Find the 2nd largest integer in array
If the array has no 2nd largest integer then return nil.
Reject all non integers elements and then find the 2nd largest integer in array

find_2nd_largest([1,2,3]) => 2

find_2nd_largest([1,1,1,1,1]) => nil
because all elements are same. Largest no. is 1. and there is no 2nd largest no.

find_2nd_largest([1,'a','2',3,3,4,5,'b']) => 4
as after rejecting non integers array will be [1,3,3,4,5]
Largest no. is 5. and 2nd largest is 4.

Return nil if there is no 2nd largest integer.
Take care of big numbers as well

Difficulty: EASY
Starter code: def find_2nd_largest(arr):
	

Solutions:
def find_2nd_largest(arr):
	arr = sorted((i for i in set(arr) if type(i) == int))
	return arr[-2] if len(arr) > 1 else None


def find_2nd_largest(arr):
	newarr = sorted(list({x for x in arr if type(x) == int}))
	return newarr[-2] if len(newarr) > 1 else None


def find_2nd_largest(arr):
	(a, b) = (float('-inf'), float('-inf'))
	for n in arr:
		if isinstance(n, int):
			if n > b:
				b = n
			if b > a:
				(a, b) = (b, a)
	return None if a == b else b


from heapq import nlargest

def find_2nd_largest(arr):
	two = nlargest(2, {v for v in arr if isinstance(v, int)})
	if len(two) == 2:
		return two[1]


def find_2nd_largest(arr):
	return next(iter(sorted((n for n in set(arr) if type(n) == int))[-2:-1]), None)


def find_2nd_largest(arr):
	res = [arr[i] for i in range(0, len(arr)) if isinstance(arr[i], int)]
	res.sort(reverse=True)
	if len(set(arr)) == 1:
		return None
	else:
		return res[1]


def find_2nd_largest(arr):
	SECOND_lARGEST = -2
	arr_set = set(arr)
	numbers = sorted((number for number in arr_set if type(number) == int))
	return numbers[SECOND_lARGEST] if len(numbers) > 1 else None


def find_2nd_largest(arr):
	from collections import defaultdict
	d = defaultdict(list)
	for elem in arr:
		d[type(elem)].append(elem)
	numbers = d[int]
	largest_num = 0
	second_largest_num = 0
	for number in numbers:
		if largest_num < number:
			largest_num = number
	for number in numbers:
		if second_largest_num < number and number < largest_num:
			second_largest_num = number
	if second_largest_num == 0:
		return None
	else:
		return second_largest_num


def find_2nd_largest(arr):
	xs = sorted({x for x in arr if isinstance(x, int)}, reverse=True)
	if len(xs) > 1 and xs[0] != xs[1]:
		return xs[1]


def filter_int(arr):
	res = []
	for i in arr:
		if type(i) == int:
			res.append(i)
		else:
			continue
	return res

def sort(arr):
	for i in range(len(arr)):
		for j in range(len(arr) - 1):
			if arr[j] > arr[j + 1]:
				(arr[j], arr[j + 1]) = (arr[j + 1], arr[j])
	return arr

def is_diff(arr):
	count = 0
	res = None
	for i in range(len(arr) - 1):
		if arr[i] != arr[i + 1]:
			count += 1
			res = True
			break
		else:
			res = False
	return res

def sec_big(a, b):
	if a > b:
		return b
	else:
		return a

def find_2nd_largest(arr):
	filtered_list = filter_int(arr)
	sorted_list = sort(filtered_list)
	if is_diff(sorted_list) == True:
		a = sorted_list[len(sorted_list) - 1]
		b = sorted_list[len(sorted_list) - 2]
		sec_largest_elem = sec_big(a, b)
		return sec_largest_elem
	else:
		return None