Complete the function so that it returns the number of seconds that have elapsed between the start and end times given.  

##### Tips:
- The start/end times are given as Date (JS/CoffeeScript), DateTime (C#), Time (Nim), datetime(Python) and Time (Ruby) instances.  
- The start time will always be before the end time.

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def elapsed_seconds(start, end):
	

Solutions:
def elapsed_seconds(start, end):
	return (end - start).total_seconds()


def elapsed_seconds(start, end):
	return (end - start).seconds


import datetime

def elapsed_seconds(start, end):
	a = end - start
	return a.total_seconds()


def elapsed_seconds(start, end):
	start = str(start).replace('-', ' ').replace(':', ' ').split(' ')
	end = str(end).replace('-', ' ').replace(':', ' ').split(' ')
	res = [int(end[i]) - int(start[i]) for i in range(6)]
	return res[2] * 86400 + res[3] * 3600 + res[4] * 60 + res[5]


def elapsed_seconds(s, e):
	ls = str(e - s).split(':')
	return int(ls[0]) * 3600 + int(ls[1]) * 60 + int(ls[2])


import datetime
import time

def elapsed_seconds(start, end):
	start = datetime.datetime.timetuple(start)
	end = datetime.datetime.timetuple(end)
	start = time.mktime(start)
	end = time.mktime(end)
	return int(end - start)


import datetime

def elapsed_seconds(start, end):
	diff = end - start
	return diff.total_seconds()


from datetime import datetime, timedelta

def elapsed_seconds(start, end):
	return (end - start).seconds