human
Minimum deletions to convert given integer to an odd number whose sum of digits is even | Set 2 | Function to find minimum count of digits required to be remove to make N odd and the sum of digits of N even.
Stores count of even digits.
Stores count of odd digits.
Iterate over the digits of N.
If current digit is even.
Update even.
Otherwise.
Update odd.
Base conditions.
Stores count of digits required to be removed to make N odd and the sum of digits of N even.
Iterate over the digits of N.
If current digit is even.
Update ans.
Otherwise ,.
Update ans.
If count of odd digits is odd.
Update ans.
Finally print the ans.
Driver code.
Input String.
Function call"
Solution in Python:

Give me a code to minimum deletions to convert given integer to an odd number whose sum of digits is even and the other steps required?
assistant
#Here is code to minimum deletions to convert given integer to an odd number whose sum of digits is even and the other steps required.
 def minOperations ( N ) :
	even = 0 ;
	odd = 0 ;
	for it in N :
		if ( int ( ord ( it ) - ord ( '0' ) ) % 2 == 0 ) :
			even += 1 ;
		else :
			odd += 1 ;
	if ( odd == 0 or odd == 1 ) :
		print ( " Not ▁ Possible " + " " ) ;
	else :
		ans = 0 ;
		for it in N :
			if ( int ( ord ( it ) - ord ( '0' ) ) % 2 == 0 ) :
				ans += 1 ;
			else :
				ans = 0 ;
		if ( odd % 2 != 0 ) :
			ans += 1 ;
		print ( ans , end = " ▁ " ) ;
if __name__ == ' _ _ main _ _ ' :
	N = "12345" ;
	minOperations ( N ) ;