What is the time complexity of Basic Calculator IV?

The optimal solution has a time complexity of O(n^3).

What is the space complexity of Basic Calculator IV?

The space complexity is O(n^2).

What data structures are used to solve Basic Calculator IV?

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

What are the key insights for solving Basic Calculator IV?

Insight 1: Polynomial representation: The key is to represent intermediate results as polynomials, which are collections of terms (coefficient and variables). Use a data structure that allows efficiently storing and manipulating these polynomials. Insight 2: Operator precedence: Handle operator precedence using a stack-based approach similar to standard calculator evaluation.

Which companies ask LeetCode 770. Basic Calculator IV in interviews?

This problem is frequently asked at Intuit, Roblox.

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

Basic Calculator IV - LeetCode 770 Solution

Basic Calculator IV - Complete Solution Guide

Basic Calculator IV is LeetCode problem 770, 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

Given an expression such as expression = "e + 8 - a + 5" and an evaluation map such as {"e": 1} (given in terms of evalvars = ["e"] and evalints = [1] ), return a list of tokens representing the simplified expression, such as ["-1*a","14"] An expression alternates chunks and symbols, with a space separating each chunk and symbol. A chunk is either an expression in parentheses, a variable, or a non-negative integer. A variable is a string of lowercase letters (not including digits.) Note that var

Detailed Explanation

The problem, Basic Calculator IV, requires you to evaluate a given algebraic expression string containing variables, integers, and operators (+, -, *). You are also provided with a mapping of variable names to integer values. The goal is to simplify the expression, substituting the known variable values and performing the calculations. The final output should be a list of terms, each representing a part of the simplified polynomial, adhering to specific formatting rules: terms are sorted by degree (number of variables), then lexicographically, with coefficients placed before the variables separated by '*'. Terms with zero coefficients should not be included in the result. The expression follows operator precedence (parentheses, multiplication, addition/subtraction), and tokens are space-separated.

Solution Approach

The solution uses a stack-based approach to evaluate the expression while representing intermediate results as polynomials. The code parses the expression token by token. Numbers are directly converted into polynomials with a constant term. Variables are either substituted with their values from the provided mapping (creating constant polynomials) or treated as single-variable polynomials. Operators are handled using two stacks: one for values (polynomials) and one for operators. The `precedence` function determines the priority of operators, ensuring correct evaluation order. Whenever a ')' is encountered or an operator with lower precedence is found on the stack, operations are applied until the correct precedence is restored. Finally, the resulting polynomial is formatted into the requested list of strings.

Step-by-Step Algorithm

Step 1: Initialize data structures: Create a `Poly` class (or equivalent in other languages) to represent polynomials as a map of variable lists to coefficients, and stacks for values (polynomials) and operators.
Step 2: Create a mapping of variable names to integer values (eval_map).
Step 3: Tokenize the expression string by splitting it at spaces.
Step 4: Iterate through each token:
Step 5: If the token is a number, create a `Poly` representing the number and push it onto the value stack.
Step 6: If the token is a variable, check if it exists in the `eval_map`. If it does, create a `Poly` representing the integer value. Otherwise, create a `Poly` representing the variable with a coefficient of 1.
Step 7: If the token is an opening parenthesis '(', push it onto the operator stack.
Step 8: If the token is a closing parenthesis ')', evaluate all operators on the operator stack until an opening parenthesis is encountered. Pop the opening parenthesis from the operator stack.
Step 9: If the token is an operator (+, -, *), evaluate all operators on the operator stack with higher or equal precedence than the current operator. Then, push the current operator onto the operator stack.
Step 10: After processing all tokens, evaluate all remaining operators on the operator stack.
Step 11: Convert the final `Poly` result to a list of strings in the specified format.
Step 12: The `Poly` class will contain methods to perform addition, subtraction, and multiplication of polynomials. Multiplication creates combinations of variables, and these combinations must be sorted and added together.
Step 13: The `toList()` method takes the polynomial and converts it into the desired string format. The terms should be sorted as follows. The number of variables in the term (degree) is most important. Terms with higher degree comes first. When the degree is the same, the terms are sorted lexicographically.
Step 14: Skip terms with zero coefficients in the final result.

Key Insights

Insight 1: Polynomial representation: The key is to represent intermediate results as polynomials, which are collections of terms (coefficient and variables). Use a data structure that allows efficiently storing and manipulating these polynomials.
Insight 2: Operator precedence: Handle operator precedence using a stack-based approach similar to standard calculator evaluation.
Insight 3: Term sorting and formatting: The output format has strict requirements, demanding careful attention to sorting the terms first by degree (number of free variables), then lexicographically.
Insight 4: Substitution: Substitute the values of known variables before processing.

Complexity Analysis

Time Complexity: O(n^3)

Space Complexity: O(n^2)

Topics

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

Companies

Asked at: Intuit, Roblox.

