human
Smallest number greater or equals to N such that it has no odd positioned bit set | Function to count the total bits.
Iterate and find the number of set bits.
Right shift the number by 1.
Function to find the nearest number.
Count the total number of bits.
To get the position.
If the last set bit is at odd position then answer will always be a number with the left bit set.
Set all the even bits which are possible.
If the number still is less than N.
Return the number by setting the next even set bit.
If we have reached this position it means tempsum > n hence turn off even bits to get the first possible number.
Turn off the bit.
If it gets lower than N then set it and return that number.
Driver code"
Solution in Python:
 Give me a code to smallest number greater or equals to N such that it has no odd positioned bit set and the other steps required?
assistant
#Here is code to smallest number greater or equals to N such that it has no odd positioned bit set and the other steps required.
 def countBits ( n ) :
	count = 0 ;
	while ( n > 0 ) :
		count += 1 ;
		n >>= 1 ;
	return count ;
def findNearestNumber ( n ) :
	cnt = countBits ( n ) ;
	cnt -= 1 ;
	if ( cnt % 2 ) :
		return 1 << ( cnt + 1 ) ;
	else :
		tempnum = 0 ;
		for i in range ( 0 , cnt + 1 , 2 ) :
			tempnum += 1 << i ;
		if ( tempnum < n ) :
			return ( 1 << ( cnt + 2 ) ) ;
		elif ( tempnum == n ) :
			return n ;
		for i in range ( 0 , cnt + 1 , 2 ) :
			tempnum -= ( 1 << i ) ;
			if ( tempnum < n ) :
				tempnum += ( 1 << i ) ;
				return tempnum ;
n = 19 ;
print ( findNearestNumber ( n ) ) ;