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The goal is to split a given string S into a Fibonacci-like sequence, where all the numbers follow the properties of the Fibonacci sequence. A sequence is Fibonacci-like if the sequence is split into F[0], F[1], ..., F[n] such that:

Return any such sequence of split numbers as a list of integers.

If it is not possible to split S into a Fibonacci-like sequence, return an empty list.

Example:

Constraints:

Clarifying Questions

  1. Can a number in the sequence begin with a ‘0’? No, unless the number itself is 0.
  2. What about the leading zeros in the string itself? Each number in the sequence should be valid, according to general numeric rules.
  3. Is an empty input string a valid input? No, S is given to have at least one digit as per constraints.

Strategy

  1. We’re going to use a recursive backtracking approach to explore different splits of the string into potential Fibonacci sequences.
  2. We can iterate through possible first and second numbers in the sequence, then generate subsequent numbers and check if they form a valid Fibonacci sequence.
  3. We will terminate early if a number with leading zeros (other than 0 itself) is formed or if the numbers grow too large (greater than the maximum integer value).

Code

def splitIntoFibonacci(S: str):
    def is_valid_number(number_str):
        # To prevent large integer issues and leading zero issues
        return len(number_str) == 1 or (number_str[0] != '0' and int(number_str) < 2**31)
    
    def backtrack(index, path):
        if index == len(S) and len(path) >= 3:
            return path
        
        current_number = 0
        for i in range(index, len(S)):
            current_number = current_number * 10 + int(S[i])
            if not is_valid_number(S[index:i+1]):
                break
            if len(path) >= 2 and current_number != path[-1] + path[-2]:
                if current_number > path[-1] + path[-2]:
                    break
                continue
            path.append(current_number)
            result = backtrack(i + 1, path)
            if result:
                return result
            path.pop()
        return []
    
    return backtrack(0, [])

# Example usage:
# print(splitIntoFibonacci("123456579")) # Outputs: [123, 456, 579]

Time Complexity

The time complexity for this backtracking approach is difficult to precisely define due to the combination of substring generation and recursive tree exploration. However, in the worst case:

A rough upper bound can be O(N^2 * 2^N), which should be acceptable given the input constraint 1 <= len(S) <= 200.

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