What is the time complexity of Parse Lisp Expression?

The optimal solution has a time complexity of O(N^2).

What data structures are used to solve Parse Lisp Expression?

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

What are the key insights for solving Parse Lisp Expression?

Insight 1: Recursion is essential to handle the nested structure of Lisp expressions. Each sub-expression can be evaluated independently, and the results combined based on the operator ('let', 'add', 'mult'). Insight 2: A dictionary (or hash map) is crucial for managing variable scopes. When evaluating a variable, we need to search for its value in the current scope and, if not found, in outer scopes.

Which companies ask LeetCode 736. Parse Lisp Expression in interviews?

This problem is frequently asked at Affirm, Attentive.

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Parse Lisp Expression - LeetCode 736 Solution

Q: What is the space complexity of Parse Lisp Expression?

The space complexity is O(N^2).

Parse Lisp Expression - Complete Solution Guide

Parse Lisp Expression is LeetCode problem 736, 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

You are given a string expression representing a Lisp-like expression to return the integer value of. The syntax for these expressions is given as follows. An expression is either an integer, let expression, add expression, mult expression, or an assigned variable. Expressions always evaluate to a single integer. (An integer could be positive or negative.) A let expression takes the form "(let v 1 e 1 v 2 e 2 ... v n e n expr)" , where let is always the string "let" , then there are one or more

Detailed Explanation

The problem asks us to evaluate a Lisp-like expression represented as a string. These expressions can be integers, 'let' expressions (variable assignments and evaluation of a final expression), 'add' expressions (sum of two expressions), or 'mult' expressions (product of two expressions). The key challenge is handling the scoping of variables within 'let' expressions, where the innermost scope takes precedence. We need to recursively parse the expression string, evaluate sub-expressions, and manage variable scopes to determine the final integer value of the entire expression.

Solution Approach

The solution uses a recursive approach to parse and evaluate the Lisp expression. The core idea is to break down the expression into smaller sub-expressions, evaluate them recursively, and combine the results according to the operator. A dictionary is used to maintain variable scopes, allowing the solution to correctly handle variable assignments and lookups. The `_parse` helper function is used to tokenize the expression string based on spaces, taking into account parenthesis balance.

Step-by-Step Algorithm

Step 1: `_parse(expression_content)`: This helper function tokenizes the expression string, splitting it into individual tokens (operators, variable names, numbers, or sub-expressions) based on spaces, while correctly handling nested parentheses.
Step 2: `_evaluate_recursive(expr, scope)`: This recursive function is the heart of the solution. It first checks if the current `expr` is a simple integer or a variable. If it's an integer, it's converted to an integer and returned. If it's a variable, its value is looked up in the current `scope` and returned.
Step 3: If the `expr` starts with '(', it's a compound expression. The `_parse` function is used to tokenize the content inside the parentheses.
Step 4: Determine the operation. If it's 'add', the two operands are recursively evaluated, and their sum is returned. If it's 'mult', the two operands are recursively evaluated, and their product is returned.
Step 5: If the operation is 'let', a new scope is created as a copy of the current scope. The 'let' expression may contain variable assignments. For each variable assignment, the expression associated with the variable is evaluated recursively within the new scope, and the variable and its value are added to the new scope.
Step 6: After processing all variable assignments in the 'let' expression, the final expression is evaluated recursively within the new scope, and the result is returned. The new scope is discarded.
Step 7: The initial call to `_evaluate_recursive` starts with the entire expression and an empty scope.
Step 8: The function handles negative numbers by checking for the '-' sign. Integers are identified using `isdigit()` (and handling the leading '-' for negative numbers).

Key Insights

Insight 1: Recursion is essential to handle the nested structure of Lisp expressions. Each sub-expression can be evaluated independently, and the results combined based on the operator ('let', 'add', 'mult').
Insight 2: A dictionary (or hash map) is crucial for managing variable scopes. When evaluating a variable, we need to search for its value in the current scope and, if not found, in outer scopes.
Insight 3: Proper parsing of the expression string is vital. Identifying tokens (integers, operators, variable names) correctly is necessary for subsequent evaluation.

Complexity Analysis

Time Complexity: O(N^2)

Space Complexity: O(N^2)

Topics

This problem involves: Hash Table, String, Stack, Recursion.

Companies

Asked at: Affirm, Attentive.

Parse Lisp Expression | LeetCode 736

Hard

Hash Table String Stack Recursion

Time: O(N^2)Space: O(N^2)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public int evaluate(String expression) {
        return _evaluate_recursive(expression, new HashMap<>());
    }

    private List<String> _parse(String expression_content) {
        List<String> tokens = new ArrayList<>();
        int balance = 0;
        int start = 0;
        for (int i = 0; i < expression_content.length(); i++) {
            char charAtI = expression_content.charAt(i);
            if (charAtI == '(') {
                balance++;
            } else if (charAtI == ')') {
                balance--;
            } else if (charAtI == ' ' && balance == 0) {
                tokens.add(expression_content.substring(start, i));
                start = i + 1;
            }
        }
        tokens.add(expression_content.substring(start));
        return tokens;
    }

    private int _evaluate_recursive(String expr, Map<String, Integer> scope) {
        if (!expr.startsWith("(")) {
            if (expr.matches("-?\\d+")) {
                return Integer.parseInt(expr);
            }
            return scope.get(expr);
        }

        String content = expr.substring(1, expr.length() - 1);
        List<String> tokens = _parse(content);
        String op = tokens.get(0);

        if (op.equals("add")) {
            int e1 = _evaluate_recursive(tokens.get(1), scope);
            int e2 = _evaluate_recursive(tokens.get(2), scope);
            return e1 + e2;
        } else if (op.equals("mult")) {
            int e1 = _evaluate_recursive(tokens.get(1), scope);
            int e2 = _evaluate_recursive(tokens.get(2), scope);
            return e1 * e2;
        } else if (op.equals("let")) {
            Map<String, Integer> let_scope = new HashMap<>(scope);
            for (int i = 1; i < tokens.size() - 2; i += 2) {
                String var_name = tokens.get(i);
                String var_expr = tokens.get(i + 1);
                int var_value = _evaluate_recursive(var_expr, let_scope);
                let_scope.put(var_name, var_value);
            }

            return _evaluate_recursive(tokens.get(tokens.size() - 1), let_scope);
        }

        return 0; 
    }
}

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Problem Info

Description

Problem 736: Parse Lisp Expression - Detailed explanation and solution approaches.

