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1106. Parsing A Boolean Expression

Difficulty Topics

Description

A boolean expression is an expression that evaluates to either true or false. It can be in one of the following shapes:

  • 't' that evaluates to true.
  • 'f' that evaluates to false.
  • '!(subExpr)' that evaluates to the logical NOT of the inner expression subExpr.
  • '&(subExpr1, subExpr2, ..., subExprn)' that evaluates to the logical AND of the inner expressions subExpr1, subExpr2, ..., subExprn where n >= 1.
  • '|(subExpr1, subExpr2, ..., subExprn)' that evaluates to the logical OR of the inner expressions subExpr1, subExpr2, ..., subExprn where n >= 1.

Given a string expression that represents a boolean expression, return the evaluation of that expression.

It is guaranteed that the given expression is valid and follows the given rules.

 

Example 1:

Input: expression = "&(|(f))"
Output: false
Explanation: 
First, evaluate |(f) --> f. The expression is now "&(f)".
Then, evaluate &(f) --> f. The expression is now "f".
Finally, return false.

Example 2:

Input: expression = "|(f,f,f,t)"
Output: true
Explanation: The evaluation of (false OR false OR false OR true) is true.

Example 3:

Input: expression = "!(&(f,t))"
Output: true
Explanation: 
First, evaluate &(f,t) --> (false AND true) --> false --> f. The expression is now "!(f)".
Then, evaluate !(f) --> NOT false --> true. We return true.

 

Constraints:

  • 1 <= expression.length <= 2 * 104
  • expression[i] is one following characters: '(', ')', '&', '|', '!', 't', 'f', and ','.

Solution

parsing-a-boolean-expression.py
class Solution:
    def parseBoolExpr(self, expression: str) -> bool:
        stack = []

        for c in expression:
            if c == ")":
                seen = set()
                while stack and stack[-1] != "(":
                    seen.add(stack.pop())
                stack.pop()
                operator = stack.pop()
                stack.append(all(seen) if operator == "&" else any(seen) if operator == "|" else not seen.pop())
            elif c != ",":
                stack.append(True if c == 't' else False if c == 'f' else c)

        return stack.pop()