What is the time complexity of String to Integer (atoi)?

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

What data structures are used to solve String to Integer (atoi)?

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

What are the key insights for solving String to Integer (atoi)?

The problem is essentially string parsing with careful handling of edge cases and potential integer overflow. We need to iterate through the string character by character, keeping track of the sign and the current number being built.

Which companies ask LeetCode 8. String to Integer (atoi) in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Databricks and 11 more companies.

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String to Integer (atoi) - LeetCode 8 Solution

String to Integer (atoi) - Complete Solution Guide

String to Integer (atoi) is LeetCode problem 8, 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

Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer. The algorithm for myAtoi(string s) is as follows: Whitespace : Ignore any leading whitespace ( " " ). Signedness : Determine the sign by checking if the next character is '-' or '+' , assuming positivity if neither present. Conversion : Read the integer by skipping leading zeros until a non-digit character is encountered or the end of the string is reached. If no digits were read, then the result is 0. R

Detailed Explanation

The problem asks us to implement a function, `myAtoi(string s)`, that converts a given string `s` into a 32-bit signed integer, following specific rules similar to the `atoi` function in C/C++. The function should first ignore any leading whitespace characters. Then, it should check for an optional sign character ('+' or '-'). After that, it should read in digits until a non-digit character is encountered. Finally, it must handle potential integer overflow by clamping the result to the 32-bit signed integer range [-2<sup>31</sup>, 2<sup>31</sup> - 1]. If no digits are read or if the string is empty after removing whitespace, the function should return 0.

Solution Approach

The solution involves iterating through the string and applying the rules outlined in the problem description. First, leading whitespace is skipped. Next, the sign is determined (defaulting to positive if no sign is present). Then, the algorithm iterates through the string, accumulating digits into a number. Before adding a digit to the number, it checks whether adding the digit would cause an integer overflow. If an overflow is detected, the function returns `INT_MAX` or `INT_MIN` accordingly. If the loop finishes without overflow, the resulting number, adjusted for the sign, is returned.

Step-by-Step Algorithm

Step 1: Initialize index `i` to 0 and iterate through the string `s` to skip leading whitespace characters. Increment `i` until a non-whitespace character is found or the end of the string is reached.
Step 2: Check for a sign character ('+' or '-') at the current index `i`. If a '+' is found, increment `i`. If a '-' is found, set the `sign` variable to -1 and increment `i`. If neither is found, the sign remains the default value of 1 (positive).
Step 3: Initialize a variable `num` to 0 to store the resulting integer.
Step 4: Iterate through the string starting from the current index `i` as long as the current character is a digit.
Step 5: Inside the loop, convert the current digit character to an integer.
Step 6: Before adding the new digit to `num`, check for potential integer overflow. If the current number `num` is greater than `INT_MAX // 10` (or `INT_MIN // 10`) or `num` is equal to `INT_MAX // 10` (or `INT_MIN // 10`) and the digit is greater than `INT_MAX % 10` (or `INT_MAX % 10 + 1` for negative numbers), return `INT_MAX` or `INT_MIN` based on the sign.
Step 7: Update `num` by multiplying it by 10 and adding the current digit: `num = num * 10 + digit`.
Step 8: Increment the index `i` to move to the next character.
Step 9: After the loop finishes (either reaching the end of the string or encountering a non-digit character), return the final result as `sign * num`.

Key Insights

The problem is essentially string parsing with careful handling of edge cases and potential integer overflow.
We need to iterate through the string character by character, keeping track of the sign and the current number being built.
The most crucial part is to check for integer overflow before multiplying the current number by 10 and adding the new digit. We need to compare the current number with `INT_MAX // 10` (or `INT_MIN // 10`) to determine if multiplying by 10 would cause an overflow.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Databricks, Goldman Sachs, Google, Infosys, Meta, Microsoft, Niantic, Nvidia, Qualcomm, Uber, Valve, Yahoo.

String to Integer (atoi) | LeetCode 8

Medium

String

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int myAtoi(String s) {
        int INT_MAX = (int)Math.pow(2, 31) - 1;
        int INT_MIN = (int)Math.pow(-2, 31);

        int i = 0;
        int n = s.length();

        while (i < n && s.charAt(i) == ' ') {
            i += 1;
        }

        int sign = 1;
        if (i < n && s.charAt(i) == '+') {
            i += 1;
        } else if (i < n && s.charAt(i) == '-') {
            sign = -1;
            i += 1;
        }

        int num = 0;
        while (i < n && Character.isDigit(s.charAt(i))) {
            int digit = s.charAt(i) - '0';
            
            if (sign == 1) {
                if (num > INT_MAX / 10 || (num == INT_MAX / 10 && digit > INT_MAX % 10)) {
                    return INT_MAX;
                }
            } else {
                if (num > INT_MAX / 10 || (num == INT_MAX / 10 && digit > (INT_MAX % 10) + 1)) {
                    return INT_MIN;
                }
            }
            
            num = num * 10 + digit;
            i += 1;
            
        }
        
        return sign * num;
    }
}

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Problem Info

Description

Problem 8: String to Integer (atoi) - Detailed explanation and solution approaches.

