Given a list of numbers, write a Python program to count Even and Odd numbers in a List.
Example:
Input: list1 = [2, 7, 5, 64, 14]
Output: Even = 3, odd = 2
Input: list2 = [12, 14, 95, 3]
Output: Even = 2, odd = 2
Example 1:
Count Even and Odd numbers from the given list using for loop Iterate each element in the list using for loop and check if num % 2 == 0, the condition to check even numbers. If the condition satisfies, then increase the even count else increase odd count.
Python3
list1 = [10, 21, 4, 45, 66, 93, 1]
even_count, odd_count = 0, 0
for num in list1:
if num % 2 == 0:
even_count += 1
else:
odd_count += 1
print("Even numbers in the list: ", even_count)
print("Odd numbers in the list: ", odd_count)
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Output:
Even numbers in the list: 3
Odd numbers in the list: 4
Time Complexity: O(N), Here N is the number of elements in the list.
Auxiliary Space: O(1), As constant extra space is used.
Example 2: Using while loop
Python3
list1 = [10, 21, 4, 45, 66, 93, 11]
even_count, odd_count = 0, 0
num = 0
while(num < len(list1)):
if list1[num] % 2 == 0:
even_count += 1
else:
odd_count += 1
num += 1
print("Even numbers in the list: ", even_count)
print("Odd numbers in the list: ", odd_count)
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Output:
Even numbers in the list: 3
Odd numbers in the list: 4
Time Complexity: O(N), Here N is the number of elements in the list.
Auxiliary Space: O(1), As constant extra space is used.
Example 3: Using Python Lambda Expressions
Python3
list1 = [10, 21, 4, 45, 66, 93, 11]
odd_count = len(list(filter(lambda x: (x%2 != 0) , list1)))
even_count = len(list(filter(lambda x: (x%2 == 0) , list1)))
print("Even numbers in the list: ", even_count)
print("Odd numbers in the list: ", odd_count)
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Output:
Even numbers in the list: 3
Odd numbers in the list: 4
Time Complexity: O(n)
Space Complexity: O(1)
Example 4: Using List Comprehension
Python3
list1 = [10, 21, 4, 45, 66, 93, 11]
only_odd = [num for num in list1 if num % 2 == 1]
odd_count = len(only_odd)
print("Even numbers in the list: ", len(list1) - odd_count)
print("Odd numbers in the list: ", odd_count)
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Output:
Even numbers in the list: 3
Odd numbers in the list: 4
Time Complexity: O(N), Here N is the number of elements in the list.
Auxiliary Space: O(1), As constant extra space is used.
Example 5: Using Recursion
Python3
even_count = 0
i = 0
odd_count = 0
def evenoddcount(lst):
global even_count
global odd_count
global i
if lst[i] % 2 == 0:
even_count += 1
else:
odd_count += 1
if i in range(len(lst)-1):
i += 1
evenoddcount(lst)
else:
print("Even numbers in the list: ", even_count)
print("Odd numbers in the list: ", odd_count)
list1 = [10, 21, 4, 45, 66, 93, 1]
evenoddcount(list1)
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Output
Even numbers in the list: 3
Odd numbers in the list: 4
Time Complexity: O(N), Here N is the number of elements in the list.
Auxiliary Space: O(1), As constant extra space is used.
Example 6: Using Bitwise XOR operator
The idea is to check whether the last bit of the number is set or not. If the last bit is set then the number is odd, otherwise even. As we know bitwise XOR Operation of the Number by 1 increments the value of the number by 1 if the number is even otherwise it decrements the value of the number by 1 if the value is odd.
CHECK IF NUMBER IS EVEN OR ODD USING XOR OPERATOR
Python3
list1 = [10, 21, 4, 45, 66, 93, 1]
even_count, odd_count = 0, 0
for num in list1:
if num ^ 1 == num + 1:
even_count += 1
else:
odd_count += 1
print("Even numbers in the list: ", even_count)
print("Odd numbers in the list: ", odd_count)
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Output
Even numbers in the list: 3
Odd numbers in the list: 4
Time Complexity: O(N), Here N is the number of elements in the list.
Auxiliary Space: O(1), As constant extra space is used.
