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+# Product of Array Except Self (Medium)
+
+## Table of Contents
+
+- [Problem Statement](#problem-statement)
+- [Examples](#examples)
+- [Constraints](#constraints)
+- [Solutions](#solutions)
+  - [Optimal Solution](#optimal-solution)
+  - [Alternative Solution](#alternative-solution)
+- [Code Explanation](#code-explanation)
+- [Complexity Analysis](#complexity-analysis)
+- [Additional Resources](#additional-resources)
+
+## Problem Statement
+
+[Product of Array Except Self](https://leetcode.com/problems/product-of-array-except-self/description/)
+
+Given an integer array `nums`, return an array `answer` such that `answer[i]` is equal to the product of all the elements of `nums` except `nums[i]`.
+
+The product of any prefix or suffix of `nums` is guaranteed to fit in a 32-bit integer.
+
+You must write an algorithm that runs in O(n) time and without using the division operation.
+
+## Examples
+
+### Example 1:
+
+```
+Input: nums = [1,2,3,4]
+Output: [24,12,8,6]
+```
+
+### Example 2:
+
+```
+Input: nums = [-1,1,0,-3,3]
+Output: [0,0,9,0,0]
+```
+
+## Constraints
+
+- 2 <= nums.length <= 10^5
+- -30 <= nums[i] <= 30
+- The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.
+
+## Solutions
+
+### Optimal Solution
+
+```python
+class Solution(object):
+    def productExceptSelf(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[int]
+        """
+        answers = []
+        leftptr = 1
+        for i in range(len(nums)):
+            answers.append(leftptr)
+            leftptr *= nums[i]
+        
+        rightptr = 1
+        for i in range(len(nums) - 1, -1, -1):
+            answers[i] *= rightptr
+            rightptr *= nums[i]
+        return answers
+```
+
+### Alternative Solution
+
+```python
+class Solution(object):
+    def productExceptSelf(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[int]
+        """
+        bef = [1] * len(nums)
+        aft = [1] * len(nums)
+        for i in range(1, len(nums)):
+            bef[i] = bef[i - 1] * nums[i - 1]
+
+        for i in range(len(nums) - 2, -1, -1):
+            aft[i] = aft[i + 1] * nums[i + 1]
+        return [bef[i] * aft[i] for i in range(len(nums))]
+```
+
+## Code Explanation
+
+### Optimal Solution
+
+The optimal solution uses a two-pass approach to calculate the product of all elements except self:
+
+1. First pass (left to right):
+   - Initialize an `answers` list to store the results.
+   - Use `leftptr` to keep track of the cumulative product from the left.
+   - For each element, append the current `leftptr` to `answers` and update `leftptr` by multiplying it with the current element.
+
+2. Second pass (right to left):
+   - Use `rightptr` to keep track of the cumulative product from the right.
+   - Iterate from right to left, multiplying each element in `answers` by `rightptr` and updating `rightptr`.
+
+This approach calculates the product of all elements to the left and right of each element without using division.
+
+### Alternative Solution
+
+The alternative solution uses two separate arrays to store the products:
+
+1. `bef` array: Stores the product of all elements to the left of each element.
+2. `aft` array: Stores the product of all elements to the right of each element.
+
+The final result is obtained by multiplying corresponding elements from `bef` and `aft`.
+
+## Complexity Analysis
+
+For the optimal solution:
+- Time Complexity: O(N), where N is the length of the input array.
+- Space Complexity: O(1), excluding the output array.
+
+## Additional Resources
+
+- [YouTube Explanation](https://youtu.be/yKZFurr4GQA?si=-wykJZfdRSw7M8UN)
+- [GitHub Solution](https://github.com/gahogg/Leetcode-Solutions/tree/main/Product%20of%20Array%20Except%20Self%20-%20Leetcode%20238)
+- [LeetCode Solution Explanation](https://leetcode.com/problems/product-of-array-except-self/solutions/5757573/solution)
+- [Personal Submission Details](https://leetcode.com/submissions/detail/1383555855/)
+
+> [!NOTE]
+> This problem is part of a larger collection following the roadmap on [algomap.io](https://algomap.io/). For more details and related problems, please refer to the AlgoMap website.