842. Split Array into Fibonacci Sequence

Given a string S of digits, such as S = "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 - 1, (that is, each integer fits 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.

Also, 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 S, or return [] if it cannot be done.

Example 1:

Input: "123456579"

Output: [123,456,579]

Example 2:

Input: "11235813"

Output: [1,1,2,3,5,8,13]

Example 3:

Input: "112358130"

Output: []

Explanation: The task is impossible.

Example 4:

Input: "0123"

Output: []

Explanation: Leading zeroes are not allowed, so "01", "2", "3" is not valid.

Example 5:

Input: "1101111"

Output: [110, 1, 111]

Explanation: The output [11, 0, 11, 11] would also be accepted.

Note:

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

Approach #1: C++.

class Solution {

public:

    vector<int> splitIntoFibonacci(string S) {

        vector<int> nums;

        helper(S, nums, 0);

        return nums;

    }

    bool helper(string S, vector<int>& nums, int start) {

        int len = S.length();

        // if we reached end of string & we have more than 2 elements

        // in our sequence then return true

        if (start >= len && nums.size() >= 3) return true;

        // since '0' in beginning is not allowed therefore if the current char is '0'

        // then we can use it alone only and cann't extend it by adding more chars at the back.

        // otherwise we make take upto 10 chars (length og MAX_INT)

        int maxLen = (S[start] == '0') ? 1 : 10;

        // Try getting a solution by forming a number with 'i' chars begging with 'start'

        for (int i = 1; i <= maxLen && start + i <= S.size(); ++i) {

            long long temp = stoll(S.substr(start, i));

            if (temp > INT_MAX) return false;

            int sz = nums.size();

            // If fibonacci property is not satisfied then we cann't get a solution

            if (sz >= 2 && nums[sz-1] + nums[sz-2] < temp) return false;

            if (sz <= 1 || nums[sz-1] + nums[sz-2] == temp) {

                nums.push_back(temp);

                if (helper(S, nums, start+i))

                    return true;

                nums.pop_back();

            }

        }

        return false;

    }

};

巴特西

842. Split Array into Fibonacci Sequence

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