Split Array into Fibonacci Sequence

Problem

You are given a string of digits num, such as "123456579". We can split it into a Fibonacci-like sequence [123, 456, 579].

Formally, a Fibonacci-like sequence is a list f of non-negative integers such that:

0 <= f[i] < 2^31, (that is, each integer fits in a 32-bit signed integer type),
f.length >= 3, and
f[i] + f[i + 1] == f[i + 2] for all 0 <= i < f.length - 2.

Note that when splitting the string into pieces, each piece must not have extra leading zeroes, except if the piece is the number 0 itself.

Return any Fibonacci-like sequence split from num, or return [] if it cannot be done.

Example 1:

Input: num = "1101111"
Output: [11,0,11,11]
Explanation: The output [110, 1, 111] would also be accepted.

Example 2:

Input: num = "112358130"
Output: []
Explanation: The task is impossible.

Example 3:

Input: num = "0123"
Output: []
Explanation: Leading zeroes are not allowed, so "01", "2", "3" is not valid.

Constraints:

1 <= num.length <= 200
num contains only digits.

Solution (Java)

class Solution {
    public List<Integer> splitIntoFibonacci(String num) {
        List<Integer> res = new ArrayList<>();
        solve(num, res, 0);
        return res;
    }

    private boolean solve(String s, List<Integer> res, int idx) {
        if (idx == s.length() && res.size() >= 3) {
            return true;
        }
        for (int i = idx; i < s.length(); i++) {
            if (s.charAt(idx) == '0' && i > idx) {
                return false;
            }
            long num = Long.parseLong(s.substring(idx, i + 1));
            if (num > Integer.MAX_VALUE) {
                return false;
            }
            int size = res.size();
            if (size >= 2 && num > res.get(size - 1) + res.get(size - 2)) {
                return false;
            }
            if (size <= 1 || num == res.get(size - 1) + res.get(size - 2)) {
                res.add((int) num);
                if (solve(s, res, i + 1)) {
                    return true;
                }
                res.remove(res.size() - 1);
            }
        }
        return false;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).