Vfrdtyky

def second_largest(numbers):

    first = 0

    second = 0

    for n in numbers:

        if n > first:

            first, second = n, first

        elif first > n > second:

            second = n

    return second or None

print(second_largest([2,2,2,-2]))

When i run this code, output is None, but i need it to be -2 and also i cant use functions as .sorted and others for array. I think that problem is in second = 0 ,but i dont know how to fix it.

Here's a few issues I see.

1. You are instantiating first and second incorrectly (what if the largest number is negative)?


2. The only time you'd want to return None is if your list size is smaller than 2.


3. Change your return condition to return second.

def second_largest(numbers):

    if len(numbers) < 2:

        return None



    first, second = numbers[0], numbers[1]

    if first < second:

        first, second = second, first



    for n in numbers[2:]:

        if n > first:

            first, second = n, first

        elif n > second:

            second = n



    return second

Not sure if this is what you're looking for, but you can basically take the largest element out of the list (or make a note of it) and then search for the second-largest element in what's left. Here, I do this by first using max() (a built-in function of python that works on any iterable object) to get the largest element of the list, then using a list comprehension to create a second list of elements that do not equal that largest element, and finally using max() again to get the second-largest element from the original list.

def second_largest(numbers):

    first = max(numbers)

    second = max([i for i in numbers if i != first])

    return second

You could use a for loop for this if you didn't want to use max() for some reason.

The code you provided does not handle cases where n == first. Changing elif first > n > second to elif n > second should be sufficient (you do not need to test if first >= n since if control can reach that line n > first must be false).

Not use sorted?

values = [2,2,2,-2]

values.sort(reverse=True)   # technically correct

second_largest = values[1]

or, less facetious

values = set([2,2,2,-2])

values.remove(max(values))

second_largest = max(values)

Or even

import heapq

heapq.nlargest(2, [2,2,2,-2])