What is the time complexity of Basic Calculator?

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

What data structures are used to solve Basic Calculator?

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

What are the key insights for solving Basic Calculator?

Insight 1: The use of a stack is crucial for handling the nested parentheses. The stack stores intermediate results and signs when encountering an opening parenthesis and retrieves them when encountering a closing parenthesis. Insight 2: We need to keep track of the current sign (positive or negative) to handle subtractions and additions correctly. The sign changes when encountering the '-' operator and needs to be applied to subsequent numbers until another operator is encountered.

Which companies ask LeetCode 224. Basic Calculator in interviews?

This problem is frequently asked at Anduril, ByteDance, Citadel, Coupang, DE Shaw and 7 more companies.

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Basic Calculator - LeetCode 224 Solution

Q: How to solve Basic Calculator?

The provided solutions use a stack to keep track of intermediate results and signs when encountering parentheses. The algorithm iterates through the string character by character. It maintains a current number, result, and sign. When a digit is encountered, it's appended to the current number. When an operator ('+' or '-') is encountered, the current number (multiplied by the current sign) is added to the result, and the number and sign are reset. When an opening parenthesis is encountered, the current result and sign are pushed onto the stack, and the result and sign are reset. When a closing parenthesis is encountered, the current number (multiplied by the current sign) is added to the result, the previous sign and result are popped from the stack, and the result is updated to reflect the value of the expression within the parentheses.

Basic Calculator - Complete Solution Guide

Basic Calculator is LeetCode problem 224, 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 a string s representing a valid expression, implement a basic calculator to evaluate it, and return the result of the evaluation . Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval() . Example 1: Input: s = "1 + 1" Output: 2 Example 2: Input: s = " 2-1 + 2 " Output: 3 Example 3: Input: s = "(1+(4+5+2)-3)+(6+8)" Output: 23 Constraints: 1 <= s.length <= 3 * 10 5 s consists of digits, '+' , '-' , '(' , ')' , and ' ' . s re

Detailed Explanation

The problem asks us to implement a basic calculator that can evaluate a given string expression. The expression consists of digits, '+', '-', '(', ')', and spaces. The calculator should return the integer result of the evaluation. We are not allowed to use built-in functions like `eval()` to directly evaluate the string.

Solution Approach

The provided solutions use a stack to keep track of intermediate results and signs when encountering parentheses. The algorithm iterates through the string character by character. It maintains a current number, result, and sign. When a digit is encountered, it's appended to the current number. When an operator ('+' or '-') is encountered, the current number (multiplied by the current sign) is added to the result, and the number and sign are reset. When an opening parenthesis is encountered, the current result and sign are pushed onto the stack, and the result and sign are reset. When a closing parenthesis is encountered, the current number (multiplied by the current sign) is added to the result, the previous sign and result are popped from the stack, and the result is updated to reflect the value of the expression within the parentheses.

Step-by-Step Algorithm

Step 1: Initialize a stack, result, sign (to 1), and num (to 0).
Step 2: Iterate through the input string character by character.
Step 3: If the character is a digit, update the current number: `num = num * 10 + int(char)`.
Step 4: If the character is '+', update the result: `result += sign * num`, reset `num = 0`, and set `sign = 1`.
Step 5: If the character is '-', update the result: `result += sign * num`, reset `num = 0`, and set `sign = -1`.
Step 6: If the character is '(', push the current `result` and `sign` onto the stack, reset `result = 0`, and set `sign = 1`.
Step 7: If the character is ')', update the result: `result += sign * num`, reset `num = 0`. Pop the previous `sign` and `result` from the stack. Update the `result` to be `prev_result + prev_sign * result`.
Step 8: After iterating through the entire string, if `num` is not 0, update the `result`: `result += sign * num`.
Step 9: Return the final `result`.

Key Insights

Insight 1: The use of a stack is crucial for handling the nested parentheses. The stack stores intermediate results and signs when encountering an opening parenthesis and retrieves them when encountering a closing parenthesis.
Insight 2: We need to keep track of the current sign (positive or negative) to handle subtractions and additions correctly. The sign changes when encountering the '-' operator and needs to be applied to subsequent numbers until another operator is encountered.
Insight 3: The problem constraints state that '+' is not a unary operator but '-' can be. This allows us to handle cases where the number is preceded by a '-'. The sign needs to be correctly managed to reflect this unary operation.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Anduril, ByteDance, Citadel, Coupang, DE Shaw, DoorDash, Oracle, Palo Alto Networks, Ripple, Rokt, Snowflake, Tesla.

