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How do I make this code much simpler Check if array has the same number of even and short?(If there is any recursive method to build this function please share)odd values in Python

Task: Complete
Complete the following function that determines if the number of even and odd values in an integer list is the same. The function would return true if the list contains 5, 1, 0, 2 (two evens and two odds), but it would return false for the list containing 5, 1, 0, 2, 11 (too many odds). The function should return true if the listof Integers is empty, since an empty list contains the same number of evens and odds (0 for both).

| In               | Out   |         Why            |
|------------------|-------|------------------------|
| [5, 1, 0, 2]     | true  | two evens and two odds |
| [5, 1, 0, 2, 11] | false | too many odds          |
| []               | true  | both have 0            |

The function doesshould not affect the contents of the list.

def balanced(lst):
    n = len(lst)
    if n % 2 != 0:
        return False
    if n % 2 == 0:
        count_1 = 0
        count_2 = 0
        for item in lst:
            if item % 2 == 0:      #even
                count_1 += 1
            if item % 2 != 0:       #odd
                count_2 += 1
    if count_1 == count_2:
        return True
    else:
        return False
def result(lst):
     if balanced(lst):
         print("Your list is successfully balanced! It has same number of evens and odds!!")
     else:
         print("Oh no! Sorry! Your list seems to be not balanced! Try another list please!")
def main():
     lst_1 = [1,2,3,5,6,8,5,9]
     lst_2 = []
     lst_3 = [2,4,5,7]
     lst_4 = [1,2,4,4]
     lst_5 = [1,2,3]
     result(lst_1)
     result(lst_2)
     result(lst_3)
     result(lst_4)
     result(lst_5)
main()

 

How do I make this code much simpler and short?(If there is any recursive method to build this function please share)

Task: Complete the following function that determines if the number of even and odd values in an integer list is the same. The function would return true if the list contains 5, 1, 0, 2 (two evens and two odds), but it would return false for the list containing 5, 1, 0, 2, 11 (too many odds). The function should return true if the list is empty, since an empty list contains the same number of evens and odds (0 for both). The function does not affect the contents of the list.

def balanced(lst):
    n = len(lst)
    if n % 2 != 0:
        return False
    if n % 2 == 0:
        count_1 = 0
        count_2 = 0
        for item in lst:
            if item % 2 == 0:      #even
                count_1 += 1
            if item % 2 != 0:       #odd
                count_2 += 1
    if count_1 == count_2:
        return True
    else:
        return False
def result(lst):
     if balanced(lst):
         print("Your list is successfully balanced! It has same number of evens and odds!!")
     else:
         print("Oh no! Sorry! Your list seems to be not balanced! Try another list please!")
def main():
     lst_1 = [1,2,3,5,6,8,5,9]
     lst_2 = []
     lst_3 = [2,4,5,7]
     lst_4 = [1,2,4,4]
     lst_5 = [1,2,3]
     result(lst_1)
     result(lst_2)
     result(lst_3)
     result(lst_4)
     result(lst_5)
main()

 

Check if array has the same number of even and odd values in Python

Task:
Complete the following function that determines if the number of even and odd values in an list of Integers is the same.

| In               | Out   |         Why            |
|------------------|-------|------------------------|
| [5, 1, 0, 2]     | true  | two evens and two odds |
| [5, 1, 0, 2, 11] | false | too many odds          |
| []               | true  | both have 0            |

The function should not affect the contents of the list.

def balanced(lst):
    n = len(lst)
    if n % 2 != 0:
        return False
    if n % 2 == 0:
        count_1 = 0
        count_2 = 0
        for item in lst:
            if item % 2 == 0:      #even
                count_1 += 1
            if item % 2 != 0:       #odd
                count_2 += 1
    if count_1 == count_2:
        return True
    else:
        return False
def result(lst):
     if balanced(lst):
         print("Your list is successfully balanced! It has same number of evens and odds!!")
     else:
         print("Oh no! Sorry! Your list seems to be not balanced! Try another list please!")
def main():
     lst_1 = [1,2,3,5,6,8,5,9]
     lst_2 = []
     lst_3 = [2,4,5,7]
     lst_4 = [1,2,4,4]
     lst_5 = [1,2,3]
     result(lst_1)
     result(lst_2)
     result(lst_3)
     result(lst_4)
     result(lst_5)
main()
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How do I make this code much simpler and short?(If there is any recursive method to build this function please share)

Task: Complete the following function that determines if the number of even and odd values in an integer list is the same. The function would return true if the list contains 5, 1, 0, 2 (two evens and two odds), but it would return false for the list containing 5, 1, 0, 2, 11 (too many odds). The function should return true if the list is empty, since an empty list contains the same number of evens and odds (0 for both). The function does not affect the contents of the list.

My code:

def balanced(lst):
    n = len(lst)
    if n % 2 != 0:
        return False
    if n % 2 == 0:
        count_1 = 0
        count_2 = 0
        for item in lst:
            if item % 2 == 0:      #even
                count_1 += 1
            if item % 2 != 0:       #odd
                count_2 += 1
    if count_1 == count_2:
        return True
    else:
        return False
def result(lst):
     if balanced(lst):
         print("Your list is successfully balanced! It has same number of evens and odds!!")
     else:
         print("Oh no! Sorry! Your list seems to be not balanced! Try another list please!")
def main():
     lst_1 = [1,2,3,5,6,8,5,9]
     lst_2 = []
     lst_3 = [2,4,5,7]
     lst_4 = [1,2,4,4]
     lst_5 = [1,2,3]
     result(lst_1)
     result(lst_2)
     result(lst_3)
     result(lst_4)
     result(lst_5)
main()