Algo2_RemoveInstance.py

def remove_instance(nums, val):
    """
    Given an array nums, and a value val, returns the new length of the array with the value removed
    i.e. the number of items in nums with val
    Input: nums=[5, 2, 2, 5, 3]  and   val = 5
    Output: 3
    """
    try:
        #check for cases of an empty array
        if len(nums) == 0:
            return 0
        else:
            count=0
            for i in nums:
                if i != val:
                    count +=1
            return count
    except:
        return 0

Hello @Julisam, thank you for participating in Week 2 of Algorithm Fridays.

This is a decent solution; I like that you handled the case if the array is empty. However, your solution assumes that nums will always be an array because if I pass None as input to your function, it would break on line 10. Ideally you want to write robust code that takes care of invalid input.

Also, your solution correctly centers around looping through the array to check for elements whose values are not equal to val. Do you think there's a faster way to loop through the array?

I've posted my solution here. Do let me know what you think.