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.
class Solution:
def splitIntoFibonacci(self, S: str) -> List[int]:
self.upper_bound = 2**31-1
self.ret = []
self.dfs(S, 0, [], [])
return self.ret
def dfs(self, S, index, ret, path):
if self.ret:
return
if index == len(S) and "".join(map(str, ret)) == S and len(ret) >= 3:
self.ret = list(ret)
return
elif index == len(S):
return
path.append(S[index])
n = int("".join(path))
if self.isFibonacci(ret, n):
self.dfs(S, index+1, ret+[n], [])
self.dfs(S, index+1, ret, path)
path.pop()
def isFibonacci(self, sequence, number):
if number < 0 or number > self.upper_bound:
return False
if len(sequence) < 2:
return True
return True if number == sequence[-1] + sequence[-2] else False