Given a natural number N, find the count of numbers from 1 to N that have an odd number of divisors.  
 
Example 1:
Input:
N = 1
Output:
1
Explanation:
1 has only one 
divisor {1}.
 
Example 2:
Input:
N = 4
Output:
2
Explanation:
4 has an odd number 
of divisors {1, 2, 4}.
 
Your Task:
You don't need to read input or print anything. Your task is to complete the function oddNumberOfDivisor() which takes an integer N and returns the count of numbers from 1 to n that have an odd number of divisors.  
 
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(1)
 
Constraints:
1 <= N <= 10^{6}

Source: geeksforgeeks | Difficulty: EASY

Starter code: #User function Template for python3

class Solution:

    def oddNumberOfDivisor (ob,N):

        # code here 

Solutions:
class Solution:

	def oddNumberOfDivisor(ob, N):
		return int(N ** (1 / 2))


class Solution:

	def oddNumberOfDivisor(ob, n):
		c = 0
		from math import sqrt
		for i in range(1, n + 1):
			s = sqrt(i)
			if s - int(s) == 0:
				c += 1
		return c


import math

class Solution:

	def oddNumberOfDivisor(ob, N):
		count = 0
		for i in range(1, N + 1):
			s = math.sqrt(i)
			if s - int(s) == 0:
				count += 1
		return count


import math

class Solution:

	def oddNumberOfDivisor(ob, N):
		return math.floor(N ** 0.5)


class Solution:

	def oddNumberOfDivisor(ob, N):
		return int(N ** 0.5)