human
Maximum profit by buying and selling a share at most K times | Greedy Approach | Function to return the maximum profit.
Find the farthest decreasing span of prices before prices rise.
Find the farthest increasing span of prices before prices fall again.
Check if the current buying price is greater than that of the previous transaction.
Store the profit.
Remove the previous transaction.
Check if the current selling price is less than that of the previous transactions.
Store the new profit.
Remove the previous transaction.
Add the new transactions.
Finally store the profits of all the transactions.
Add the highest K profits.
Return the maximum profit.
Driver code"
How can the above be solved in Python?

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Give me a code to maximum profit by buying and selling a share at most K times and the other steps required?
assistant
#Here is code to maximum profit by buying and selling a share at most K times and the other steps required.
 def maxProfit ( n , k , prices ) :
	ans = 0
	buy = 0
	sell = 0
	transaction = [ ]
	profits = [ ]
	while ( sell < n ) :
		buy = sell
		while ( buy < n - 1 and prices [ buy ] >= prices [ buy + 1 ] ) :
			buy += 1
		sell = buy + 1
		while ( sell < n and prices [ sell ] >= prices [ sell - 1 ] ) :
			sell += 1
		while ( len ( transaction ) != 0 and prices [ buy ] < prices [ transaction [ len ( transaction ) - 1 ] [ 0 ] ] ) :
			p = transaction [ len ( transaction ) - 1 ]
			profits . append ( prices [ p [ 1 ] - 1 ] - prices [ p [ 0 ] ] )
			transaction . remove ( transaction [ len ( transaction ) - 1 ] )
		profits . sort ( reverse = True )
		while ( len ( transaction ) != 0 and prices [ sell - 1 ] > prices [ transaction [ len ( transaction ) - 1 ] [ 1 ] - 1 ] ) :
			p = transaction [ len ( transaction ) - 1 ]
			profits . append ( prices [ p [ 1 ] - 1 ] - prices [ buy ] )
			buy = p [ 0 ]
			transaction . remove ( transaction [ len ( transaction ) - 1 ] )
		transaction . append ( [ buy , sell ] )
	profits . sort ( reverse = True )
	while ( len ( transaction ) != 0 ) :
		profits . append ( prices [ transaction [ len ( transaction ) - 1 ] [ 1 ] - 1 ] - prices [ transaction [ len ( transaction ) - 1 ] [ 0 ] ] )
		transaction . remove ( transaction [ len ( transaction ) - 1 ] )
	profits . sort ( reverse = True )
	while ( k != 0 and len ( profits ) != 0 ) :
		ans += profits [ 0 ]
		profits . remove ( profits [ 0 ] )
		k -= 1
	return ans
if __name__ == ' _ _ main _ _ ' :
	k = 3
	prices = [ 1 , 12 , 10 , 7 , 10 , 13 , 11 , 10 , 7 , 6 , 9 ]
	n = len ( prices )
	print ( " Maximum ▁ profit ▁ is " , maxProfit ( n , k , prices ) )