human
Count all disjoint pairs having absolute difference at least K from a given array | ''Function to count distinct pairs with absolute difference atleast K.
'' Track the element that  have been paired.
'' Stores count of distinct pairs.
'' Pick all elements one by one.
'' If already visited.
'' If already visited.
'' If difference is at least K.
'' Mark element as visited and  increment the count.
'' Print the final count.
''Driver Code.
'' Given arr[].
'' Size of array.
'' Given difference K.
'' Function Call"
Solution in Python:
assistant
#Here is code to count all disjoint pairs having absolute difference at least K from a given array and the other steps required.
 def countPairsWithDiffK ( arr , N , K ) :
	vis = [ 0 ] * N
	count = 0
	for i in range ( N ) :
		if ( vis [ i ] == 1 ) :
			continue
		for j in range ( i + 1 , N ) :
			if ( vis [ j ] == 1 ) :
				continue
			if ( abs ( arr [ i ] - arr [ j ] ) >= K ) :
				count += 1
				vis [ i ] = 1
				vis [ j ] = 1
				break
	print ( count )
if __name__ == ' _ _ main _ _ ' :
	arr = [ 1 , 3 , 3 , 5 ]
	N = len ( arr )
	K = 2
	countPairsWithDiffK ( arr , N , K )