Basic Calculator | LeetCode 224

Hard

Math String Stack Recursion

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

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public int calculate(String s) {
        Stack<Integer> stack = new Stack<>();
        int result = 0;
        int sign = 1;
        int num = 0;

        for (char charr : s.toCharArray()) {
            if (Character.isDigit(charr)) {
                num = num * 10 + (charr - '0');
            } else if (charr == '+') {
                result += sign * num;
                num = 0;
                sign = 1;
            } else if (charr == '-') {
                result += sign * num;
                num = 0;
                sign = -1;
            } else if (charr == '(') {
                stack.push(result);
                stack.push(sign);
                result = 0;
                sign = 1;
            } else if (charr == ')') {
                result += sign * num;
                num = 0;

                int prev_sign = stack.pop();
                int prev_result = stack.pop();

                result = prev_result + prev_sign * result;
            }
        }

        if (num != 0) {
            result += sign * num;
        }

        return result;
    }
}

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Problem Info

Description

Problem 224: Basic Calculator - Detailed explanation and solution approaches.

Topics

Math String Stack Recursion

Companies

AndurilByteDanceCitadelCoupangDE ShawDoorDashOraclePalo Alto Networks

Stats

Difficulty:Hard

Languages:4

Companies:12

Problem #:224

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 a stack, result, sign (to 1), and num (to 0).
2
Step 2: Iterate through the input string character by character.
3
Step 3: If the character is a digit, update the current number: `num = num * 10 + int(char)`.
4
Step 4: If the character is '+', update the result: `result += sign * num`, reset `num = 0`, and set `sign = 1`.
5
Step 5: If the character is '-', update the result: `result += sign * num`, reset `num = 0`, and set `sign = -1`.
6
Step 6: If the character is '(', push the current `result` and `sign` onto the stack, reset `result = 0`, and set `sign = 1`.
7
Step 7: If the character is ')', update the result: `result += sign * num`, reset `num = 0`. Pop the previous `sign` and `result` from the stack. Update the `result` to be `prev_result + prev_sign * result`.
8
Step 8: After iterating through the entire string, if `num` is not 0, update the `result`: `result += sign * num`.
9
Step 9: Return 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)

👍 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 is processed in constant time. The space complexity is determined by the stack, which stores intermediate results and signs for nested parentheses. In the worst case, the number of opening parentheses can be proportional to the length of the string, resulting in a linear space complexity.

📖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

Recursion

A recursive approach can also be used to solve this problem. The idea is to recursively evaluate the expression within each set of parentheses. You'd need an index to keep track of the current position in the string.

Time: O(n)Space: O(n) in the worst case due to recursion depth (nested parentheses), but in average cases might be lower.

Trade-offs: Recursion might be more intuitive for some, but it can be less efficient due to function call overhead and may lead to stack overflow for very deep nested expressions. The iterative approach with a stack is generally preferred.

✨

Key Insights

1Insight 1: The use of a stack is crucial for handling the nested parentheses. The stack stores intermediate results and signs when encountering an opening parenthesis and retrieves them when encountering a closing parenthesis.
2Insight 2: We need to keep track of the current sign (positive or negative) to handle subtractions and additions correctly. The sign changes when encountering the '-' operator and needs to be applied to subsequent numbers until another operator is encountered.
3Insight 3: The problem constraints state that '+' is not a unary operator but '-' can be. This allows us to handle cases where the number is preceded by a '-'. The sign needs to be correctly managed to reflect this unary operation.

📚

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: Empty string or string with only whitespace. The provided solution correctly handles these cases as the loop will simply not execute and return 0.
Edge case 2: Leading or trailing spaces. The solution handles these because it iterates char by char, skipping whitespace. Therefore, extra whitespace characters simply do nothing and do not affect the outcome.
Edge case 3: A single number such as "123". The logic correctly parses the number, stores it, and returns it after the main loop.
Edge case 4: Nested parentheses, such as "(1+(4+5+2)-3)+(6+8)". The use of the stack handles nested parentheses correctly by storing intermediate results and signs.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the sign correctly, especially with subtraction. Forgetting to apply the sign to the number before adding it to the result can lead to incorrect calculations.

✗Common mistake 2: Incorrectly handling parentheses. Forgetting to store and retrieve the previous result and sign when encountering parentheses can lead to incorrect evaluation of nested expressions.

✗Common mistake 3: Not resetting the number after processing an operator. This will cause the number to be accumulated across multiple terms, leading to incorrect results.

✗Common mistake 4: Assuming maximum stack size to be low and not handling possible overflows in C implementation.

✅

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: Evaluating mathematical expressions in calculators or programming languages. This algorithm can be used as a foundation for implementing more complex calculators or expression parsers.

🏦Real-world scenario 2: Parsing and interpreting command-line arguments or configuration files. Many configuration files use expressions that need to be evaluated, and this algorithm can be used for that purpose.

🎮Real-world scenario 3: Data analysis and scientific computing where formulas must be interpreted for calculation.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

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

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary 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: July 15, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

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

@DevExpert92Jul 7

❓

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.

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"?

For this hard-level Math 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?