What is the time complexity of Basic Calculator II?

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

What data structures are used to solve Basic Calculator II?

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

What are the key insights for solving Basic Calculator II?

Insight 1: The key insight is that we can handle multiplication and division immediately as we encounter them, due to their higher precedence. Addition and subtraction can be handled later. Insight 2: Using a stack helps to store the intermediate results and operators. Specifically, we can push numbers onto the stack based on the current operator, and then sum the stack at the end.

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Basic Calculator II - LeetCode 227 Solution

Q: Which companies ask LeetCode 227. Basic Calculator II in interviews?

This problem is frequently asked at Accenture, Airbnb, Anduril, ByteDance, Coupang and 15 more companies.

Basic Calculator II - Complete Solution Guide

Basic Calculator II is LeetCode problem 227, a Medium 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 which represents an expression, evaluate this expression and return its value . The integer division should truncate toward zero. You may assume that the given expression is always valid. All intermediate results will be in the range of [-2 31 , 2 31 - 1] . Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval() . Example 1: Input: s = "3+2*2" Output: 7 Example 2: Input: s = " 3/2 " Output: 1 Example 3: Input: s

Detailed Explanation

The problem asks us to implement a basic calculator that can evaluate a string expression containing non-negative integers and the operators +, -, *, and /. The integer division should truncate towards zero. We are not allowed to use built-in functions like `eval()`. The expression is guaranteed to be valid, and intermediate results will be within the range of a 32-bit integer.

Solution Approach

The solution iterates through the input string `s`. It maintains a `current_number` and an `operator`. When a digit is encountered, it's used to build the current number. When an operator or the end of the string is reached, the `current_number` is processed based on the last encountered `operator`. If the operator is `+` or `-`, the `current_number` (or its negation) is pushed onto the stack. If the operator is `*` or `/`, the top element of the stack is popped, the corresponding operation is performed with `current_number`, and the result is pushed back onto the stack. Finally, the sum of all elements in the stack is returned.

Step-by-Step Algorithm

Step 1: Initialize an empty stack, `current_number` to 0, and `operator` to '+'.
Step 2: Iterate through the input string `s`.
Step 3: If the current character is a digit, update `current_number` by multiplying it by 10 and adding the digit.
Step 4: If the current character is an operator or it is the end of the string, process the `current_number` based on the `operator`:
Step 5: If `operator` is '+', push `current_number` onto the stack.
Step 6: If `operator` is '-', push `-current_number` onto the stack.
Step 7: If `operator` is '*', pop the top element from the stack, multiply it by `current_number`, and push the result back onto the stack.
Step 8: If `operator` is '/', pop the top element from the stack, divide it by `current_number` (integer division), and push the result back onto the stack.
Step 9: Update `operator` to the current character and reset `current_number` to 0.
Step 10: After iterating through the entire string, sum up all the elements in the stack and return the result.

Key Insights

Insight 1: The key insight is that we can handle multiplication and division immediately as we encounter them, due to their higher precedence. Addition and subtraction can be handled later.
Insight 2: Using a stack helps to store the intermediate results and operators. Specifically, we can push numbers onto the stack based on the current operator, and then sum the stack at the end.
Insight 3: We need to handle spaces properly and consider the last character of the string when it's a digit.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Math, String, Stack.

Companies

Asked at: Accenture, Airbnb, Anduril, ByteDance, Coupang, DE Shaw, DoorDash, Google, Highspot, IXL, NetApp, Rokt, ServiceNow, Snap, Snowflake, Tesla, The Trade Desk, Verkada, Zoho, Zoox.