Topics

Hash Table String Stack Recursion

Companies

AffirmAttentive

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:736

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

8 steps

1
Step 1: `_parse(expression_content)`: This helper function tokenizes the expression string, splitting it into individual tokens (operators, variable names, numbers, or sub-expressions) based on spaces, while correctly handling nested parentheses.
2
Step 2: `_evaluate_recursive(expr, scope)`: This recursive function is the heart of the solution. It first checks if the current `expr` is a simple integer or a variable. If it's an integer, it's converted to an integer and returned. If it's a variable, its value is looked up in the current `scope` and returned.
3
Step 3: If the `expr` starts with '(', it's a compound expression. The `_parse` function is used to tokenize the content inside the parentheses.
4
Step 4: Determine the operation. If it's 'add', the two operands are recursively evaluated, and their sum is returned. If it's 'mult', the two operands are recursively evaluated, and their product is returned.
5
Step 5: If the operation is 'let', a new scope is created as a copy of the current scope. The 'let' expression may contain variable assignments. For each variable assignment, the expression associated with the variable is evaluated recursively within the new scope, and the variable and its value are added to the new scope.
6
Step 6: After processing all variable assignments in the 'let' expression, the final expression is evaluated recursively within the new scope, and the result is returned. The new scope is discarded.
7
Step 7: The initial call to `_evaluate_recursive` starts with the entire expression and an empty scope.
8
Step 8: The function handles negative numbers by checking for the '-' sign. Integers are identified using `isdigit()` (and handling the leading '-' for negative numbers).

💡

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^2)

💾

Space Complexity

O(N^2)

📝

Why?

The time complexity is O(N^2) where N is the length of the expression. The `_parse` function iterates through the string, and the `_evaluate_recursive` function can call `_parse` and itself recursively. In the worst-case scenario, the length of the string to parse or evaluate halves in each level of recursion, resulting in O(N) levels of recursion, and `_parse` take O(N) time. Also, the `let` expression can lead to the creation of `N` new scopes, contributing O(N) complexity to each recursion. The space complexity is O(N^2) because in the worst case, the scope can have O(N) variables, and at each of O(N) recursion levels, a new copy of the scope is created for let expressions. Hence, the complexity becomes O(N*N).

📖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

Iterative Approach with Stack

An iterative approach using a stack to manage the scopes and the parsing process could be implemented. This would avoid the overhead of recursive function calls.

Time: O(N^2)Space: O(N^2)

Trade-offs: The iterative approach might be more complex to implement and understand than the recursive approach, especially when dealing with the nested structure of the expressions. However, it could be more efficient in terms of space usage if the stack is managed carefully.

✨

Key Insights

1Insight 1: Recursion is essential to handle the nested structure of Lisp expressions. Each sub-expression can be evaluated independently, and the results combined based on the operator ('let', 'add', 'mult').
2Insight 2: A dictionary (or hash map) is crucial for managing variable scopes. When evaluating a variable, we need to search for its value in the current scope and, if not found, in outer scopes.
3Insight 3: Proper parsing of the expression string is vital. Identifying tokens (integers, operators, variable names) correctly is necessary for subsequent evaluation.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

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: Negative integers. The code correctly handles negative integers by checking for the '-' sign before digits and using `atoi` (C) or `stoi` (C++) and `Integer.parseInt` (Java) or by checking the first character (Python).
Edge case 2: Empty expression. The problem statement specifies that there are no leading or trailing spaces, and all tokens are separated by a single space. The code handles the general structure without explicit handling of empty strings as input, because the constraints prevent the case. However, it is important to have error handling in real-world applications.
Edge case 3: Deeply nested let expressions. The recursive approach handles deeply nested 'let' expressions correctly because each 'let' expression creates a new scope, ensuring that variables are resolved in the correct order.
Edge case 4: Variables shadowing variables from outer scopes. 'let' expressions create new scopes, shadowing variables from outer scopes when a variable name is redefined. The recursive calls ensure that variables are looked up in the innermost scope first.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly parsing the expression string. Failing to handle spaces and parentheses correctly will lead to incorrect tokenization and evaluation.

✗Common mistake 2: Not handling variable scoping correctly. Failing to create a new scope for each 'let' expression will result in incorrect variable lookups and assignments.

✗Common mistake 3: Not handling integer conversions carefully. Incorrectly converting strings to integers or not considering negative integers will lead to wrong results.

✗Common mistake 4: Forgetting to copy the scope when entering a 'let' block will cause variables from the outer scope to be modified.

✅

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: Interpreters and compilers for programming languages use similar techniques to parse and evaluate expressions. This problem provides a simplified version of the core logic involved in these systems.

🏦Real-world scenario 2: Configuration file parsing. Configuration files often have a hierarchical structure and variable assignments. A similar parsing and evaluation process can be used to interpret these files.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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 17, 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

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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.

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 "Parse Lisp Expression"?

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

🤖 Need help with this problem?