human
Maximum number of mangoes that can be bought | ''Function to check if mid number of mangoes can be bought.
'' Store the coins.
'' If watermelons needed are greater  than given watermelons.
'' Store remaining watermelons if vl  watermelons are used to buy mangoes.
'' Store the value of coins if these  watermelon get sold.
'' Increment coins by ex.
'' Number of mangoes that can be buyed  if only x coins needed for one mango.
'' If the condition is satisfied,  return true.
'' Otherwise return false.
''Function to find the maximum number of mangoes that can be bought by selling watermelons.
'' Initialize the boundary values.
'' Store the required result.
'' Binary Search.
'' Store the mid value.
'' Check if it is possible to  buy mid number of mangoes.
'' Otherwise, update r to mid -1.
'' Return the result.
''Driver Code.
'' Given Input.
'' Function Call"
How can the above be solved in Python?

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Can you provide a code to maximum number of mangoes that can be bought and the other steps required?
assistant
#Here is code to maximum number of mangoes that can be bought and the other steps required.
 def check ( n , m , x , y , vl ) :
	temp = m
	if ( vl > n ) :
		return False
	ex = n - vl
	ex *= y
	temp += ex
	cr = temp // x
	if ( cr >= vl ) :
		return True
	return False
def maximizeMangoes ( n , m , x , y ) :
	l = 0
	r = n
	ans = 0
	while ( l <= r ) :
		mid = l + ( r - l ) // 2
		if ( check ( n , m , x , y , mid ) ) :
			ans = mid
			l = mid + 1
		else :
			r = mid - 1
	return ans
if __name__ == ' _ _ main _ _ ' :
	W = 4
	C = 8
	x = 4
	y = 4
	print ( maximizeMangoes ( W , C , x , y ) )