human
Write me a code to boolean Parenthesization Problem and the other steps required given the following:
Boolean Parenthesization Problem | DP | CountWays function.
Base Condition.
Count number of True in left Partition.
Count number of False in left Partition.
Count number of True in right Partition.
Count number of False in right Partition.
Evaluate AND operation.
Evaluate OR operation.
Evaluate XOR operation.
Driver code.
We obtain the string T | T & F ^ T.
There are 4 ways ( ( T T ) & ( F ^ T ) ) , ( T | ( T & ( F ^ T ) ) ) , ( ( ( T T ) & F ) ^ T ) and ( T | ( ( T & F ) ^ T ) )"
How can the above be solved in Python?
assistant
#Here is code to boolean Parenthesization Problem and the other steps required.
 def countWays ( N , S ) :
	dp = [ [ [ - 1 for k in range ( 2 ) ] for i in range ( N + 1 ) ] for j in range ( N + 1 ) ]
	return parenthesis_count ( S , 0 , N - 1 , 1 , dp )
def parenthesis_count ( Str , i , j , isTrue , dp ) :
	if ( i > j ) :
	return 0
	if ( i == j ) :
	if ( isTrue == 1 ) :
		return 1 if Str [ i ] == ' T ' else 0
	else :
		return 1 if Str [ i ] == ' F ' else 0
	if ( dp [ i ] [ j ] [ isTrue ] != - 1 ) :
	return dp [ i ] [ j ] [ isTrue ]
	temp_ans = 0
	for k in range ( i + 1 , j , 2 ) :
	if ( dp [ i ] [ k - 1 ] [ 1 ] != - 1 ) :
		leftTrue = dp [ i ] [ k - 1 ] [ 1 ]
	else :
		leftTrue = parenthesis_count ( Str , i , k - 1 , 1 , dp )
	if ( dp [ i ] [ k - 1 ] [ 0 ] != - 1 ) :
		leftFalse = dp [ i ] [ k - 1 ] [ 0 ]
	else :
		leftFalse = parenthesis_count ( Str , i , k - 1 , 0 , dp )
	if ( dp [ k + 1 ] [ j ] [ 1 ] != - 1 ) :
		rightTrue = dp [ k + 1 ] [ j ] [ 1 ]
	else :
		rightTrue = parenthesis_count ( Str , k + 1 , j , 1 , dp )
	if ( dp [ k + 1 ] [ j ] [ 0 ] != - 1 ) :
		rightFalse = dp [ k + 1 ] [ j ] [ 0 ]
	else :
		rightFalse = parenthesis_count ( Str , k + 1 , j , 0 , dp )
	if ( Str [ k ] == ' & ' ) :
		if ( isTrue == 1 ) :
		temp_ans = temp_ans + leftTrue * rightTrue
		else :
		temp_ans = temp_ans + leftTrue * rightFalse + leftFalse * rightTrue + leftFalse * rightFalse
	elif ( Str [ k ] == ' ▁ ' ) :
		if ( isTrue == 1 ) :
		temp_ans = temp_ans + leftTrue * rightTrue + leftTrue * rightFalse + leftFalse * rightTrue
		else :
		temp_ans = temp_ans + leftFalse * rightFalse
	elif ( Str [ k ] == ' ^ ' ) :
		if ( isTrue == 1 ) :
		temp_ans = temp_ans + leftTrue * rightFalse + leftFalse * rightTrue
		else :
		temp_ans = temp_ans + leftTrue * rightTrue + leftFalse * rightFalse
	dp [ i ] [ j ] [ isTrue ] = temp_ans
	return temp_ans
symbols = " TTFT "
operators = " | & ^ "
S = " "
j = 0
for i in range ( len ( symbols ) ) :
	S = S + symbols [ i ]
	if ( j < len ( operators ) ) :
		S = S + operators [ j ]
		j += 1
N = len ( S )
print ( countWays ( N , S ) )