What is the time complexity of Additive Number?

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

What data structures are used to solve Additive Number?

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

What are the key insights for solving Additive Number?

Insight 1: The problem can be solved by systematically trying all possible combinations for the first two numbers in the sequence. Insight 2: Backtracking can be implicitly used by checking the validity of the sequence formed by the first two numbers and recursively checking if the rest of the string matches the subsequent sum of the two numbers.

Which companies ask LeetCode 306. Additive Number in interviews?

This problem is frequently asked at Epic Systems.

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

Additive Number - LeetCode 306 Solution

Additive Number - Complete Solution Guide

Additive Number is LeetCode problem 306, 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

An additive number is a string whose digits can form an additive sequence . A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two. Given a string containing only digits, return true if it is an additive number or false otherwise. Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid. Example 1: Input: "112358" Output: tru

Detailed Explanation

The problem asks us to determine if a given string of digits can form an additive sequence. An additive sequence consists of at least three numbers where, except for the first two, each subsequent number is the sum of the preceding two. The string should be parsed such that leading zeros in any number of the sequence are not allowed, unless the number is zero itself. We need to return `true` if such a sequence exists within the given string, and `false` otherwise.

Solution Approach

The provided solution uses a nested loop approach to iterate through all possible starting positions for the first two numbers of a potential additive sequence. For each combination, it checks if the remaining part of the string can be formed by repeatedly summing the last two numbers and comparing the sum as a string with the remaining substring. The 'startsWith' or equivalent method is used to check if the substring equals to the new sum string.

Step-by-Step Algorithm

Step 1: Iterate through all possible lengths for the first number in the sequence using a loop (outer loop with index i).
Step 2: Check for leading zeros in the first number. If leading zeros exist (and the number is not 0), the sequence is invalid, and we continue to the next potential first number.
Step 3: Iterate through all possible lengths for the second number in the sequence (inner loop with index j).
Step 4: Check for leading zeros in the second number. If leading zeros exist (and the number is not 0), the sequence is invalid, and we continue to the next potential second number.
Step 5: Convert the first and second numbers (substrings) to integers (or long long to handle larger numbers).
Step 6: Use a while loop to check if the rest of the string forms an additive sequence. Calculate the sum of the first and second numbers and check if the remaining string starts with this sum string.
Step 7: If the remaining string does not start with the expected sum, the current combination is invalid and continue to the next combination (first number and second number pair).
Step 8: If the remaining string does start with the sum, update the indices and the first and second number variables and repeat.
Step 9: If the while loop completes (meaning the entire string has been parsed successfully), then return True because a valid additive sequence was found.
Step 10: If the loops complete without finding any valid sequence, return False.

Key Insights

Insight 1: The problem can be solved by systematically trying all possible combinations for the first two numbers in the sequence.
Insight 2: Backtracking can be implicitly used by checking the validity of the sequence formed by the first two numbers and recursively checking if the rest of the string matches the subsequent sum of the two numbers.
Insight 3: Leading zeros are invalid unless the number is just zero. Handling this case is crucial for correctness.

Complexity Analysis

Time Complexity: O(n^3)

Space Complexity: O(n)

Topics

This problem involves: String, Backtracking.

Companies

Asked at: Epic Systems.

Additive Number | LeetCode 306

Medium

String Backtracking

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

Solution Code

Available in:

class Solution {
    public boolean isAdditiveNumber(String num) {
        int n = num.length();
        
        // i is the exclusive end index of the first number
        for (int i = 1; i < n; i++) {
            String num1_str = num.substring(0, i);
            // A number with a leading zero is only valid if it's "0" itself.
            // If the first digit is '0', any first number longer than 1 is invalid.
            // Since we iterate i increasing, we can break.
            if (num1_str.length() > 1 && num1_str.charAt(0) == '0') {
                break;
            }

            // j is the exclusive end index of the second number
            for (int j = i + 1; j < n; j++) {
                String num2_str = num.substring(i, j);
                // Same leading zero check for the second number.
                // If the first digit of num2 is '0', any num2 longer than 1 is invalid.
                if (num2_str.length() > 1 && num2_str.charAt(0) == '0') {
                    break;
                }

                long n1 = Long.parseLong(num1_str);
                long n2 = Long.parseLong(num2_str);
                int k = j;  // k is the start index for the rest of the string
                
                boolean is_valid_sequence = true;
                while (k < n) {
                    long next_sum = n1 + n2;
                    String next_sum_str = String.valueOf(next_sum);
                    
                    // Check if the rest of the string starts with the calculated sum
                    if (!num.startsWith(next_sum_str, k)) {
                        is_valid_sequence = false;
                        break;
                    }
                    
                    // Move to the next part of the string and update the numbers
                    k += next_sum_str.length();
                    long temp = n2;
                    n2 = next_sum;
                    n1 = temp;
                }
                
                // If the loop completed and consumed the entire string
                // Since j starts from i+1 and goes up to n-1, k starts at a position
                // where there is at least one character left, guaranteeing we check
                // for at least a third number.
                if (is_valid_sequence) {
                    return true;
                }
            }
        }
        
        return false;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 306: Additive Number - Detailed explanation and solution approaches.

