human
Count all pairs of an array which differ in K bits | Function to calculate K bit different pairs in array.
Get the maximum value among all array elemensts.
Set the count array to 0 , count [ ] stores the total frequency of array elements.
Initialize result.
For 0 bit answer will be total count of same number.
if count [ i ] is 0 , skip the next loop as it will not contribute the answer.
Update answer if k differ bit found.
Driver code"
Solution in Python:

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Can you provide a code to count all pairs of an array which differ in K bits and the other steps required?
assistant
#Here is code to count all pairs of an array which differ in K bits and the other steps required.
 def kBitDifferencePairs ( arr , n , k ) :
	MAX = max ( arr )
	count = [ 0 for i in range ( MAX + 1 ) ]
	for i in range ( n ) :
		count [ arr [ i ] ] += 1
	ans = 0
	if ( k == 0 ) :
		for i in range ( MAX + 1 ) :
			ans += ( count [ i ] * ( count [ i ] - 1 ) ) // 2
		return ans
	for i in range ( MAX + 1 ) :
		if ( count [ i ] == 0 ) :
			continue
		for j in range ( i + 1 , MAX + 1 ) :
			if ( bin ( i ^ j ) . count ( '1' ) == k ) :
				ans += count [ i ] * count [ j ]
	return ans
k = 2
arr = [ 2 , 4 , 1 , 3 , 1 ]
n = len ( arr )
print ( " Total ▁ pairs ▁ for ▁ k ▁ = " , k , " are " , kBitDifferencePairs ( arr , n , k ) )
k = 3
print ( " Total ▁ pairs ▁ for ▁ k ▁ = " , k , " are " , kBitDifferencePairs ( arr , n , k ) )