Update 238_product_of_array_except_self.md

Author: Joseph Luce

leetcode/medium/238_product_of_array_except_self.md

@@ -35,3 +35,33 @@ class Solution:
             curr_prod *= n
         return result
 ```
+
+## Best Solution (Constant Space)
+- Runtime: O(N)
+- Space: O(1)
+- N = Number of elements in array
+
+Similar to the solution above, instead of creating two arrays, we can first create an array for the products to the right of each number.
+This array can then be used as the final result, we can then calculate the left products as we modify the first array from left to right.
+This is constant space, assuming the returned result array is not considered extra space.
+
+```
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        if len(nums) == 0:
+            return []
+        result = self.calc_left_product(nums[::-1])[::-1] # right products
+        curr_prod = 1
+        for index, n in enumerate(nums):
+            result[index] *= curr_prod
+            curr_prod *= n
+        return result
+        
+    def calc_left_product(self, nums):
+        result = [1]*len(nums)
+        curr_prod = nums[0]
+        for index, n in enumerate(nums[1:], 1):
+            result[index] = curr_prod
+            curr_prod *= n
+        return result
+```