From ea19d09d97bbd1854d2f715ecc9584f41871d0df Mon Sep 17 00:00:00 2001
From: duncanmichel <25329325+duncanmichel@users.noreply.github.com>
Date: Thu, 28 Feb 2019 12:20:41 -0500
Subject: [PATCH 1/2] Create ProductArrayExceptSelf.py

---
 LeetCode/ProductArrayExceptSelf.py | 46 ++++++++++++++++++++++++++++++
 1 file changed, 46 insertions(+)
 create mode 100644 LeetCode/ProductArrayExceptSelf.py

diff --git a/LeetCode/ProductArrayExceptSelf.py b/LeetCode/ProductArrayExceptSelf.py
new file mode 100644
index 0000000..ad341eb
--- /dev/null
+++ b/LeetCode/ProductArrayExceptSelf.py
@@ -0,0 +1,46 @@
+"""
+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
+"""
+
+
+
+"""
+My Solution:
+
+"""
+
+"""
+Fastest Solution:
+
+"""

From bf15a36cc6e9483ac140c20ed5b455bccc44605a Mon Sep 17 00:00:00 2001
From: duncanmichel <25329325+duncanmichel@users.noreply.github.com>
Date: Thu, 28 Feb 2019 16:14:13 -0500
Subject: [PATCH 2/2] Update ProductArrayExceptSelf.py

---
 LeetCode/ProductArrayExceptSelf.py | 40 ++++++++++++++++++++++++++++--
 1 file changed, 38 insertions(+), 2 deletions(-)

diff --git a/LeetCode/ProductArrayExceptSelf.py b/LeetCode/ProductArrayExceptSelf.py
index ad341eb..8b016b5 100644
--- a/LeetCode/ProductArrayExceptSelf.py
+++ b/LeetCode/ProductArrayExceptSelf.py
@@ -33,14 +33,50 @@ def prod (l,r):
         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
 """