What is the time complexity of Parsing A Boolean Expression?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Parsing A Boolean Expression?

This problem uses String, Stack, Recursion. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Parsing A Boolean Expression?

Insight 1: The problem is naturally recursive because the sub-expressions within the parentheses are also boolean expressions that need to be evaluated. Insight 2: Using a stack data structure simplifies the parsing process by allowing us to keep track of the operators and operands as we traverse the expression string.

Which companies ask LeetCode 1106. Parsing A Boolean Expression in interviews?

This problem is frequently asked at HiLabs.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Parsing A Boolean Expression - LeetCode 1106 Solution

Parsing A Boolean Expression - Complete Solution Guide

Parsing A Boolean Expression is LeetCode problem 1106, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 . '&(subExpr 1 , subExpr 2 , ..., subExpr n )' that evaluates to the logical AND of the inner expressions subExpr 1 , subExpr 2 , ..., subExpr n where n >= 1 . '|(subExpr 1 , subExpr 2 , ..., subExpr n )' that evaluates to the logical OR of

Detailed Explanation

The problem requires us to parse a boolean expression string and evaluate it to either true or false. The boolean expression can be a simple 't' (true) or 'f' (false), or it can be a more complex expression involving logical NOT ('!'), AND ('&'), or OR ('|') operators. The operators can have multiple sub-expressions as operands, enclosed in parentheses and separated by commas. The goal is to recursively evaluate the expression, taking into account the operator precedence and the values of the sub-expressions, and return the final boolean result.

Solution Approach

The solution employs a stack-based approach to parse and evaluate the boolean expression. The algorithm iterates through the expression string character by character. When it encounters a closing parenthesis ')', it pops elements from the stack until it finds the corresponding opening parenthesis '('. The popped elements represent the operands for the operator. After popping the '(', the operator itself is popped from the stack. Based on the operator type ('!', '&', or '|'), the algorithm performs the corresponding logical operation on the operands and pushes the result ('t' or 'f') back onto the stack. Characters other than ')', and ',' are pushed directly onto the stack. Finally, the stack will contain a single element, which is the result of the entire expression.

Step-by-Step Algorithm

Step 1: Initialize an empty stack to store characters encountered while parsing the expression.
Step 2: Iterate through the expression string character by character.
Step 3: If the current character is not a closing parenthesis ')' and not a comma ',', push it onto the stack.
Step 4: If the current character is a closing parenthesis ')', pop elements from the stack until an opening parenthesis '(' is encountered. Store these popped elements as operands in a set.
Step 5: Pop the opening parenthesis '('.
Step 6: Pop the operator ('!', '&', or '|') from the stack.
Step 7: Perform the boolean operation based on the operator and the operands collected.
Step 8: Push the result ('t' or 'f') back onto the stack.
Step 9: After processing all characters, the final result will be the only element remaining on the stack. Return true if it is 't', and false otherwise.

Key Insights

Insight 1: The problem is naturally recursive because the sub-expressions within the parentheses are also boolean expressions that need to be evaluated.
Insight 2: Using a stack data structure simplifies the parsing process by allowing us to keep track of the operators and operands as we traverse the expression string.
Insight 3: The ',' character acts as a separator, allowing for multiple operands to the same operator. The implementation ignores the comma and pushes other characters onto the stack.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Stack, Recursion.

Companies

Asked at: HiLabs.

Parsing A Boolean Expression | LeetCode 1106

Hard

String Stack Recursion

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.Stack;
import java.util.HashSet;
import java.util.Set;

class Solution {
    public boolean parseBoolExpr(String expression) {
        Stack<Character> stack = new Stack<>();
        for (char charr : expression.toCharArray()) {
            if (charr == ')') {
                Set<Character> operands = new HashSet<>();
                while (stack.peek() != '(') {
                    operands.add(stack.pop());
                }
                
                // Pop '('
                stack.pop();
                
