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Formatted question description: https://leetcode.ca/all/842.html

# 842. Split Array into Fibonacci Sequence

Medium

## Description

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: []

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

## Solution

First check the maximum end index of the first number and the second number. Suppose the string contains only three numbers that form a Fibonacci-like sequence, then the third number can’t be shorter than either the first number of the second number. Suppose the length of string S is length, then the maximum end index of the first number is length / 2 - 1 and the maximum end index of the second number is length * 2 / 3 - 1.

For each possible choice of the first number and the second number, try to form a Fibonacci-like sequence starting from the first number and the second number. If one choice can lead to a Fibonacci-like sequence that equals S, return a list containing all the numbers in the sequence. Otherwise, return an empty list.

• class Solution {
public List<Integer> splitIntoFibonacci(String S) {
List<Integer> list = new ArrayList<Integer>();
if (S == null || S.length() == 0)
return list;
int length = S.length();
int firstMax = Math.min(length / 2, 10), secondMax = Math.min(length * 2 / 3, 20);
for (int i = 1; i <= firstMax; i++) {
if (i > 1 && S.charAt(0) == '0' || Long.parseLong(S.substring(0, i)) > Integer.MAX_VALUE)
break;
int secondMaxUpper = Math.min(secondMax, i + 11);
for (int j = i + 1; j <= secondMaxUpper; j++) {
List<Integer> splitIndices = new ArrayList<Integer>();
String firstNumStr = S.substring(0, i);
String secondNumStr = S.substring(i, j);
if (secondNumStr.length() > 1 && secondNumStr.charAt(0) == '0')
break;
StringBuffer sb = new StringBuffer(firstNumStr + secondNumStr);
int prevIndex = j;
while (sb.length() < length) {
long firstNum = Long.parseLong(firstNumStr);
long secondNum = Long.parseLong(secondNumStr);
long sum = firstNum + secondNum;
if (sum > Integer.MAX_VALUE)
break;
String sumStr = String.valueOf(sum);
sb.append(sumStr);
int index = prevIndex + sumStr.length();
if (index < length)
prevIndex = index;
firstNumStr = secondNumStr;
secondNumStr = sumStr;
}
if (sb.toString().equals(S)) {
int size = splitIndices.size();
for (int k = size - 1; k >= 0; k--) {
int splitIndex = splitIndices.get(k);
sb.insert(splitIndex, ',');
}
String[] array = sb.toString().split(",");
int sequenceLength = array.length;
for (int k = 0; k < sequenceLength; k++)
return list;
}
}
}
return list;
}
}

• // OJ: https://leetcode.com/problems/split-array-into-fibonacci-sequence/
// 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;
}
};

• class Solution(object):
def splitIntoFibonacci(self, S):
"""
:type S: str
:rtype: List[int]
"""
res = []
self.dfs(S, [], res)
return res

def dfs(self, num_str, path, res):
if len(path) >= 3 and  path[-1] != path[-2] + path[-3]:
return False
if not num_str and len(path) >= 3:
res.extend(path)
return True
for i in range(len(num_str)):
curr = num_str[:i+1]
if (curr[0] == '0' and len(curr) != 1) or int(curr) >= 2**31:
continue
if self.dfs(num_str[i+1:], path + [int(curr)], res):
return True
return False

• 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
}