What is the time complexity of Add Strings?

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

What is the space complexity of Add Strings?

The space complexity is O(max(n, m)).

What data structures are used to solve Add Strings?

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

What are the key insights for solving Add Strings?

Insight 1: The core idea is to simulate the manual addition process, starting from the least significant digits (rightmost characters) of the input strings. Insight 2: Maintaining a `carry` variable is essential to handle cases where the sum of two digits exceeds 9.

Which companies ask LeetCode 415. Add Strings in interviews?

This problem is frequently asked at Airbnb, Avito, Capital One, Jane Street, Wayfair.

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Add Strings - LeetCode 415 Solution

Q: How to solve Add Strings?

The provided solutions simulate the addition of two numbers represented as strings. They iterate through the strings from right to left, adding the corresponding digits along with any carry from the previous addition. The resulting digits are then appended to a string builder (or similar data structure) to form the final sum string. The key is to handle the carry correctly and to ensure that shorter strings are treated as if they are padded with leading zeros.

Add Strings - Complete Solution Guide

Add Strings is LeetCode problem 415, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given two non-negative integers, num1 and num2 represented as string, return the sum of num1 and num2 as a string . You must solve the problem without using any built-in library for handling large integers (such as BigInteger ). You must also not convert the inputs to integers directly. Example 1: Input: num1 = "11", num2 = "123" Output: "134" Example 2: Input: num1 = "456", num2 = "77" Output: "533" Example 3: Input: num1 = "0", num2 = "0" Output: "0" Constraints: 1 <= num1.length, num2.length

Detailed Explanation

The problem requires us to add two non-negative integers, `num1` and `num2`, which are given as strings. We must return the sum of these two numbers also as a string. The crucial constraint is that we cannot use built-in libraries for handling large integers or directly convert the input strings to integers. This forces us to simulate the addition process digit by digit, similar to how we perform addition manually.

Solution Approach

The provided solutions simulate the addition of two numbers represented as strings. They iterate through the strings from right to left, adding the corresponding digits along with any carry from the previous addition. The resulting digits are then appended to a string builder (or similar data structure) to form the final sum string. The key is to handle the carry correctly and to ensure that shorter strings are treated as if they are padded with leading zeros.

Step-by-Step Algorithm

Step 1: Initialize pointers `i` and `j` to the last index of `num1` and `num2`, respectively. Also, initialize `carry` to 0 and create an empty string or string builder to store the result.
Step 2: Iterate as long as either `i` or `j` is non-negative or the `carry` is greater than 0.
Step 3: In each iteration, extract the digits from `num1` and `num2` at the current indices `i` and `j`. If an index is out of bounds (i.e., `i < 0` or `j < 0`), treat the digit as 0.
Step 4: Calculate the sum of the two digits and the `carry`.
Step 5: Calculate the new `carry` by dividing the sum by 10. The remainder of the division is the digit to be appended to the result.
Step 6: Append the digit (remainder) to the result string/string builder.
Step 7: Decrement `i` and `j` to move to the next digits (from right to left).
Step 8: If using a string builder, reverse the resulting string before returning it. Otherwise, if concatenating to a string, prepend to the result.

Key Insights

Insight 1: The core idea is to simulate the manual addition process, starting from the least significant digits (rightmost characters) of the input strings.
Insight 2: Maintaining a `carry` variable is essential to handle cases where the sum of two digits exceeds 9.
Insight 3: The length of the result can be at most one more than the maximum length of the two input strings (due to a possible carry-over at the end).

Complexity Analysis

Time Complexity: O(max(n, m))

Space Complexity: O(max(n, m))

Topics

This problem involves: Math, String, Simulation.

Companies

Asked at: Airbnb, Avito, Capital One, Jane Street, Wayfair.

Add Strings | LeetCode 415

Easy

Math String Simulation

Time: O(max(n, m))Space: O(max(n, m))

Solution Code

Available in:

class Solution {
    public String addStrings(String num1, String num2) {
        StringBuilder sb = new StringBuilder();
        int i = num1.length() - 1;
        int j = num2.length() - 1;
        int carry = 0;
        while (i >= 0 || j >= 0 || carry > 0) {
            int digit1 = i >= 0 ? num1.charAt(i) - '0' : 0;
            int digit2 = j >= 0 ? num2.charAt(j) - '0' : 0;
            int sum = digit1 + digit2 + carry;
            sb.append(sum % 10);
            carry = sum / 10;
            i--;
            j--;
        }
        return sb.reverse().toString();
    }
}

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Problem Info

Description

Problem 415: Add Strings - Detailed explanation and solution approaches.

