How do you solve Additive Number on LeetCode?

Additive Number (LeetCode #306) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: The sequence must have at least three numbers.

What is the brute force solution for Additive Number?

The brute force approach for Additive Number is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible pair of starting numbers and check if the subsequent numbers follow the additive sequence rule. This approach is simple but can be inefficient due to the number of combinations we need to check. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Additive Number?

Additive Number uses the following concepts: String, Backtracking. The recommended patterns to study are: Backtracking, String Manipulation.

What are the interview tips for Additive Number?

Explain your thought process clearly while coding. Consider edge cases like leading zeros early on. Practice writing recursive functions to get comfortable with backtracking.

What are common mistakes when solving Additive Number?

Not handling leading zeros correctly. Failing to check all combinations of the first two numbers.

#306

Additive Number

Medium

StringBacktrackingBacktrackingString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

We can optimize the brute force approach by using backtracking to validate the sequence more efficiently. Instead of checking all combinations, we will directly build the sequence and validate it as we go.

⚙️

Algorithm

3 steps

1Step 1: Define a recursive function that takes the current index, the first two numbers, and the string.
2Step 2: Check if the current index has reached the end of the string and if we have at least three numbers in the sequence.
3Step 3: Generate the next number by adding the first two numbers and check if it matches the substring starting from the current index.

solution.py24 lines

1# Full working Python code
2
3def isAdditiveNumber(num):
4    def backtrack(index, first, second):
5        if index == len(num):
6            return True
7        next_num = str(int(first) + int(second))
8        if not num.startswith(next_num, index):
9            return False
10        return backtrack(index + len(next_num), second, next_num)
11
12    n = len(num)
13    for i in range(1, n):
14        for j in range(i + 1, n):
15            first = num[:i]
16            second = num[i:j]
17            if (len(first) > 1 and first[0] == '0') or (len(second) > 1 and second[0] == '0'):
18                continue
19            if backtrack(j, first, second):
20                return True
21    return False
22
23# Example usage
24print(isAdditiveNumber('112358'))  # Output: True

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops for selecting the first two numbers, but the recursive backtracking reduces unnecessary checks. The space complexity is O(n) because of the recursive call stack.

1The sequence must have at least three numbers.
2Leading zeros are not allowed unless the number is '0'.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.