Given a number **n**, return the number of positive odd numbers below **n**, EASY!

Expect large Inputs!

Difficulty: introductory

Starter code:

def odd_count(n):
	

Solutions:
def odd_count(n):
    return len(range(1, n, 2))

def odd_count(n):
    if n < 1:
        return 0

    return int(n / 2)
    


def odd_count(n):
    #Return number of positive odd numbers below N
    #Find numbers below N -> Add +1 to count 
    #More of a math problem
    #Try to find a way besides checking every individual integer.
    
    return(n // 2)
    
    """
    Look at the top of the range: if it is odd then add 1, if even leave alone.

Look at the bottom of the range: if it is odd then subtract 1, if even leave alone.

Take the difference between the new even top and bottom; then divide by two.

So for the range [1,100] you get 100−02=50 odd integers.
"""

import math
def odd_count(n):
    return math.floor(n / 2)

def odd_count(n): return n>>1

odd_count=lambda n:len(range(1,n,2))

def odd_count(n):
    return int(n>>1)

def odd_count(n):
    print(n)
    return (n-1)/2 if not n % 2 == 0 else n/2

odd_count = (2).__rfloordiv__

def odd_count(n):
    count = 0
    if n % 2 == 0:
        count = n - (n / 2)
    else:
        count = ((n + 1) - (n + 1)/2) - 1
    return count

def odd_count(n):
    res=len(list(range(n-1)))/2
    if res.is_integer():
        return res 
    else:
        return res + 0.5
    
    


import math
def odd_count(n):
    if n % 2 != 0:
        return (n-1)/2
    else:
        return math.ceil((n-1)/2)


def odd_count(n):
    return ((n-1)/2,n/2)[n%2==0]

def odd_count(n):
    result = 0
    temp = n%2
    if temp==1:
        return n/2-.5
    return n/2


def odd_count(n):
    if n % 2 == 1:
        return n//2
    else:
        return n/2

def odd_count(n: int) -> int:
    """This functions returns the number of positive odd numbers below 'n'."""
    if n <= 0:
        return 0
    return n // 2

def odd_count(n):
    return (n - 1) // 2 + (n % 2 == 0)

def odd_count(n):
    return n.__divmod__(2)[(1).__invert__()]

import math
def odd_count(n):
#     c = 0
#     for i in range(n):
#         if i%2!=0:
#             c += 1
#   return c
    return math.floor(n/2)

def odd_count(n):
    if n < 1:
        return 0
    else:
        return n // 2

import math
def odd_count(n):     
    if (n<=0):
        return 0
    return math.ceil((n-1)/2)

import math as m
def odd_count(n):
    pass
#     count
#     for i in range(n):
    return int(n/2)


import math
def odd_count(n):
    return math.ceil((n-1)/2)#len([i for i in range(n) if (i % 2) == 1])

def odd_count(n):
   # if n%2 != 0: 
     #   return range(0,n, 2
    return len((range(1, n, 2)))

def odd_count(n):
    #return len([x for x in range(1,n,2)])
    return n//2 

odd_count = lambda n: ((n-1)//2)+1 if (n-1)%2!=0 else (n-1)//2

def odd_count(n):
#     list = []
#     for i in range(n):
#         if i % 2 != 0:
#             list.append(i)
#     return len(list)
    return n // 2
            


def odd_count(n):
    return round(int(n/2),0)

def odd_count(n):
    return (n-1)//2 if n%2==1 else  (n-1)//2+1

'''Large inputs are given because if you use a list comprehension iteration approach'''
#then your asymptomatic runtime will exceed the server threshold to stop the request
#Instead, treat this as a math problem to allow it to maintain 0(1)
def odd_count(n):
    return (n-1)/2 if n%2!=0 else n/2

def odd_count(n):
    return 1 if n == 1 else int(n / 2)

def odd_count(n):
    return len(list(range(1, n, 2)))
    # return the length of the number range with odd numbers


def odd_count(n):
    return (n - 1 )/ 2 if n % 2 == 1 else n / 2
    # it returns half of the amount of occurences from 1 to n. The first part is for odd numbers and the second for even numbers.


def odd_count(n):
    print(n)
    if n % 2:
        return ((n+1)/2)-1
    return n/2

def odd_count(n):
    return int(round((n/2), 1))


def odd_count(n):
    if n%2==0:
        return n*0.5
    else:
        return (n-1)*0.5

def odd_count(n):
    diff = n/2
    return int(diff)


def odd_count(n: int) -> int:
    """Return the number of positive odd numbers below n."""
    
    if n % 2 == 0:
        odd_count = n / 2
    else:
        odd_count = n // 2
    
    return odd_count

def odd_count(n):
#     odds = []
#     for i in range(1, n, 2):
#         odds.append(i)
            