Topics

Math String Simulation

Companies

AirbnbAvitoCapital OneJane StreetWayfair

Stats

Difficulty:Easy

Languages:4

Companies:5

Problem #:415

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: Initialize pointers `i` and `j` to the last index of `num1` and `num2`, respectively. Also, initialize `carry` to 0 and create an empty string or string builder to store the result.
2
Step 2: Iterate as long as either `i` or `j` is non-negative or the `carry` is greater than 0.
3
Step 3: In each iteration, extract the digits from `num1` and `num2` at the current indices `i` and `j`. If an index is out of bounds (i.e., `i < 0` or `j < 0`), treat the digit as 0.
4
Step 4: Calculate the sum of the two digits and the `carry`.
5
Step 5: Calculate the new `carry` by dividing the sum by 10. The remainder of the division is the digit to be appended to the result.
6
Step 6: Append the digit (remainder) to the result string/string builder.
7
Step 7: Decrement `i` and `j` to move to the next digits (from right to left).
8
Step 8: If using a string builder, reverse the resulting string before returning it. Otherwise, if concatenating to a string, prepend to 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(max(n, m))

💾

Space Complexity

O(max(n, m))

📝

Why?

The algorithm iterates through each digit of both numbers performing constant-time operations (addition, modulo, division). The length of the resulting string is directly proportional to the length of the input strings, hence the 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

Convert to Integer (Not Allowed)

A naive approach would be to convert both strings to integers, add them, and then convert the result back to a string. However, this is prohibited by the problem constraints because the inputs can be very large numbers exceeding the typical integer limits.

Time: Potentially O(n) or O(m) for conversion, O(1) for addition, and O(log(result)) for string conversion, but not applicable due to the problem constraints.Space: O(1) for the addition, O(log(result)) for result string conversion, but not applicable due to problem constraints.

Trade-offs: This approach is simpler to implement if allowed, but it's not suitable for large numbers and violates the problem's constraints.

✨

Key Insights

1Insight 1: The core idea is to simulate the manual addition process, starting from the least significant digits (rightmost characters) of the input strings.
2Insight 2: Maintaining a `carry` variable is essential to handle cases where the sum of two digits exceeds 9.
3Insight 3: The length of the result can be at most one more than the maximum length of the two input strings (due to a possible carry-over at the end).

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: One of the numbers is "0". The solution correctly handles this as the loop continues based on the length of the other number and any remaining carry.
Edge case 2: Both numbers are "0". The solution correctly returns "0" in this case.
Edge case 3: One number is significantly longer than the other. The solution gracefully handles this by treating digits beyond the end of the shorter number as 0.
Edge case 4: The sum results in a carry-over at the most significant digit. The solution correctly handles this by continuing the loop until the carry becomes 0, effectively adding an extra digit to the result (e.g., 999 + 1 = 1000).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the carry. The carry is crucial for correctly adding the digits, especially when the sum of two digits is greater than or equal to 10.

✗Common mistake 2: Incorrectly handling strings of different lengths. It's important to treat the shorter string as if it's padded with leading zeros.

✗Common mistake 3: Forgetting to reverse the string (for solutions that append digits to the result). Since the digits are processed from right to left, the result needs to be reversed to obtain the correct order.

✗Common mistake 4: Not handling the edge case where the sum of the most significant digits results in a carry-over. The algorithm must continue processing until the carry becomes 0.

✅

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: Arbitrary-precision arithmetic in scientific computing, where very large numbers need to be handled without loss of precision.

🏦Real-world scenario 2: Cryptography, where large numbers are used for encryption and decryption, and algorithms for arithmetic operations on these numbers are essential.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

This is one of the first problems I solved when learning Math. It's a great foundation for more complex problems.

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 26, 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 14

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

@AlgoMasterJul 7

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

@InterviewProJun 30

❓

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Add Strings"?

For this easy-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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