Topics

String

Companies

AdobeAmazonAppleBloombergDatabricksGoldman SachsGoogleInfosys

Stats

Difficulty:Medium

Languages:4

Companies:16

Problem #:8

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 index `i` to 0 and iterate through the string `s` to skip leading whitespace characters. Increment `i` until a non-whitespace character is found or the end of the string is reached.
2
Step 2: Check for a sign character ('+' or '-') at the current index `i`. If a '+' is found, increment `i`. If a '-' is found, set the `sign` variable to -1 and increment `i`. If neither is found, the sign remains the default value of 1 (positive).
3
Step 3: Initialize a variable `num` to 0 to store the resulting integer.
4
Step 4: Iterate through the string starting from the current index `i` as long as the current character is a digit.
5
Step 5: Inside the loop, convert the current digit character to an integer.
6
Step 6: Before adding the new digit to `num`, check for potential integer overflow. If the current number `num` is greater than `INT_MAX // 10` (or `INT_MIN // 10`) or `num` is equal to `INT_MAX // 10` (or `INT_MIN // 10`) and the digit is greater than `INT_MAX % 10` (or `INT_MAX % 10 + 1` for negative numbers), return `INT_MAX` or `INT_MIN` based on the sign.
7
Step 7: Update `num` by multiplying it by 10 and adding the current digit: `num = num * 10 + digit`.
8
Step 8: Increment the index `i` to move to the next character.
9
Step 9: After the loop finishes (either reaching the end of the string or encountering a non-digit character), return the final result as `sign * num`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the number of characters we potentially iterate through in the string. The space complexity is constant because the space required does not scale with the size of the input string. No additional data structures are created that depend on the size of input string.

📖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

Using Regular Expressions

Regular expressions can be used to extract the relevant part of the string before converting to an integer. This approach might be more concise but potentially less efficient.

Time: O(n) in the worst case, where n is the length of the string (due to regex matching). However, the performance of regex can vary depending on the engine and the complexity of the pattern.Space: O(1) or O(n), depending on the regex engine and whether it creates a new string as a result of the match. Some engines might operate in-place.

Trade-offs: Regular expressions can make the code shorter, but they can also be less readable and potentially less performant than the explicit iteration approach. Use this when conciseness is preferred, and performance is not critical.

✨

Key Insights

1The problem is essentially string parsing with careful handling of edge cases and potential integer overflow.
2We need to iterate through the string character by character, keeping track of the sign and the current number being built.
3The most crucial part is to check for integer overflow before multiplying the current number by 10 and adding the new digit. We need to compare the current number with `INT_MAX // 10` (or `INT_MIN // 10`) to determine if multiplying by 10 would cause an overflow.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Empty string: The initial whitespace skipping handles the case where the input string is empty or contains only whitespace, resulting in a return value of 0.
String with only whitespace: Similar to the empty string case, the solution correctly returns 0.
String starting with non-digit characters (e.g., 'words and 987'): The algorithm stops parsing when it encounters the first non-digit character, so it correctly handles cases where the string doesn't start with a number, resulting in a return value of 0.
String with '+' or '-' but no following digits (e.g., '+', '-'): Because the digit parsing part requires an `isdigit()` check, no number will be formed, and the result will be 0.
Integer overflow: The solution includes a check for potential integer overflow before each digit is added, ensuring that the result is always within the 32-bit signed integer range.
Leading zeros: The algorithm correctly handles leading zeros by ignoring them during the digit accumulation process. The result is not affected by leading zeros like '00042'.

⚠️

Common Mistakes to Avoid

✗Not handling leading whitespace: Forgetting to skip leading whitespace characters can lead to incorrect results.

✗Not handling the sign correctly: Failing to properly determine and apply the sign can cause incorrect conversion to a positive or negative number.

✗Integer overflow: Not checking for potential integer overflow can result in incorrect results or runtime errors. The critical step is checking before multiplication and addition in each iteration.

✗Assuming that the string will always contain a valid integer: The algorithm needs to handle cases where the string doesn't contain a valid integer representation (e.g., empty string, string with only non-digit characters).

✗Incorrectly using INT_MAX and INT_MIN in the overflow check. It is vital to check `num > INT_MAX // 10` or `num == INT_MAX // 10 and digit > INT_MAX % 10` to avoid potential overflows.

✅

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?

💼Data parsing: This algorithm is used in many applications that involve parsing numerical data from strings, such as reading configuration files, processing user input, or extracting data from text documents.

🏦Web development: Converting user input from text fields into numerical values for calculations or database storage often requires similar string-to-integer conversion logic.

🎮Financial applications: Processing financial data, such as stock prices or account balances, often involves converting strings to numbers with careful attention to precision and range.

💡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

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

⚠️ 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 9, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 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 techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "String to Integer (atoi)"?

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

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