Parsing A Boolean Expression in Python

Suppose we have a boolean expression, we have to find the result after evaluating that expression.
An expression can either be −
"t", evaluating to True;
"t", evaluating to True;
"f", evaluating to False;
"f", evaluating to False;
"!(expression)", evaluating to the logical NOT of the inner expression;
"!(expression)", evaluating to the logical NOT of the inner expression;
"&(expr1,expr2,...)", evaluating to the logical AND of 2 or more inner expressions;
"&(expr1,expr2,...)", evaluating to the logical AND of 2 or more inner expressions;
"|(expr1,expr2,...)", evaluating to the logical OR of 2 or more inner expressions;
"|(expr1,expr2,...)", evaluating to the logical OR of 2 or more inner expressions;
So, if the input is like "|(!(t),&(t,f,t))", then the output will be fasle, this is because !(t) is false, then &(t,f,t) is also false, so the OR of all false values will be false.
To solve this, we will follow these steps −
define solve(), this will take e, i
define solve(), this will take e, i
if e[i] is same as "f", then −return (False, i + 1)
if e[i] is same as "f", then −
return (False, i + 1)
return (False, i + 1)
Otherwise when e[i] is same as "t" −
Otherwise when e[i] is same as "t" −
return (True,i + 1)
return (True,i + 1)
op := e[i], i := i + 2
op := e[i], i := i + 2
Define one stack
Define one stack
while e[i] is not closing parentheses, do −if e[i] is same as ",", then −i := i + 1Ignore following part, skip to the next iterationres,i := solve(e, i)push res into stack
while e[i] is not closing parentheses, do −
if e[i] is same as ",", then −i := i + 1Ignore following part, skip to the next iteration
if e[i] is same as ",", then −
i := i + 1
i := i + 1
Ignore following part, skip to the next iteration
Ignore following part, skip to the next iteration
res,i := solve(e, i)
res,i := solve(e, i)
push res into stack
push res into stack
if op is same as "&", then −return true when all elements are true in stack, otherwise false, i + 1
if op is same as "&", then −
return true when all elements are true in stack, otherwise false, i + 1
return true when all elements are true in stack, otherwise false, i + 1
otherwise when op is same as " OR " −return true when at least one elements is true in stack, otherwise false, i + 1
otherwise when op is same as " OR " −
return true when at least one elements is true in stack, otherwise false, i + 1
return true when at least one elements is true in stack, otherwise false, i + 1
return (inverse of stack[0], i + 1)
return (inverse of stack[0], i + 1)
From the main method, do the following −
From the main method, do the following −
s,y := solve(expression, 0)
s,y := solve(expression, 0)
return s
return s
Let us see the following implementation to get better understanding −
 Live Demo
class Solution(object):
   def parseBoolExpr(self, expression):
      s,y = self.solve(expression,0)
      return s
   def solve(self,e,i):
      if e[i] =="f":
         return False,i+1
      elif e[i] == "t":
         return True,i+1
      op = e[i]
      i = i+2
      stack = []
      while e[i]!=")":
         if e[i] == ",":
            i+=1
            continue
         res,i = self.solve(e,i)
         stack.append(res)
      if op == "&":
         return all(stack),i+1
      elif op == "|":
         return any(stack),i+1
      return not stack[0],i+1
ob = Solution()
print(ob.parseBoolExpr("|(!(t),&(t,f,t))"))
"|(!(t),&(t,f,t))"
False