Parsing A Boolean Expression in Python

PythonServer Side ProgrammingProgramming

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;

  • "f", evaluating to False;

  • "!(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 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

  • if e[i] is same as "f", then −

    • return (False, i + 1)

  • Otherwise when e[i] is same as "t" −

  • return (True,i + 1)

  • op := e[i], i := i + 2

  • Define one stack

  • while e[i] is not closing parentheses, do −

    • if e[i] is same as ",", then −

      • i := i + 1

      • Ignore following part, skip to the next iteration

    • res,i := solve(e, i)

    • push res into stack

  • if op is same as "&", then −

    • 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

  • return (inverse of stack[0], i + 1)

  • From the main method, do the following −

  • s,y := solve(expression, 0)

  • return s

Let us see the following implementation to get better understanding −

Example

 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))"))

Input

"|(!(t),&(t,f,t))"

Output

False
raja
Published on 04-Jun-2020 14:47:31
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