Product of array except self#

 1class Solution(object):
 2    def productExceptSelf(self, nums):
 3        """
 4        :type nums: List[int]
 5        :rtype: List[int]
 6        """
 7        out = [1 for idx in range(len(nums))]
 8        for idx, _ in enumerate(nums):
 9            outp = 1
10            for jdx, n in enumerate(nums):
11                if idx != jdx:
12                    outp *= n
13            out[idx] = outp
14        return out
15
16
17class Solution2(object):
18    def productExceptSelf(self, nums):
19
20        # The length of the input array
21        length = len(nums)
22
23        # The left and right arrays as described in the algorithm
24        L, R, answer = [0] * length, [0] * length, [0] * length
25
26        # L[i] contains the product of all the elements to the left
27        # Note: for the element at index '0', there are no elements to the left,
28        # so the L[0] would be 1
29        L[0] = 1
30        for i in range(1, length):
31
32            # L[i - 1] already contains the product of elements to the left of 'i - 1'
33            # Simply multiplying it with nums[i - 1] would give the product of all
34            # elements to the left of index 'i'
35            L[i] = nums[i - 1] * L[i - 1]
36
37        # R[i] contains the product of all the elements to the right
38        # Note: for the element at index 'length - 1', there are no elements to the right,
39        # so the R[length - 1] would be 1
40        R[length - 1] = 1
41        for i in reversed(range(length - 1)):
42
43            # R[i + 1] already contains the product of elements to the right of 'i + 1'
44            # Simply multiplying it with nums[i + 1] would give the product of all
45            # elements to the right of index 'i'
46            R[i] = nums[i + 1] * R[i + 1]
47
48        # Constructing the answer array
49        for i in range(length):
50            # For the first element, R[i] would be product except self
51            # For the last element of the array, product except self would be L[i]
52            # Else, multiple product of all elements to the left and to the right
53            answer[i] = L[i] * R[i]
54
55        return answer
56
57
58in_arrs = [
59    [2, 1, 2, 1, 0, 1, 2],
60    [3, 3, 5, 0, 0, 3, 1, 4],
61    [3, 5, 0, 1, 4],
62    [1, 2, 3, 1],
63    [1, 2, 3, 4],
64    [1, 1, 1, 3, 3, 4, 3, 2, 4, 2],
65]
66if __name__ == "__main__":
67
68    sol = Solution2()
69    for nin in in_arrs:
70        r = sol.productExceptSelf(nin)
71        print(r)