diff --git a/0238_product-of-array-except-self/NOTES.md b/0238_product-of-array-except-self/NOTES.md
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+## What if we were allowed to use division?
+
+Well, one might think that you can then find product(nums) and divide by nums[i] in the loop. But the problem with this approach is when there are 0s in the array. That causes the entire product to become zero.
+
+So, we probably need to keep track of zero count and the first zero index.
+
+1. If there are >1 zeroes, then the result would be all zeroes so.
+2. If there's only 1 zero, then we need to keep product excluding the zero to be kept. Then fill the result with all zeroes except for the 0 element's index which would equal the product.
diff --git a/0238_product-of-array-except-self/README.md b/0238_product-of-array-except-self/README.md
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+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.
+
+**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 <= 105`
+*   `-30 <= nums[i] <= 30`
+*   The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.
+
+**Follow up:** Can you solve the problem in `O(1)` extra space complexity? (The output array **does not** count as extra space for space complexity analysis.)
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diff --git a/0238_product-of-array-except-self/python3/solution.py b/0238_product-of-array-except-self/python3/solution.py
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+from functools import reduce
+
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        # We can divide this problem by finding the prefix and
+        # postfix products of a given element and then multiplying
+        # them together
+        result = [0] * len(nums)
+        
+        # Find prefix products, i.e, for ith element, store
+        # product of 0 to a[i - 1]
+        prefix = 1
+        for i, num in enumerate(nums):
+            result[i] = prefix
+            prefix *= num
+        
+        # Find suffix products by iterating from the last, and
+        # combine them with prefix in the same step to reduce cost
+        suffix = 1
+        for i in range(len(nums) - 1, -1, -1):
+            result[i] *= suffix
+            suffix *= nums[i]
+        
+        return result
+            
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