Basic Calculator IV | LeetCode 770

Hard

Hash Table Math String Stack Recursion

Time: O(n^3)Space: O(n^2)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<String> basicCalculatorIV(String expression, String[] evalvars, int[] evalints) {
        class Poly extends HashMap<List<String>, Integer> {
            public Poly() {
                super();
            }

            public Poly(Poly other) {
                super(other);
            }

            public Poly add(Poly other) {
                Poly res = new Poly(this);
                for (Map.Entry<List<String>, Integer> entry : other.entrySet()) {
                    List<String> v = entry.getKey();
                    int c = entry.getValue();
                    res.put(v, res.getOrDefault(v, 0) + c);
                }
                return res;
            }

            public Poly subtract(Poly other) {
                Poly res = new Poly(this);
                for (Map.Entry<List<String>, Integer> entry : other.entrySet()) {
                    List<String> v = entry.getKey();
                    int c = entry.getValue();
                    res.put(v, res.getOrDefault(v, 0) - c);
                }
                return res;
            }

            public Poly multiply(Poly other) {
                Poly res = new Poly();
                for (Map.Entry<List<String>, Integer> entry1 : this.entrySet()) {
                    List<String> v1 = entry1.getKey();
                    int c1 = entry1.getValue();
                    for (Map.Entry<List<String>, Integer> entry2 : other.entrySet()) {
                        List<String> v2 = entry2.getKey();
                        int c2 = entry2.getValue();

                        List<String> new_vars = new ArrayList<>();
                        new_vars.addAll(v1);
                        new_vars.addAll(v2);
                        Collections.sort(new_vars);

                        res.put(new_vars, res.getOrDefault(new_vars, 0) + c1 * c2);
                    }
                }
                return res;
            }

            public List<String> toList() {
                List<Map.Entry<List<String>, Integer>> terms = new ArrayList<>();
                for (Map.Entry<List<String>, Integer> entry : this.entrySet()) {
                    if (entry.getValue() != 0) {
                        terms.add(entry);
                    }
                }

                terms.sort((a, b) -> {
                    if (a.getKey().size() != b.getKey().size()) {
                        return -Integer.compare(a.getKey().size(), b.getKey().size());
                    } else {
                        return compareLists(a.getKey(), b.getKey());
                    }
                });

                List<String> result = new ArrayList<>();
                for (Map.Entry<List<String>, Integer> entry : terms) {
                    List<String> v = entry.getKey();
                    int c = entry.getValue();
                    if (v.isEmpty()) {
                        result.add(String.valueOf(c));
                    } else {
                        StringBuilder sb = new StringBuilder();
                        sb.append(c).append("*");
                        for (int i = 0; i < v.size(); i++) {
                            sb.append(v.get(i));
                            if (i < v.size() - 1) {
                                sb.append("*");
                            }
                        }
                        result.add(sb.toString());
                    }
                }
                return result;
            }
        }

        Map<String, Integer> eval_map = new HashMap<>();
        for (int i = 0; i < evalvars.length; i++) {
            eval_map.put(evalvars[i], evalints[i]);
        }

        List<Poly> values = new ArrayList<>();
        List<String> ops = new ArrayList<>();

        Function<String, Integer> precedence = op -> {
            if (op.equals("+") || op.equals("-")) return 1;
            if (op.equals("*")) return 2;
            return 0;
        };

        Runnable apply_op = () -> {
            String op = ops.remove(ops.size() - 1);
            Poly right = values.remove(values.size() - 1);
            Poly left = values.remove(values.size() - 1);
            if (op.equals("+")) values.add(left.add(right));
            else if (op.equals("-")) values.add(left.subtract(right));
            else if (op.equals("*")) values.add(left.multiply(right));
        };

        for (String token : expression.split(" ")) {
            if (token.matches("\\d+")) {
                Poly p = new Poly();
                p.put(new ArrayList<>(), Integer.parseInt(token));
                values.add(p);
            } else if (token.matches("[a-zA-Z]+")) {
                if (eval_map.containsKey(token)) {
                    Poly p = new Poly();
                    p.put(new ArrayList<>(), eval_map.get(token));
                    values.add(p);
                } else {
                    Poly p = new Poly();
                    List<String> varList = new ArrayList<>();
                    varList.add(token);
                    p.put(varList, 1);
                    values.add(p);
                }
            } else if (token.equals("(")) {
                ops.add(token);
            } else if (token.equals(")")) {
                while (!ops.get(ops.size() - 1).equals("(")) {
                    apply_op.run();
                }
                ops.remove(ops.size() - 1);
            } else { // operator
                while (!ops.isEmpty() && !ops.get(ops.size() - 1).equals("(") &&
                        precedence.apply(ops.get(ops.size() - 1)) >= precedence.apply(token)) {
                    apply_op.run();
                }
                ops.add(token);
            }
        }

        while (!ops.isEmpty()) {
            apply_op.run();
        }

        return values.isEmpty() ? new ArrayList<>() : values.get(0).toList();
    }
    
    private int compareLists(List<String> list1, List<String> list2) {
        for (int i = 0; i < Math.min(list1.size(), list2.size()); i++) {
            int comparison = list1.get(i).compareTo(list2.get(i));
            if (comparison != 0) {
                return comparison;
            }
        }
        return Integer.compare(list1.size(), list2.size());
    }

    interface Function<T, R> {
        R apply(T t);
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 770: Basic Calculator IV - Detailed explanation and solution approaches.

