human
Count of pairs with difference at most K with no element repeating | Function to count the number of pairs whose difference is atmost K in an array.
Sorting the Array.
Variable to store the count of pairs whose difference is atmost K.
Loop to consider the consecutive pairs of the array.
if Pair found increment the index by 2.
Driver Code.
Function Call"
How can the above be solved in Python?
 ... What is a code to count of pairs with difference at most K with no element repeating and the other steps required?
assistant
#Here is code to count of pairs with difference at most K with no element repeating and the other steps required.
 def countPairs ( arr , k ) :
	arr . sort ( )
	pair = 0
	index = 0
	while ( index < len ( arr ) - 1 ) :
		if arr [ index + 1 ] - arr [ index ] <= k :
			pair += 1
			index += 2
		else :
			index += 1
	return pair
if __name__ == " _ _ main _ _ " :
	arr = [ 1 , 4 , 3 , 7 , 5 ]
	k = 2
	count = countPairs ( arr , k )
	print ( count )