human
Check if N numbers with Even Sum can be selected from a given Array | Function to check if an odd sum can be made using N integers from the array.
Initialize odd and even counts.
Iterate over the array to count the no . of even and odd integers.
If element is odd.
If element is even.
Check if even_freq is more than N.
If odd_freq is odd.
Consider even count of odd.
Calculate even required.
If even count is less than required count.
Calculate even required.
If even count is less than required count.
Driver Code"
Solution in Python:
 ... Give me a code to check if N numbers with Even Sum can be selected from a given Array and the other steps required?
assistant
#Here is code to check if N numbers with Even Sum can be selected from a given Array and the other steps required.
 def checkEvenSum ( arr , N , size ) :
	even_freq , odd_freq = 0 , 0
	for i in range ( size ) :
		if ( arr [ i ] & 1 ) :
			odd_freq += 1
		else :
			even_freq += 1
	if ( even_freq >= N ) :
		return True
	else :
		if ( odd_freq & 1 ) :
			taken = odd_freq - 1
			req = N - taken
			if ( even_freq < req ) :
				return False
			else :
				return True
		else :
			taken = odd_freq
			req = N - taken
			if ( even_freq < req ) :
				return False
			else :
				return True
	return False
if __name__ == " _ _ main _ _ " :
	arr = [ 9 , 2 , 3 , 4 , 18 , 7 , 7 , 6 ]
	size = len ( arr )
	N = 5
	if ( checkEvenSum ( arr , N , size ) ) :
		print ( " Yes " )
	else :
		print ( " No " )