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Parsing a Boolean Expression - Step by Step Explanation

This repository contains multiple implementations of the solution for the LeetCode problem Parsing a Boolean Expression. The problem requires parsing and evaluating a Boolean expression based on the given rules for logical operators AND, OR, and NOT. Below is a step-by-step explanation for how the solution is designed and implemented in each language.

Problem Recap

We are given a boolean expression that evaluates either to true or false. The expression can be:

  1. t which evaluates to true.
  2. f which evaluates to false.
  3. !(subExpr) which evaluates to the logical NOT of the inner expression subExpr.
  4. &(subExpr1, subExpr2, ..., subExprn) which evaluates to the logical AND of the inner expressions.
  5. |(subExpr1, subExpr2, ..., subExprn) which evaluates to the logical OR of the inner expressions.

Our goal is to evaluate the expression and return the result as either true or false.

Approach Overview

  1. Stack-based Solution:

    • We traverse the expression character by character.
    • When we encounter a closing parenthesis ), it signals the end of an inner expression. We pop all the elements inside the parentheses, apply the logical operator (!, &, or |), and push the result back onto the stack.
    • We continue this process until we finish processing the entire string.

  2. Operators and Logic:

    • ! (NOT): Inverts the truth value of the sub-expression.
    • & (AND): Returns true if all the sub-expressions are true, otherwise returns false.
    • | (OR): Returns true if at least one sub-expression is true.

Step-by-Step Walkthrough for Each Language

C++ Code

  1. Initialize the Stack:

    • A stack is used to keep track of the characters in the expression.
    • We process each character from the input expression string one by one.

  2. Process Characters:

    • If a closing parenthesis ) is encountered, start collecting the sub-expressions within the parentheses until the opening parenthesis ( is found.
    • Determine the operator (!, &, or |) immediately before the parentheses.

  3. Evaluate Sub-expressions:

    • For !: Apply the NOT operation on the single sub-expression.
    • For &: Apply the AND operation across all sub-expressions.
    • For |: Apply the OR operation across all sub-expressions.

  4. Push Results Back to Stack:

    • After evaluating, push the result (t or f) back onto the stack.

  5. Final Result:

    • After the entire expression is processed, the top of the stack contains the final result, which is either true or false.

Java Code

  1. Stack Initialization:

    • A stack is initialized to store characters during the traversal of the input expression.

  2. Handling Parentheses:

    • When a closing parenthesis ) is encountered, we collect the characters from the stack until the corresponding opening parenthesis ( is found.
    • The operator that precedes the opening parenthesis determines how the sub-expressions should be evaluated.

  3. Logical Operations:

    • For ! (NOT), we invert the truth value of the sub-expression.
    • For & (AND), we check if all sub-expressions evaluate to true; otherwise, the result is false.
    • For | (OR), we return true if at least one sub-expression evaluates to true.

  4. Final Push:

    • After evaluating the sub-expressions, we push the result back onto the stack for further use.

  5. Returning the Result:

    • Once the entire expression is processed, the top of the stack gives the final result of the Boolean expression.

JavaScript Code

  1. Use of Stack:

    • A stack is used to track the characters and sub-expressions within the Boolean expression as we iterate through it.

  2. Detecting Sub-expressions:

    • When encountering a ), gather all the sub-expressions inside the parentheses and determine the operator (!, &, or |).

  3. Evaluating the Expression:

    • Apply the operator to the collected sub-expressions:
      • For !, invert the truth value.
      • For &, ensure all sub-expressions are true to result in true.
      • For |, check if any sub-expression is true.

  4. Update Stack:

    • Push the evaluation result back to the stack.

  5. Return Final Boolean:

    • After traversing the entire input, the Boolean result is at the top of the stack and is returned.

Python Code

  1. Stack for Processing:

    • Initialize a stack to manage the characters of the expression.
    • Each character is processed sequentially.

  2. Sub-expression Handling:

    • Upon encountering ), gather the sub-expressions within the parentheses and then evaluate them based on the operator right before (.

  3. Boolean Logic Application:

    • Apply the corresponding operator:
      • For !, invert the value.
      • For &, check that all sub-expressions are true.
      • For |, return true if at least one sub-expression is true.

  4. Push Result:

    • After evaluation, push the resulting value back to the stack.

  5. Final Result:

    • At the end of the iteration, the stack will contain the final result of the Boolean expression, which is returned.

Go Code

  1. Stack Initialization:

    • Use a stack to keep track of characters and sub-expressions as we parse through the Boolean expression.

  2. Processing Characters:

    • When a ) is found, pop elements from the stack until we reach the corresponding (.
    • Use the operator found before ( to evaluate the sub-expressions.

  3. Evaluate Expression:

    • Based on the operator:
      • For !, apply NOT on the single sub-expression.
      • For &, apply AND across all sub-expressions.
      • For |, apply OR across all sub-expressions.

  4. Push Back Result:

    • After evaluating, push the result (t or f) back to the stack.

  5. Final Output:

    • The result of the Boolean expression is found at the top of the stack at the end of the process.

Conclusion

In all the implementations, the key steps involve:

  1. Stack-based parsing: Using a stack to manage characters and sub-expressions.
  2. Operator handling: Determining the logical operation to apply (!, &, |).
  3. Sub-expression evaluation: Processing sub-expressions within parentheses and pushing the results back to the stack.
  4. Final result: The top of the stack contains the Boolean result after processing the entire expression.

Time Complexity

  • Time Complexity: $$O(n)$$, where (n) is the length of the expression. Each character in the string is processed once, and the sub-expressions are evaluated linearly.

Space Complexity

  • Space Complexity: $$O(n)$$, as we use a stack to store intermediate characters and results during the parsing process.

By following this step-by-step guide, you will understand how the Boolean expression is parsed and evaluated using a stack-based approach across multiple programming languages.