LeetCode #842 — MEDIUM

Split Array into Fibonacci Sequence

Move from brute-force thinking to an efficient approach using backtracking strategy.

The Problem

Problem Statement

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: []
Explanation: The task is impossible.

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Backtracking

Example 1

"1101111"

Example 2

"112358130"

Example 3

"0123"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #842: Split Array into Fibonacci Sequence
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)) {
                ans.add((int) x);
                if (dfs(j + 1)) {
                    return true;
                }
                ans.remove(ans.size() - 1);
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #842: Split Array into Fibonacci Sequence
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
}

# Accepted solution for LeetCode #842: Split Array into Fibonacci Sequence
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

// Accepted solution for LeetCode #842: Split Array into Fibonacci Sequence
struct Solution;

impl Solution {
    fn split_into_fibonacci(s: String) -> Vec<i32> {
        let mut cur = vec![];
        let mut res = vec![];
        Self::dfs(0, &mut cur, &mut res, &s, s.len());
        res
    }
    fn dfs(start: usize, cur: &mut Vec<i32>, res: &mut Vec<i32>, s: &str, end: usize) {
        if start == end && cur.len() > 2 {
            *res = cur.to_vec();
        } else {
            for i in 1..=(end - start) {
                let t = &s[start..start + i];
                if &s[start..start + 1] == "0" && i > 1 {
                    break;
                }
                if let Ok(x) = t.parse::<i32>() {
                    let n = cur.len();
                    if n < 2 {
                        cur.push(x);
                        Self::dfs(start + i, cur, res, s, end);
                        cur.pop();
                    } else {
                        if x > cur[n - 1] + cur[n - 2] {
                            break;
                        }
                        if x == cur[n - 1] + cur[n - 2] {
                            cur.push(x);
                            Self::dfs(start + i, cur, res, s, end);
                            cur.pop();
                        }
                    }
                } else {
                    break;
                }
            }
        }
    }
}

#[test]
fn test() {
    let s = "123456579".to_string();
    let res = vec![123, 456, 579];
    assert_eq!(Solution::split_into_fibonacci(s), res);
    let s = "11235813".to_string();
    let res = vec![1, 1, 2, 3, 5, 8, 13];
    assert_eq!(Solution::split_into_fibonacci(s), res);
    let s = "112358130".to_string();
    let res: Vec<i32> = vec![];
    assert_eq!(Solution::split_into_fibonacci(s), res);
    let s = "0123".to_string();
    let res: Vec<i32> = vec![];
    assert_eq!(Solution::split_into_fibonacci(s), res);
    let s = "1101111".to_string();
    let res = vec![110, 1, 111];
    assert_eq!(Solution::split_into_fibonacci(s), res);
}

// Accepted solution for LeetCode #842: Split Array into Fibonacci Sequence
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #842: Split Array into Fibonacci Sequence
// 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)) {
//                 ans.add((int) x);
//                 if (dfs(j + 1)) {
//                     return true;
//                 }
//                 ans.remove(ans.size() - 1);
//             }
//         }
//         return false;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.