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Parsing A Boolean Expression

Can you solve this real interview question? Parsing A Boolean Expression - 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.

class Solution:
    def parseBoolExpr(self, expression: str) -> bool:
        st = deque()
        for c in expression:
            if c == "," or c == "(":
                continue
            if c in ["t", "f", "!", "&", "|"]:
                st.append(c)
            elif c == ")":
                has_true = False
                has_false = False
                while st[-1] not in ["!", "&", "|"]:
                    top_value = st.pop()
                    if top_value == "t":
                        has_true = True
                    elif top_value == "f":
                        has_false = True
                op = st.pop()
                if op == "!":
                    st.append("t" if not has_true else "f")
                elif op == "&":
                    st.append("f" if has_false else "t")
                else:
                    st.append("t" if has_true else "f")
        return st[-1] == "t"