AlgorithmFriday_Week4

get_product.py

def get_product(nums=[]):
    """
    Given an integer array nums,
    returns an arry products, such that each entry at position i, in products is a produc of all the other lements in nms except num[i]
    
    Examle: nums=[4,5,10,2]
    Output: [100,80,40,200]
    
    """
    try:
        #check for cases of Null array
        if nums==None:
            return []

        #check for cases of an empty array
        elif len(nums) == 0:
            return []

        #check for cases of one-element array
        elif len(nums) == 1:
            return [1]

        #check for cases of two-element (just swap element)
        elif len(nums) == 2:
            return nums[::-1]

        else:
            arr_prod=1
            for i in nums:
                arr_prod *=i

            start = 0
            end = len(nums)-1
            while(start<=end):
                if (start != end):
                    nums[start] = arr_prod//nums[start]
                    nums[end] = arr_prod//nums[end]
                else:
                    nums[start] = arr_prod//nums[start]
                # move pointer forward and backward
                start += 1
                end -= 1
            return nums
    except:
        return []

Hello @Julisam, thank you for participating in Week 4 of Algorithm Fridays.

This is a really neat solution, one I like very much - from your optimization on lines 20 and 24 to your use of the two-pointer concept. Very clean!

The one test case your solution doesn't pass however, is when one of the entries of the array is 0.

get_product([4, 3, 0]); ❌ // should return [0, 0, 12] but yours returns []

Another not-so-serious thing I'd say is you can combine the conditions on line 12 and 16 into one, if (nums == None) || len(nums) == 0, but apart from that, this is a great attempt!

Thank you for your honest review. I really appreciate it.