Formatted question description: https://leetcode.ca/all/842.html
842. Split Array into Fibonacci Sequence
Level
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 all0 <= 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.
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);
splitIndices.add(i);
splitIndices.add(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)
splitIndices.add(index);
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++)
list.add(Integer.parseInt(array[k]));
return list;
}
}
}
return list;
}
}