                // Pop operator
                char op = stack.pop();
                
                if (op == '!') {
                    // NOT operation: one operand
                    // If 't' was the operand, 't' is in operands, result is 'f'.
                    // If 'f' was the operand, 't' is not in operands, result is 't'.
                    if (operands.contains('t')) {
                        stack.push('f');
                    } else {
                        stack.push('t');
                    }
                } else if (op == '&') {
                    // AND operation
                    // If any operand is 'f', the result is 'f'
                    if (operands.contains('f')) {
                        stack.push('f');
                    } else {
                        stack.push('t');
                    }
                } else if (op == '|') {
                    // OR operation
                    // If any operand is 't', the result is 't'
                    if (operands.contains('t')) {
                        stack.push('t');
                    } else {
                        stack.push('f');
                    }
            }
            } else if (charr != ',') {
                stack.push(charr);
            }
        }
        
        // The final result is the only item left on the stack
        return stack.pop() == 't';
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1106: Parsing A Boolean Expression - Detailed explanation and solution approaches.

Topics

String Stack Recursion

Companies

HiLabs

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1106

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Initialize an empty stack to store characters encountered while parsing the expression.
2
Step 2: Iterate through the expression string character by character.
3
Step 3: If the current character is not a closing parenthesis ')' and not a comma ',', push it onto the stack.
4
Step 4: If the current character is a closing parenthesis ')', pop elements from the stack until an opening parenthesis '(' is encountered. Store these popped elements as operands in a set.
5
Step 5: Pop the opening parenthesis '('.
6
Step 6: Pop the operator ('!', '&', or '|') from the stack.
7
Step 7: Perform the boolean operation based on the operator and the operands collected.
8
Step 8: Push the result ('t' or 'f') back onto the stack.
9
Step 9: After processing all characters, the final result will be the only element remaining on the stack. Return true if it is 't', and false otherwise.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because each character in the input expression is visited and processed at most a constant number of times. The space complexity is linear because the stack's size can grow proportionally to the length of the input expression, especially in deeply nested expressions.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Descent Parser

A recursive descent parser could be used to directly translate the grammar of the boolean expression into recursive functions. Each function would be responsible for parsing a specific part of the expression (e.g., an operator, a sub-expression).

Time: O(n), where n is the length of the expression string.Space: O(d), where d is the maximum depth of the expression tree (due to recursion stack). In the worst case, d can be n if the expression is deeply nested linearly.

Trade-offs: Recursive descent parsing might be more readable and easier to understand conceptually for some, as it directly mirrors the grammar. However, it can be less efficient in terms of memory usage due to the recursion stack, especially for deeply nested expressions, compared to the iterative stack-based approach.

✨

Key Insights

1Insight 1: The problem is naturally recursive because the sub-expressions within the parentheses are also boolean expressions that need to be evaluated.
2Insight 2: Using a stack data structure simplifies the parsing process by allowing us to keep track of the operators and operands as we traverse the expression string.
3Insight 3: The ',' character acts as a separator, allowing for multiple operands to the same operator. The implementation ignores the comma and pushes other characters onto the stack.

📚

Key Terms Explained

For Students

Recursion

A function that calls itself to solve smaller versions of the same problem.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Simple expression like 't' or 'f'. The algorithm correctly pushes the value onto the stack and returns the corresponding boolean value.
Edge case 2: Expression with nested operators, like '&(|(f))'. The algorithm correctly parses the inner expression first, reducing the complexity as it moves outwards.
Edge case 3: Expression with multiple operands, such as '&(f,t,f)'. The algorithm uses a set to efficiently handle any number of operands for a given operator.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling operator precedence. It is crucial to evaluate sub-expressions within parentheses before applying operators outside of them.

✗Common mistake 2: Failing to properly handle different numbers of operands for the AND and OR operators. The solution must correctly evaluate these operators regardless of how many operands are present.

✗Common mistake 3: Not using a Set to collect operands which impacts the time complexity of AND and OR evaluation.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Compilers use similar parsing techniques to evaluate complex boolean conditions in if-else statements or while loops.

🏦Real-world scenario 2: Configuration files often use boolean expressions to define settings or rules. A parser is needed to interpret these expressions and apply the corresponding configurations.

🎮Real-world scenario 3: Game development uses boolean expressions for AI decision making and conditional logic for game events.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 27, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Parsing A Boolean Expression"?

For this hard-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?