human
Minimum swaps so that binary search can be applied | Function to find minimum swaps ..
Here we are getting number of smaller and greater elements of k ..
Here we are calculating the actual position of k in the array ..
Implementing binary search as per the above - discussed cases ..
If we find the k ..
If we need minimum element swap ..
Else the element is at the right position ..
If we need maximum element swap ..
Else element is at the right position.
Calculating the required swaps ..
If it is impossible ..
Driver function"
How can the above be solved in Python?

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Write me a code to minimum swaps so that binary search can be applied and the other steps required?
assistant
#Here is code to minimum swaps so that binary search can be applied and the other steps required.
 def findMinimumSwaps ( arr , n , k ) :
	num_min = num_max = need_minimum = 0
	need_maximum = swaps = 0
	for i in range ( n ) :
		if ( arr [ i ] < k ) :
			num_min += 1
		elif ( arr [ i ] > k ) :
			num_max += 1
	for i in range ( n ) :
		if ( arr [ i ] == k ) :
			pos = i
			break
	left = 0
	right = n - 1
	while ( left <= right ) :
		mid = ( left + right ) // 2
		if ( arr [ mid ] == k ) :
			break
		elif ( arr [ mid ] > k ) :
			if ( pos > mid ) :
				need_minimum += 1
			else :
				num_min -= 1
			left = mid + 1
		else :
			if ( pos < mid ) :
				need_maximum += 1
			else :
				num_max -= 1
			right = mid - 1
	if ( need_minimum > need_maximum ) :
		swaps = swaps + need_maximum
		num_min = num_min - need_maximum
		need_minimum = ( need_minimum - need_maximum )
		need_maximum = 0
	else :
		swaps = swaps + need_minimum
		num_max = num_max - need_minimum
		need_maximum = ( need_maximum - need_minimum )
		need_minimum = 0
	if ( need_maximum > num_max or need_minimum > num_min ) :
		return - 1
	else :
		return ( swaps + need_maximum + need_minimum )
if __name__ == " _ _ main _ _ " :
	arr = [ 3 , 10 , 6 , 7 , 2 , 5 , 4 ]
	k = 4
	n = len ( arr )
	print ( findMinimumSwaps ( arr , n , k ) )