Example 7: Using Bitwise AND operator
The idea is to check whether the last bit of the number is set or not. If the last bit is set then the number is odd, otherwise even.
As we know bitwise AND Operation of the Number by 1 will be 1, If it is odd because the last bit will be already set. Otherwise, it will give 0 as output.
Python3
list1 = [10, 21, 4, 45, 66, 93, 1]
even_count, odd_count = 0, 0
for num in list1:
if not num & 1:
even_count += 1
else:
odd_count += 1
print("Even numbers in the list: ", even_count)
print("Odd numbers in the list: ", odd_count)
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Output
Even numbers in the list: 3
Odd numbers in the list: 4
Time Complexity: O(N), Here N is the number of elements in the list.
Auxiliary Space: O(1), As constant extra space is used.
Example 8: Using Bitwise OR operator
The idea is to check whether the last bit of the number is set or not. If the last bit is set then the number is odd, otherwise even. As we know bitwise OR Operation of the Number by 1 increment the value of the number by 1 if the number is even otherwise it will remain unchanged. So, if after OR operation of number with 1 gives a result which is greater than the number then it is even and we will return true otherwise it is odd and we will return false.
Python3
list1 = [10, 21, 4, 45, 66, 93, 1]
even_count, odd_count = 0, 0
for num in list1:
if num | 1 > num:
even_count += 1
else:
odd_count += 1
print("Even numbers in the list: ", even_count)
print("Odd numbers in the list: ", odd_count)
|
Output
Even numbers in the list: 3
Odd numbers in the list: 4
Time Complexity: O(N), Here N is the number of elements in the list.
Auxiliary Space: O(1), As constant extra space is used.
Method: Using the enumerate function
Python3
lst = [12, 14, 95, 3];c=0;c1=0
for i,a in enumerate(lst):
if a%2==0:
c+=1
else:
c1+=1
print("even number count",c,"odd number count",c1)
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Output
even number count 2 odd number count 2
Time Complexity: O(N), Here N is the number of elements in the list.
Auxiliary Space: O(1), As constant extra space is used.
Method: Using Numpy.Array :
Python
import numpy as np
List = [10, 21, 4, 45, 66, 93, 11]
list1 = np.array(List)
Even_list = list1[list1 % 2 == 0]
print("Even numbers in the list: ", len(Even_list))
print("Odd numbers in the list: ", len(list1)-len(Even_list))
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Output:
Even numbers in the list: 3
Odd numbers in the list: 4
Time Complexity: O(N), Here N is the number of elements in the list.
Auxiliary Space: O(1), As constant extra space is used.
Method: Using Numpy.where() function :
note: install numpy module using command “pip install numpy”
Algorithm:
Convert the given list to a numpy array.
Use numpy.where() function to find the indices of even and odd numbers in the array.
Count the number of indices using the len() function.
Print the counts.
Here’s the Python program to count Even and Odd numbers in a List using numpy.where() function:
Python3
import numpy as np
list1 = [2, 7, 5, 64, 14]
arr = np.array(list1)
even_count = len(np.where(arr % 2 == 0)[0])
odd_count = len(np.where(arr % 2 == 1)[0])
print("Even numbers in the list: ", even_count)
print("Odd numbers in the list: ", odd_count)
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Output:
Even numbers in the list: 3
Odd numbers in the list: 2
Time Complexity: O(N), where N is the number of elements in the list.
Space Complexity: O(N), as we create a numpy array of size N.
Method: Using Sum and len function
Approach:
- Initialize list.
- Initialize even_count and odd_count variables to store numbers.
- Use Sum method which counts 1 if the number in the list is odd and store the result in odd_count.
- Subtract the total length list to odd_count and store the value in even_count.
- Print event_count and odd_count.
Python3
list1 = [10, 21, 4, 45, 66, 93, 1]
even_count, odd_count = 0, 0
odd_count = sum(1 for i in list1 if i&1)
even_count = len(list1) - odd_count
print("Even numbers in the list: ", even_count)
print("Odd numbers in the list: ", odd_count)
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Output:
Even numbers in the list: 3
Odd numbers in the list: 4
Time Complexity: O(n), where n is the length of the list.
Space Complexity: O(1), Because no extra space is used.
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