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Shortest path length between two given nodes such that adjacent nodes are at bit difference 2 | Function to count set bits in a number.
Stores count of set bits in xo.
Iterate over each bits of xo.
If current bit of xo is 1.
Update count.
Update xo.
Function to find length of shortest path between the nodes a and b.
Stores XOR of a and b.
Stores the count of set bits in xorVal.
If cnt is an even number.
Driver Code.
Given N.
Given a and b.
Function call"
Solution in Python:
 What is a code to shortest path length between two given nodes such that adjacent nodes are at bit difference 2 and the other steps required?
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#Here is code to shortest path length between two given nodes such that adjacent nodes are at bit difference 2 and the other steps required.
 def countbitdiff ( xo ) :
	count = 0
	while ( xo ) :
		if ( xo % 2 == 1 ) :
			count += 1
		xo = xo // 2
	return count
def shortestPath ( n , a , b ) :
	xorVal = a ^ b
	cnt = countbitdiff ( xorVal )
	if ( cnt % 2 == 0 ) :
		print ( cnt // 2 )
	else :
		print ( " - 1" )
if __name__ == ' _ _ main _ _ ' :
	n = 15
	a , b = 15 , 3
	shortestPath ( n , a , b )