human
Count all distinct pairs with difference equal to k | Standard binary search function.
Returns count of pairs with difference k in arr [ ] of size n ..
Sort array elements.
Pick a first element point.
Driver Code"
Solution in Python:

Can you provide a code to count all distinct pairs with difference equal to k and the other steps required?
assistant
#Here is code to count all distinct pairs with difference equal to k and the other steps required.
 def binarySearch ( arr , low , high , x ) :
	if ( high >= low ) :
		mid = low + ( high - low ) // 2
		if x == arr [ mid ] :
			return ( mid )
		elif ( x > arr [ mid ] ) :
			return binarySearch ( arr , ( mid + 1 ) , high , x )
		else :
			return binarySearch ( arr , low , ( mid - 1 ) , x )
	return - 1
def countPairsWithDiffK ( arr , n , k ) :
	count = 0
	arr . sort ( )
	for i in range ( 0 , n - 2 ) :
		if ( binarySearch ( arr , i + 1 , n - 1 , arr [ i ] + k ) != - 1 ) :
			count += 1
	return count
arr = [ 1 , 5 , 3 , 4 , 2 ]
n = len ( arr )
k = 3
print ( " Count ▁ of ▁ pairs ▁ with ▁ given ▁ diff ▁ is ▁ " , countPairsWithDiffK ( arr , n , k ) )