1106. Parsing A Boolean Expression - Detailed Explanation

Complexity Analysis

• Time Complexity:
  The recursive DFS solution processes each character in the expression exactly once. Hence, the time complexity is O(n), where n is the length of the expression.

• Space Complexity:
  The space used is mainly due to the recursion stack, which in the worst case can be O(n) if the expression is deeply nested.

Approach 2: Brute Force Using Iterative Replacement

Idea

A brute force method involves repeatedly finding the innermost sub-expression (one that does not contain any further nested parentheses), evaluating it, and then replacing that sub-expression in the string with its Boolean result (t or f). Continue this process until the entire expression reduces to a single literal.

Step-by-Step Walkthrough

1. Find the Innermost Expression:
   Scan the string to locate a pair of matching parentheses that does not contain any other parentheses inside.

2. Evaluate the Sub-Expression:
   Based on the operator preceding the parentheses, evaluate the sub-expression.

   • For !, simply invert the value.
   • For &, check if all values are true.
   • For |, check if at least one value is true.

3. Replace the Expression:
   Replace the entire segment (operator, parentheses, and contents) with the resulting literal.

4. Repeat:
   Continue this process until no parentheses remain.

Drawbacks

• Inefficiency:
  This method may require multiple scans of the string, leading to a worse-case time complexity of O(n²) for large expressions.
• String Manipulation Overhead:
  Repeatedly modifying the string can be expensive.

Pseudocode Outline

while expression contains '(':
    for each character in expression:
         if found an innermost parenthesis:
              evaluate it based on the operator before it
              replace that substring with 't' or 'f'
return the final literal as the result

Note: Due to its inefficiency, this approach is not recommended for production-level interview problems, but it can help in understanding the parsing process.

Common Mistakes and Edge Cases

Common Mistakes

• Mishandling Nested Expressions:
  Not correctly matching parentheses when expressions are deeply nested.

• Incorrect Index Management:
  Losing track of the current position while using recursion or a stack can lead to out-of-bound errors.

• Operator Misinterpretation:
  Mixing up the behavior of !, &, and | can result in wrong answers.

Edge Cases

• Single Literal Input:
  The expression is just "t" or "f".

• Deep Nesting:
  Expressions that are nested many layers deep (e.g., multiple levels of negations).

• Multiple Arguments:
  Ensure that the solution correctly handles operators with more than two operands.

Variations

• Additional Operators:
  Problems might introduce other operators (e.g., XOR) or modify the format of the Boolean expressions.

• Arithmetic Expressions:
  Similar techniques are used in parsing arithmetic expressions (Basic Calculator problems).

Related Problems for Further Practice

• Evaluate Reverse Polish Notation:
  Use a stack to evaluate expressions written in postfix notation.

• Basic Calculator I / II / III:
  These problems involve parsing and evaluating mathematical expressions.

• Construct Binary Expression Tree:
  Convert an infix expression into a binary tree and then evaluate it.

• Infix to Postfix Conversion:
  Practice converting infix expressions to postfix notation before evaluation.