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.

Difficulty: introductory

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

def elapsed_seconds(start, end):
    s =  end - start
    return s.total_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()

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

from datetime import datetime, timedelta
def elapsed_seconds(start, end):
    return (end - start).seconds