Topics

Hash Table Math String Stack Recursion

Companies

IntuitRoblox

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:770

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

14 steps

1
Step 1: Initialize data structures: Create a `Poly` class (or equivalent in other languages) to represent polynomials as a map of variable lists to coefficients, and stacks for values (polynomials) and operators.
2
Step 2: Create a mapping of variable names to integer values (eval_map).
3
Step 3: Tokenize the expression string by splitting it at spaces.
4
Step 4: Iterate through each token:
5
Step 5: If the token is a number, create a `Poly` representing the number and push it onto the value stack.
6
Step 6: If the token is a variable, check if it exists in the `eval_map`. If it does, create a `Poly` representing the integer value. Otherwise, create a `Poly` representing the variable with a coefficient of 1.
7
Step 7: If the token is an opening parenthesis '(', push it onto the operator stack.
8
Step 8: If the token is a closing parenthesis ')', evaluate all operators on the operator stack until an opening parenthesis is encountered. Pop the opening parenthesis from the operator stack.
9
Step 9: If the token is an operator (+, -, *), evaluate all operators on the operator stack with higher or equal precedence than the current operator. Then, push the current operator onto the operator stack.
10
Step 10: After processing all tokens, evaluate all remaining operators on the operator stack.
11
Step 11: Convert the final `Poly` result to a list of strings in the specified format.
12
Step 12: The `Poly` class will contain methods to perform addition, subtraction, and multiplication of polynomials. Multiplication creates combinations of variables, and these combinations must be sorted and added together.
13
Step 13: The `toList()` method takes the polynomial and converts it into the desired string format. The terms should be sorted as follows. The number of variables in the term (degree) is most important. Terms with higher degree comes first. When the degree is the same, the terms are sorted lexicographically.
14
Step 14: Skip terms with zero coefficients in the final result.

💡

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^3)

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The polynomial multiplication dominates the time complexity. Each term in a polynomial can consist of multiple variables, and sorting/manipulating those variable lists during multiplication adds to the overall complexity. Space complexity is dominated by the size of the polynomials being stored. In the worst case, an intermediate polynomial might have a number of terms that is proportional to n^2.

📖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

Instead of using stacks, a recursive descent parser can be implemented. This approach defines functions for each grammar rule (expression, term, factor), recursively parsing the expression. Polynomials are still the core data structure for results.

Time: O(n^3)Space: O(n^2)

Trade-offs: Recursive descent parsers can be more readable and easier to understand for some, but can be more difficult to implement correctly due to potential stack overflow issues in deeply nested expressions. Space complexity for recursive calls also should be considered.

✨

Key Insights

1Insight 1: Polynomial representation: The key is to represent intermediate results as polynomials, which are collections of terms (coefficient and variables). Use a data structure that allows efficiently storing and manipulating these polynomials.
2Insight 2: Operator precedence: Handle operator precedence using a stack-based approach similar to standard calculator evaluation.
3Insight 3: Term sorting and formatting: The output format has strict requirements, demanding careful attention to sorting the terms first by degree (number of free variables), then lexicographically.
4Insight 4: Substitution: Substitute the values of known variables before processing.

📚

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: Empty expression: The code should gracefully handle an empty expression string by returning an empty list.
Edge case 2: Expression with only a constant: The code should correctly evaluate an expression consisting of only a number, such as "5", and return a list containing only the number.
Edge case 3: Expression with only a variable: The code must handle cases where the expression is a single variable, and properly substitute or represent it in the output.
Edge case 4: Terms with zero coefficients: The code must correctly omit terms with zero coefficients from the final output.
Edge case 5: Expressions with unbalanced parentheses: Problem statement guarantees that the expression is valid, so the solutions do not need to explicitly handle unbalanced parentheses.
Edge case 6: Overflow: The problem guarantees that intermediate results will be in the range of [-2^31, 2^31 -1], so solutions don't have to handle overflow.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect operator precedence: Failing to handle operator precedence correctly can lead to incorrect evaluation of the expression.

✗Common mistake 2: Incorrect term sorting: Not sorting the terms by degree and lexicographically as required will lead to incorrect output format.

✗Common mistake 3: Including terms with zero coefficients: The code must explicitly filter out terms with zero coefficients from the final result.

✗Common mistake 4: Memory management in C: Forgetting to `free` dynamically allocated memory in C implementations leads to memory leaks.

✗Common mistake 5: Failing to copy terms in C: Make sure to allocate space and copy strings when creating new terms from existing terms rather than copying the pointer. This could lead to double frees, etc.

✅

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: Symbolic computation systems: This algorithm is a simplified version of what is used in symbolic computation systems like Mathematica or Maple for simplifying and manipulating algebraic expressions.

🏦Real-world scenario 2: Optimizing compilers: Compilers use similar techniques for constant folding and algebraic simplification to optimize code.

🎮Real-world scenario 3: Financial modeling: Similar techniques can be used for simplifying complex financial models involving multiple variables and formulas.

💡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

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

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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: July 21, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

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

@AlgoMasterJul 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 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 "Basic Calculator IV"?

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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?