mp.find(target - nums[i]) != mp.end())
+                return {i, mp[target - nums[i]]};
+            mp[nums[i]] = i;
+        }
+        return {0, 0};
+    }
+};