842

842. Split Array into Fibonacci Sequence (Medium)

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] < 231, (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 = "123456579"
Output: [123,456,579]

Example 2:

Input: num = "11235813"
Output: [1,1,2,3,5,8,13]

Example 3:

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

Example 4:

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

Example 5:

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

 

Constraints:

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

Companies:
Amazon

Related Topics:
String, Backtracking

Similar Questions:

• Additive Number (Medium)
• Fibonacci Number (Easy)

Solution 1. DFS

// OJ: https://leetcode.com/problems/split-array-into-fibonacci-sequence/
// Author: github.com/lzl124631x
// Time: O(N!)
// Space: O(N)
class Solution {
    vector<int> ans;
    bool dfs(string &s, int i) {
        if (i == s.size()) return ans.size() >= 3;
        for (int j = i; j < s.size(); ++j) {
            long n = stol(s.substr(i, j - i + 1));
            if (n > INT_MAX) return false;
            if (ans.size() >= 2) {
                long target = (long)ans.back() + ans[ans.size() - 2];
                if (target > INT_MAX) return false;
                if (n == target) {
                    ans.push_back(n);
                    if (dfs(s, j + 1)) return true;
                    ans.pop_back();
                    break;
                }
                if (n > target) break;
            } else {
                ans.push_back(n);
                if (dfs(s, j + 1)) return true;
                ans.pop_back();
            }
            if (s[i] == '0') return false;
        }
        return false;
    }
public:
    vector<int> splitIntoFibonacci(string s) {
        dfs(s, 0);
        return ans;
    }
};