Topics

String Backtracking

Companies

Epic Systems

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:306

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: Iterate through all possible lengths for the first number in the sequence using a loop (outer loop with index i).
2
Step 2: Check for leading zeros in the first number. If leading zeros exist (and the number is not 0), the sequence is invalid, and we continue to the next potential first number.
3
Step 3: Iterate through all possible lengths for the second number in the sequence (inner loop with index j).
4
Step 4: Check for leading zeros in the second number. If leading zeros exist (and the number is not 0), the sequence is invalid, and we continue to the next potential second number.
5
Step 5: Convert the first and second numbers (substrings) to integers (or long long to handle larger numbers).
6
Step 6: Use a while loop to check if the rest of the string forms an additive sequence. Calculate the sum of the first and second numbers and check if the remaining string starts with this sum string.
7
Step 7: If the remaining string does not start with the expected sum, the current combination is invalid and continue to the next combination (first number and second number pair).
8
Step 8: If the remaining string does start with the sum, update the indices and the first and second number variables and repeat.
9
Step 9: If the while loop completes (meaning the entire string has been parsed successfully), then return True because a valid additive sequence was found.
10
Step 10: If the loops complete without finding any valid sequence, return False.

💡

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)

📦 Linear space - Memory grows with input size.

📝

Why?

The nested loops iterate through almost all possible pairs of first and second numbers, which creates an n^2 factor. Within the inner loop, we have a while loop to check if the sequence is valid which, in the worst case, needs to parse almost the entire string again, yielding an n factor. Therefore, the overall complexity is O(n^2 * n) = O(n^3).

📖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

Backtracking with explicit recursion

An alternative approach is to explicitly use recursion to check for the additive sequence. The function will recursively call itself after finding the first two numbers, reducing the problem to a smaller substring.

Time: O(n^3)Space: O(n) due to recursion depth

Trade-offs: Recursion can be more readable for some, but might be slightly slower due to function call overhead. It also faces potential stack overflow issues with extremely long inputs (although with input string constraint of length 35, it's not much of a problem).

✨

Key Insights

1Insight 1: The problem can be solved by systematically trying all possible combinations for the first two numbers in the sequence.
2Insight 2: Backtracking can be implicitly used by checking the validity of the sequence formed by the first two numbers and recursively checking if the rest of the string matches the subsequent sum of the two numbers.
3Insight 3: Leading zeros are invalid unless the number is just zero. Handling this case is crucial for correctness.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: Leading zeros. The solution explicitly checks for leading zeros and handles them correctly. Numbers with leading zeros, except for '0' itself, are considered invalid.
Edge case 2: Very large numbers. The solution uses `long long` (or `Long` in Java) to handle potential overflow. The problem constraints mentions num length <= 35, so we can use long long, which handles numbers up to 2^63 - 1. If numbers could become arbitrarily large, a string-based addition would be needed.
Edge case 3: Numbers '00', '01', etc. These should all be handled correctly as they will fail the leading zero checks in Step 2 and Step 4.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle leading zeros correctly. This leads to incorrect sequences being considered valid.

✗Common mistake 2: Integer overflow. If the numbers become too large, standard integer types can overflow, leading to incorrect sums and invalid results. Need to use `long long` or `Long`.

✗Common mistake 3: Not checking the condition that the sequence should have at least three numbers. The provided solutions check this implicitly due to loop constraints.

✅

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: Data compression algorithms. Identifying patterns in sequences is crucial for data compression.

🏦Real-world scenario 2: Cryptography. Analyzing patterns in encrypted data is used to break codes.

🎮Real-world scenario 3: Financial analysis. Identifying patterns in time series data to forecast future stock prices or economic trends.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: October 5, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 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 is the difference between backtracking and brute force?

Backtracking prunes invalid paths early using constraints, while brute force explores all possibilities. In backtracking, you "choose" an option, "explore" it recursively, then "unchoose" (backtrack) if it doesn't lead to a solution. This eliminates entire branches of the search tree.

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 "Additive Number"?

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

🤖 Need help with this problem?