Basic Calculator II | LeetCode 227

Medium

Math String Stack

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 currentNumber = 0;
        char operator = '+';
        int sLen = s.length();

        for (int i = 0; i < sLen; i++) {
            char currentChar = s.charAt(i);

            if (Character.isDigit(currentChar)) {
                currentNumber = currentNumber * 10 + (currentChar - '0');
            }

            if ((!Character.isDigit(currentChar) && !Character.isWhitespace(currentChar)) || i == sLen - 1) {
                if (operator == '+') {
                    stack.push(currentNumber);
                } else if (operator == '-') {
                    stack.push(-currentNumber);
                } else if (operator == '*') {
                    int lastNum = stack.pop();
                    stack.push(lastNum * currentNumber);
                } else if (operator == '/') {
                    int lastNum = stack.pop();
                    stack.push(lastNum / currentNumber);
                }

                operator = currentChar;
                currentNumber = 0;
            }
        }

        int sum = 0;
        while (!stack.isEmpty()) {
            sum += stack.pop();
        }

        return sum;
    }
}

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Problem Info

Description

Problem 227: Basic Calculator II - Detailed explanation and solution approaches.

Topics

Math String Stack

Companies

AccentureAirbnbAndurilByteDanceCoupangDE ShawDoorDashGoogle

Stats

Difficulty:Medium

Languages:4

Companies:20

Problem #:227

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

10 steps

1
Step 1: Initialize an empty stack, `current_number` to 0, and `operator` to '+'.
2
Step 2: Iterate through the input string `s`.
3
Step 3: If the current character is a digit, update `current_number` by multiplying it by 10 and adding the digit.
4
Step 4: If the current character is an operator or it is the end of the string, process the `current_number` based on the `operator`:
5
Step 5: If `operator` is '+', push `current_number` onto the stack.
6
Step 6: If `operator` is '-', push `-current_number` onto the stack.
7
Step 7: If `operator` is '*', pop the top element from the stack, multiply it by `current_number`, and push the result back onto the stack.
8
Step 8: If `operator` is '/', pop the top element from the stack, divide it by `current_number` (integer division), and push the result back onto the stack.
9
Step 9: Update `operator` to the current character and reset `current_number` to 0.
10
Step 10: After iterating through the entire string, sum up all the elements in the stack and return the 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 directly proportional to the length of the input string because each character is processed at most a constant number of times. The space complexity is determined by the maximum size of the stack, which depends on the structure of the input expression. An expression with many consecutive additions or subtractions will result in a larger stack.

📖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

Two stacks approach

Use two stacks: one for numbers and one for operators. Handle precedence of operators while pushing onto the operator stack. When reaching the end of the string or encountering a lower precedence operator, perform the operations on the top of the stack until the precedence condition is met.

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

Trade-offs: This approach is conceptually closer to operator precedence rules but can be more complex to implement correctly.

✨

Key Insights

1Insight 1: The key insight is that we can handle multiplication and division immediately as we encounter them, due to their higher precedence. Addition and subtraction can be handled later.
2Insight 2: Using a stack helps to store the intermediate results and operators. Specifically, we can push numbers onto the stack based on the current operator, and then sum the stack at the end.
3Insight 3: We need to handle spaces properly and consider the last character of the string when it's a digit.

📚

Key Terms Explained

For Students

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: Input string contains only a single number. The solution correctly handles this because `current_number` will be built, and the loop finishes. The stack will contain only that number.
Edge case 2: Leading or trailing spaces in the string. The `isspace()` check ignores spaces.
Edge case 3: Consecutive operators are not allowed as the problem stated that the expression is valid.
Edge case 4: Division by zero is avoided due to problem constraints saying all integers are non-negative and the expression is valid.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the last number in the string if it's not followed by an operator. The solution handles this by including the condition `i == s_len - 1` in the check to process `current_number`.

✗Common mistake 2: Incorrectly handling operator precedence. Failing to perform multiplication and division before addition and subtraction will lead to incorrect results. The solution avoids this by performing multiplication and division immediately when encountered.

✗Common mistake 3: Not handling spaces correctly can lead to parsing errors. The solution incorporates an `isspace()` check to skip spaces.

✗Common mistake 4: Integer overflow during intermediate calculations. This problem specifies that results will fit in a 32-bit integer.

✅

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: Implementing compilers or interpreters for programming languages requires parsing and evaluating expressions.

🏦Real-world scenario 2: Scientific calculators and spreadsheet applications need to evaluate mathematical expressions.

🎮Real-world scenario 3: Evaluating mathematical formulas entered by users in various software applications.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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 18, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Basic Calculator II"?

For this medium-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.

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