842. Split Array into Fibonacci Sequence

Explanation:

To solve this problem, we can use a backtracking approach to recursively try out different combinations of numbers to form the Fibonacci-like sequence. We start by choosing the first two numbers, then recursively check if the next number in the sequence can be formed from the sum of the previous two numbers. If we reach the end of the input string and have a valid Fibonacci-like sequence, we return it.

We iterate through different combinations of the first two numbers.
For each pair of numbers, we try to build the Fibonacci-like sequence.
We check if the next number in the sequence can be formed from the sum of the previous two numbers.
Recursively continue this process until we reach the end of the input string.
If a valid Fibonacci-like sequence is found, return it. Otherwise, return an empty list.

Time Complexity: O(2^N), where N is the length of the input string num. In the worst-case scenario, we explore all possible combinations of numbers.

Space Complexity: O(N), where N is the length of the input string num. The space is used for the recursion stack.

:

View Leetcode Problem

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

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

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Explanation:

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