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Cost to Balance the parentheses | Python 3 code to calculate the minimum cost to make the given parentheses balanced.
To store absolute count of balanced and unbalanced parenthesis.
o ( open bracket ) stores count of ' ( ' and c ( close bracket ) stores count of ').
Driver code"
Solution in Python:
 Give me a code to cost to Balance the parentheses and the other steps required?
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#Here is code to cost to Balance the parentheses and the other steps required.
 def costToBalance ( s ) :
	if ( len ( s ) == 0 ) :
		print ( 0 )
	ans = 0
'
	o = 0
	c = 0
	for i in range ( len ( s ) ) :
		if ( s [ i ] == ' ( ' ) :
			o += 1
		if ( s [ i ] == ' ) ' ) :
			c += 1
	if ( o != c ) :
		return - 1
	a = [ 0 for i in range ( len ( s ) ) ]
	if ( s [ 0 ] == ' ( ' ) :
		a [ 0 ] = 1
	else :
		a [ 0 ] = - 1
	if ( a [ 0 ] < 0 ) :
		ans += abs ( a [ 0 ] )
	for i in range ( 1 , len ( s ) ) :
		if ( s [ i ] == ' ( ' ) :
			a [ i ] = a [ i - 1 ] + 1
		else :
			a [ i ] = a [ i - 1 ] - 1
		if ( a [ i ] < 0 ) :
			ans += abs ( a [ i ] )
	return ans
if __name__ == ' _ _ main _ _ ' : s = " ) ) ) ( ( ( "
	print ( costToBalance ( s ) ) s = " ) ) ( ( "
	print ( costToBalance ( s ) )