Given a string. Count the number of Camel Case characters in it.
Example 1:
Input:
S = "ckjkUUYII"
Output: 5
Explanation: Camel Case characters present:
U, U, Y, I and I.
Ã¢â¬â¹Example 2:
Input: 
S = "abcd"
Output: 0
Explanation: No Camel Case character
present.
Your Task:
You don't need to read input or print anything. Your task is to complete the function countCamelCase() which takes the string S as input and returns the count of the camel case characters in the string.
Expected Time Complexity: O(|S|).
Expected Auxiliary Space: O(1).
Constraints:
1<=|S|<=10^{5}

Difficulty: EASY
Tags: String algorithms

Data structures
Skill types: Data structures
Expected Time Complexity: O(|S|).
Expected Auxiliary Space: O(1).

Starter code:
#User function Template for python3

class Solution:
    def countCamelCase (self,s):
        # your code here

Solutions:
class Solution:

	def countCamelCase(self, s):
		a = 0
		for i in s:
			if i.isupper():
				a = a + 1
		return a


class Solution:

	def countCamelCase(self, s):
		count = 0
		for char in s:
			if char.isupper():
				count = count + 1
		return count


class Solution:

	def countCamelCase(self, s):
		c = 0
		for i in s:
			if i >= 'A' and i <= 'Z':
				c += 1
		return c


class Solution:

	def countCamelCase(self, s):
		answer = 0
		for i in range(len(s)):
			if s[i].isupper():
				answer += 1
		return answer


class Solution:

	def countCamelCase(self, s):
		count = 0
		for i in s:
			if ord(i) >= 65 and ord(i) <= 90:
				count += 1
		return count


class Solution:

	def countCamelCase(self, s):
		a = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
		c = 0
		for i in range(len(s)):
			if s[i] in a:
				c += 1
			else:
				continue
		return c


class Solution:

	def countCamelCase(self, s):
		c = 0
		for i in range(len(s)):
			if ord(s[i]) >= 65 and ord(s[i]) <= 91:
				c += 1
		return c


class Solution:

	def countCamelCase(self, s):
		total = 0
		for e in s:
			ascii_no = ord(e)
			if ascii_no > 64 and ascii_no < 91:
				total = total + 1
		return total


class Solution:

	def countCamelCase(self, s):
		s_str_list = list(str(s))
		count = 0
		for i in s_str_list:
			if i.isupper():
				count += 1
		return count


class Solution:

	def countCamelCase(self, s):
		count = 0
		for i in range(len(s)):
			if 'A' <= s[i] <= 'Z':
				count += 1
		return count


class Solution:

	def countCamelCase(self, s):
		c = 0
		for i in s:
			if i == i.upper():
				c += 1
		return c


class Solution:

	def countCamelCase(self, s):
		result = [i for i in s if i.isupper()]
		return len(result)


class Solution:

	def countCamelCase(self, s):
		l = len(s)
		ctr = 0
		for i in range(0, l):
			if ord(s[i]) >= 65 and ord(s[i]) <= 90:
				ctr += 1
		return ctr


class Solution:

	def countCamelCase(self, s):
		c = 0
		for i in s:
			c += i.isupper()
		return c


class Solution:

	def countCamelCase(self, s):
		return sum([1 for i in s if i.isupper()])


class Solution:

	def countCamelCase(self, s):
		count = 0
		for i in s:
			if ord(i) >= ord('A') and ord(i) <= ord('Z'):
				count += 1
		return count


class Solution:

	def countCamelCase(self, s):
		(n, count) = (len(s), 0)
		for i in range(n):
			if s[i].isupper():
				count += 1
		return count


class Solution:

	def countCamelCase(self, s):
		return len([x for x in s if x.isupper()])


class Solution:

	def countCamelCase(self, s):
		c = 0
		for i in s:
			if i.isupper():
				c += 1
			else:
				continue
		return c


class Solution:

	def countCamelCase(self, s):
		q = ['Q', 'W', 'E', 'R', 'T', 'Y', 'U', 'I', 'O', 'P', 'L', 'K', 'J', 'H', 'G', 'F', 'D', 'S', 'A', 'Z', 'X', 'C', 'V', 'B', 'N', 'M']
		count = 0
		for i in s:
			if i == i.upper():
				count = count + 1
		return count


class Solution:

	def countCamelCase(self, s):
		lst = list(s)
		c = 0
		for i in lst:
			if i.isupper():
				c += 1
		return c


class Solution:

	def countCamelCase(self, s):
		count = 0
		for i in s:
			if i != i.lower():
				count += 1
		return count


class Solution:

	def countCamelCase(self, s):
		count = 0
		import string
		alphabet = list(string.ascii_uppercase)
		for i in s:
			if i.islower():
				pass
			else:
				count += 1
		return count


class Solution:

	def countCamelCase(self, s):
		c = 0
		a = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
		for i in s:
			if i in a:
				c += 1
		return c


class Solution:

	def countCamelCase(self, s):
		count = sum((1 for c in s if c.isupper()))
		return count


class Solution:

	def countCamelCase(self, s):
		c = 0
		for i in s:
			if i.isupper() == True:
				c = c + 1
		return c


import string

class Solution:

	def countCamelCase(self, s):
		c = 0
		for i in s:
			if i in string.ascii_uppercase:
				c = c + 1
		return c


class Solution:

	def countCamelCase(self, s):
		count = 0
		camel_case = 'ASDFGHJKLZXCVBNMQWERTYUIOP'
		for i in s:
			if i in camel_case:
				count += 1
		return count


class Solution:

	def countCamelCase(self, s):
		count = 0
		l = len(s)
		c1 = 0
		for i in s:
			if i.isupper():
				count += 1
			else:
				c1 += 1
		return count


class Solution:

	def countCamelCase(self, s):
		count = 0
		for i in range(len(s)):
			if s[i] in 'ABCDEFGHIJKLMNOPQRSTUVWXYZ':
				count += 1
		return count


class Solution:

	def countCamelCase(self, S):
		count = 0
		for i in S:
			if i == i.upper():
				count = count + 1
		return count


class Solution:

	def countCamelCase(self, s):
		sl = []
		for i in s:
			if i.upper() == i:
				sl.append(i)
		return len(sl)


class Solution:

	def countCamelCase(self, s):
		count = 0
		for elem in s:
			if 65 <= ord(elem) <= 90:
				count += 1
		return count