human
Triplets in array with absolute difference less than k | Return the lower bound i . e smallest index of element having value greater or equal to value.
Return the number of triplet indices satisfies the three constraints.
sort the array.
for each element from index 2 to n - 1..
finding the lower bound of arr [ i ] - k ..
If there are at least two elements between lower bound and current element ..
increment the count by lb - i C 2..
Driver code"
How can the above be solved in Python?

Write me a code to triplets in array with absolute difference less than k and the other steps required?
assistant
#Here is code to triplets in array with absolute difference less than k and the other steps required.
 def binary_lower ( value , arr , n ) :
	start = 0
	end = n - 1
	ans = - 1
	while ( start <= end ) :
		mid = ( start + end ) // 2
		if ( arr [ mid ] >= value ) :
			end = mid - 1
			ans = mid
		else :
			start = mid + 1
	return ans
def countTriplet ( arr , n , k ) :
	count = 0
	arr . sort ( )
	for i in range ( 2 , n ) :
		cur = ( binary_lower ( arr [ i ] - k , arr , n ) )
		if ( cur <= i - 2 ) :
			count += ( ( i - cur ) * ( i - cur - 1 ) ) // 2
	return count
if __name__ == " _ _ main _ _ " :
	arr = [ 1 , 1 , 2 , 2 , 3 ]
	k = 1
	n = len ( arr )
	print ( countTriplet ( arr , n , k ) )