diff --git a/LeetCode/ProductArrayExceptSelf.py b/LeetCode/ProductArrayExceptSelf.py
new file mode 100644
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+++ b/LeetCode/ProductArrayExceptSelf.py
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+"""
+Given an array nums of n integers where n > 1,  return an array output such that output[i] is equal to the product of all the elements 
+of nums except nums[i].
+Example:
+Input:  [1,2,3,4]
+Output: [24,12,8,6]
+Note: Please solve it without division and in O(n).
+Follow up:
+Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity 
+analysis.)
+"""
+
+"""
+Draft Solution (too slow):
+from functools import reduce
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        def prod (l,r):
+            return reduce((lambda a,b: a*b),nums[l:r])
+        
+        if nums is None:
+            return 0
+        
+        ans = nums.copy()
+        
+        for i in range(0,len(nums)):
+            if i == 0:
+                ans[i] = prod(i+1,len(nums))
+            elif i == len(nums)-1:
+                ans[i] = prod(0,i)
+            else:
+                ans[i] = prod(0,i) * prod(i+1,len(nums))
+        return ans
+"""
+
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        if nums is None:
+            return 0
+        
+        temp = 1
+        ans = []
+        
+        #first time going through the array, it's beginning to the end. p keeps a running total of the product, and each element will equal the running total of the products of the elements
+        for i in range (0,len(nums)): #
+            ans.append(temp)
+            temp *= nums[i]
+        
+        temp = 1
+        
+        #the 2nd time going through the array, you're doing the same process, but backwards, finishing off the result by multiplying the elements that came after.
+        for i in range(len(nums)-1,-1,-1):
+            ans[i] *= temp
+            temp *= nums[i]
+        
+        return ans
+
+
+"""
+My Solution:
+Runtime: 120 ms, faster than 32.48% of Python3 online submissions for Product of Array Except Self.
+Memory Usage: 20.8 MB, less than 5.12% of Python3 online submissions for Product of Array Except Self.
+"""
+
+"""
+Fastest Solution:
+class Solution:
+    # @param {integer[]} nums
+    # @return {integer[]}
+    def productExceptSelf(self, nums):
+        p = 1
+        n = len(nums)
+        output = []
+        for i in range(0,n):
+            output.append(p)
+            p = p * nums[i]
+        p = 1
+        for i in range(n-1,-1,-1):
+            output[i] = output[i] * p
+            p = p * nums[i]
+        return output
+"""