# 842. Split Array into Fibonacci Sequence

## Description

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


Example 2:

Input: num = "112358130"
Output: []


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.

## Solutions

• class Solution {
private List<Integer> ans = new ArrayList<>();
private String num;

public List<Integer> splitIntoFibonacci(String num) {
this.num = num;
dfs(0);
return ans;
}

private boolean dfs(int i) {
if (i == num.length()) {
return ans.size() >= 3;
}
long x = 0;
for (int j = i; j < num.length(); ++j) {
if (j > i && num.charAt(i) == '0') {
break;
}
x = x * 10 + num.charAt(j) - '0';
if (x > Integer.MAX_VALUE
|| (ans.size() >= 2 && x > ans.get(ans.size() - 1) + ans.get(ans.size() - 2))) {
break;
}
if (ans.size() < 2 || x == ans.get(ans.size() - 1) + ans.get(ans.size() - 2)) {
if (dfs(j + 1)) {
return true;
}
ans.remove(ans.size() - 1);
}
}
return false;
}
}

• class Solution {
public:
vector<int> splitIntoFibonacci(string num) {
int n = num.size();
vector<int> ans;
function<bool(int)> dfs = [&](int i) -> bool {
if (i == n) {
return ans.size() > 2;
}
long long x = 0;
for (int j = i; j < n; ++j) {
if (j > i && num[i] == '0') {
break;
}
x = x * 10 + num[j] - '0';
if (x > INT_MAX || (ans.size() > 1 && x > (long long) ans[ans.size() - 1] + ans[ans.size() - 2])) {
break;
}
if (ans.size() < 2 || x == (long long) ans[ans.size() - 1] + ans[ans.size() - 2]) {
ans.push_back(x);
if (dfs(j + 1)) {
return true;
}
ans.pop_back();
}
}
return false;
};
dfs(0);
return ans;
}
};

• class Solution:
def splitIntoFibonacci(self, num: str) -> List[int]:
def dfs(i):
if i == n:
return len(ans) > 2
x = 0
for j in range(i, n):
if j > i and num[i] == '0':
break
x = x * 10 + int(num[j])
if x > 2**31 - 1 or (len(ans) > 2 and x > ans[-2] + ans[-1]):
break
if len(ans) < 2 or ans[-2] + ans[-1] == x:
ans.append(x)
if dfs(j + 1):
return True
ans.pop()
return False

n = len(num)
ans = []
dfs(0)
return ans


• func splitIntoFibonacci(num string) []int {
n := len(num)
ans := []int{}
var dfs func(int) bool
dfs = func(i int) bool {
if i == n {
return len(ans) > 2
}
x := 0
for j := i; j < n; j++ {
if j > i && num[i] == '0' {
break
}
x = x*10 + int(num[j]-'0')
if x > math.MaxInt32 || (len(ans) > 1 && x > ans[len(ans)-1]+ans[len(ans)-2]) {
break
}
if len(ans) < 2 || x == ans[len(ans)-1]+ans[len(ans)-2] {
ans = append(ans, x)
if dfs(j + 1) {
return true
}
ans = ans[:len(ans)-1]
}
}
return false
}
dfs(0)
return ans
}