Solution Explanation for Additive Number

This problem asks whether a given string num can be partitioned into at least three numbers forming an additive sequence. An additive sequence means that each number (after the first two) is the sum of the previous two. Numbers cannot have leading zeros.

The solution employs a depth-first search (DFS) approach, which is a type of backtracking. It explores all possible partitions of the input string to check if they satisfy the additive sequence condition.

Approach

1. Iteration through possible first two numbers: The outer loops iterate through all possible lengths for the first two numbers in the sequence. i represents the length of the first number, and j represents the length of the second number. The loops ensure that these lengths are within bounds and handle the case where numbers can't have leading zeros.

2. Parsing first two numbers: The first two numbers (a and b) are parsed from the input string based on the lengths i and j - i.

3. Recursive DFS (dfs function): The dfs function recursively checks if the rest of the string forms an additive sequence based on the current a and b.

   • Base case: If the remaining string is empty (num is empty), it means a valid additive sequence has been found, so true is returned.
   • Leading zero check: If a + b is greater than 0 and the first digit of the remaining string is '0', it's an invalid sequence (leading zeros are not allowed), so false is returned.
   • Iteration through possible next numbers: The inner loop iterates through possible lengths of the next number. It tries to parse a number from the beginning of the remaining string.
   • Sum check: If the parsed number equals a + b, the dfs function is recursively called with the updated a and b (the second and third numbers become the new first and second).
   • Return value: If the recursive call returns true, it means a valid additive sequence has been found. Otherwise, the loop continues to try different lengths for the next number. If no valid next number is found, it returns false.

4. Return value: The main function returns true if a valid additive sequence is found during the iteration; otherwise, it returns false.

Time and Space Complexity Analysis

• Time Complexity: O(N^3), where N is the length of the input string. The nested loops have O(N^2) complexity. The recursive dfs function also iterates, in the worst case, through all the remaining characters in the string, adding another O(N) factor, resulting in the overall O(N^3) complexity.

• Space Complexity: O(N) in the worst case. This is due to the recursive calls in dfs, which can have a maximum depth of N in the worst case (a sequence where each number is a single digit). The space used is mainly for the call stack during recursion.

Code Improvements (Especially for Java and C++)

The provided Java and C++ solutions include a min(..., 19) constraint within their loops. This helps prevent stack overflow issues for extremely long input strings. This limitation means that the solution might miss some valid additive sequences if those sequences require numbers with more than 18 digits. Without this precaution, it is possible to hit stack overflow errors during recursion.

The Python code is slightly more concise because it implicitly handles potential integer overflow using Python's arbitrary-precision integers. The Java and C++ code uses long which still has a limitation of integer size. To handle potentially larger numbers, you would need to use a Big Integer library.