# Split Array Into Fibonacci Sequence Problem

## Description

LeetCode Problem 842.

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:

``````1
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Input: num = "123456579"
Output: [123,456,579]
``````

Example 2:

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

Example 3:

``````1
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Input: num = "112358130"
Output: []
``````

Example 4:

``````1
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Input: num = "0123"
Output: []
Explanation: Leading zeroes are not allowed, so "01", "2", "3" is not valid.
``````

Example 5:

``````1
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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.

## Sample C++ Code

``````1
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class Solution {
public:
vector<int> ans;
string origS;

void backtrack(vector<int>& seq, int idx) {
int len = seq.size();

if (idx == origS.size() && len >= 3 &&
seq[len-1] == seq[len-2]+seq[len-3]) {
ans = seq;
}
if (idx >= origS.size())
return;

long long num = 0;
for (int i = idx; i < origS.size(); i ++) {
num = num * 10 + origS[i] - '0';
if (num > INT_MAX) break;
if (origS[idx] == '0' && i > idx) break;
if (len >= 2) {
if (num == (unsigned)(seq[len-1]) + (unsigned)(seq[len-2])) {
seq.push_back(num);
backtrack(seq, i+1);
seq.pop_back();
} else if (num > (unsigned)(seq[len-1]) + (unsigned)(seq[len-2])) {
break;
}
} else {
seq.push_back(num);
backtrack(seq, i+1);
seq.pop_back();
}
}
}

vector<int> splitIntoFibonacci(string S) {
origS = S;
vector<int> seq;
backtrack(seq, 0);
return ans;
}
};
``````