#     return len(odds)
    return n // 2

def odd_count(n):
    if n % 2 == 0:
        return  round(n/2) 
    return  round(n-1)/2

def odd_count(z):
    return z//2

def odd_count(n):
    return int((n - 1) / 2) if n % 2 != 0 else int((n - 1) / 2 + 1)



def odd_count(n):
    n = round(n)
    if not n%2:
        return n / 2
    else:
        return (n - 1) / 2

def odd_count(n):
    c = range(1, n, 2)
    for num in c:
        return len(c)

def odd_count(n):
    nums = n//2
    return nums

def odd_count(n):
    c = 0
    if n % 2 == 0:
        c = n / 2
    else :
        c = (n / 2) - 0.5
    return (c)

import math
def odd_count(n):
    return int(math.floor(n/2))

def odd_count(n):
    return n//2
    count_odd= len(count_odd)
    return(count_odd)

def odd_count(n):
    return n // 2
    
    # did not work
    #count = 0
    #for x in range(n):
        #if x % 2 != 0:
            #count = count + 1
    #return count        


def odd_count(n):
    print(n)
    return (n-1)/2 if n % 2 != 0 else (n)/2


import math
def odd_count(n):
    n /= 2
    return math.floor(n)

def odd_count(n):
    return n // 2
    oddlist = []
    for eachnumber in range(n-1,0,-1):
        if eachnumber % 2 == 0:
            continue
        else:
            oddlist.append(eachnumber)
    return len(oddlist)

def odd_count(n):
    l = list(range(1,n))
    e = l[0:n:2]
    return len(e)



def odd_count(n):
    # This work on test but takes too much time. 
    #counter = 0
    #for num in range(0, n): 
    #     if num % 2 != 0: 
    #        counter += 1
    return len(range(1, n,2))

from math import ceil, floor
def odd_count(n):
    return ceil(n/2) if n % 2 == 0 else floor(n/2)

def odd_count(n):
    return n/2-0.5 if n%2 else n/2

def odd_count(n):
    return (n - 1 if n % 2 else n) // 2

def odd_count(n):
#     d = 0
#     for i in range(n):
#         if i %2 != 0:
#             d+=1
#         else:
#             continue
            
    return n//2

def odd_count(n):
    even_list = range(0, n + 1)[0%2::2] 
    return len(even_list) - 1

def odd_count(n):
    if n % 2 > 0:
        return (n - 1) / 2
    if n % 2 == 0:
        return n / 2

def odd_count(n):
    if n % 2 == 1:
        count = (n-1)/2
    else:
        count = n/2
    return count


def odd_count(n):
    return n//2
    #return len(list)
    #pass


def odd_count(n):
    q = n // 2
    return q

def odd_count(n):
#     lst = []
#     count = 0
#     for x in range(1,n):
#         if x % 2 == 1 and x < n:
#             lst.append(x)
#             count += 1
#     return count
  #  count = [x for x in range(1,n) if x%2==1]
    return n // 2

def odd_count(n):
    return int(n / 2) if n % 2 == 0 else int((n - 1) / 2)

odd_count = lambda n: (n-1)//2 if n%2 else n//2

def odd_count(n):
    return (n-1)/2 if n % 2 != 0 else round((n-1)/2+0.5)

def odd_count(n):
    if n%2 == 0:
        return (round(n/2))
    if n%2 != 0:
        return ((n-1)/2)  


def odd_count(n):
   return (n+1/2)//2

def odd_count(n):
    f = n//2
    return f

def odd_count(n):
    return int(n - (n/2 + 1) + 1)

def odd_count(n):
#     return len([i for i in range(n) if i%2==1])
    return n // 2

def odd_count(n):
    if n % 2 == 0: # Odd number
        return n//2
    else:
        return (n-1)//2

def odd_count(n):
    new = n / 2
    return int(new)

import math as MATHEMATICA
def odd_count(n):
    return MATHEMATICA.ceil(n/2)-1 if n%2 != 0 else MATHEMATICA.ceil(n/2)

def odd_count(n):
    if n%2==0:
        return n/2
    else:
        return (n+1)/2 -1

import unittest


def odd_count(n):
    return n // 2


class TestOddCount(unittest.TestCase):
    def test_should_return_7_when_given_n_is_15(self):
        n = 15
        actual = odd_count(n)
        self.assertEqual(actual, 7)

    def test_should_return_7511_when_given_n_is_15023(self):
        n = 15023
        actual = odd_count(n)
        self.assertEqual(actual, 7511)


def odd_count(n):
    if n % 2 == 1:   #odd
        return (n - 1) / 2
    else:            #even
        return n / 2

def odd_count(n):
    return len(range(0,n-1,2))

import math
def odd_count(n):
    return math.floor(n/2+1)-1

def odd_count(n):
    return round((n-1)/2) if n % 2 == 1 else n/2

def odd_count(n):
#     count = 0
#     for i in range(1,n):
#         if  i % 2 != 0:
#             count +=1
            
    return int(n/2)      

def odd_count(n):
    return n//2 #len(list(range(1,n))[::2])

def odd_count(n):
    oddListLen = len(range(1,n,2))
    return oddListLen

def odd_count(n):
    return n//2 if n%2!=0 else n/2



def odd_count(n):
    return len(range(n-1)[::2])

def odd_count(n):
  g=0
  if n%2!=0:
      g=(n-1)/2
      g+1
  else:
       g=n/2
  return g

def odd_count(a):
  return int(a/2)   

def odd_count(n):
    return (n/2) - .5 if n %2 != 0 else n/2

def odd_count(n):
    return 0.5*(n-1) if n%2 else n/2

def odd_count(n):
    if not n % 2:
        return int(n / 2)
    else:
        return(int((n-1) / 2))
        


def odd_count(n):
    if n % 2 != 0:
        return n / 2 -.5
    else:
        return n / 2

def odd_count(n):
    '''count = 0
    for x in range(n):
        if(x % 2 != 0):
            count += 1'''
    return len(list(range(1, n, 2)))


def odd_count(n):
    return round((n-1)/2,0) if n%2>0 else n/2