Split a String into a Fibonacci Sequence
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Problem Statement Solution 1: Recursive Backtracking Solution 2: Two-Pointer Approach Solution 3: Recursive Depth-First Search Solution 4: Dynamic Programming Approach Solution 5: Greedy Approach with Early Termination Solution 6: String Manipulation with Validation Solution 7: Iterative Approach with Early Exit More Problems Solutions
Problem Statement
Imagine you have a string made up of digits, like "123456579". Your goal is to split this string into a sequence of numbers that follow the Fibonacci sequence rules.
A Fibonacci-like sequence is a list of numbers where:
Each number is less than 2312^{31}231 (i.e., it fits within a 32-bit signed integer).
There are at least three numbers in the sequence.
Each number in the sequence, starting from the third one, is the sum of the two preceding numbers.
For instance, in a sequence [a, b, c], it must hold that a + b = c, and this rule should apply to all subsequent numbers.
When you split the string into numbers, each number must not have unnecessary leading zeroes (e.g., "01" is not allowed, but "0" is fine).
Your task is to determine if you can split the string into such a sequence. If it is possible, return any valid sequence. If not, return an empty list.
Example 1
Input: str = "1101111"
Output: [11, 0, 11, 11]
Example 2
Input: str = "112358130"
Output: []
Example 3
Input: str = "0123"
Output: []
Solution 1: Recursive Backtracking
function backtrackingFibonacciSequence(num) { // Backtracking Approach
const sequence = [];
function explore(index) {
if (index === num.length && sequence.length >= 3) return true;
for (let length = 1; length <= num.length - index; length++) {
if (length > 10) break; // Prevent large numbers
const segment = num.slice(index, index + length);
if (segment.length > 1 && segment[0] === '0') break; // No leading zeros
const value = parseInt(segment, 10);
if (value > 2**31 - 1) break; // 32-bit limit
if (sequence.length >= 2) {
const [first, second] = [sequence[sequence.length - 2], sequence[sequence.length - 1]];
if (first + second !== value) continue;
}
sequence.push(value);
if (explore(index + length)) return true;
sequence.pop();
}
return false;
}
explore(0);
return sequence;
}
const str1 = "1101111";
backtrackingFibonacciSequence(str1); //output: [11, 0, 11, 11]
const str2 = "112358130";
backtrackingFibonacciSequence(str2); //output: []
const str3 = "0123";
backtrackingFibonacciSequence(str3); //output: []
Solution 2: Two-Pointer Approach
function twoPointerFibonacciSequence(num) { // Two-Pointer Approach
const length = num.length;
for (let firstEnd = 1; firstEnd < length / 2; firstEnd++) {
for (let secondEnd = firstEnd + 1; secondEnd < length; secondEnd++) {
const firstPart = num.slice(0, firstEnd);
const secondPart = num.slice(firstEnd, secondEnd);
if (firstPart.length > 1 && firstPart[0] === '0' || secondPart.length > 1 && secondPart[0] === '0') continue;
const fibSeq = [parseInt(firstPart), parseInt(secondPart)];
let pos = secondEnd;
while (pos < length) {
const next = fibSeq[fibSeq.length - 1] + fibSeq[fibSeq.length - 2];
const nextStr = next.toString();
if (num.slice(pos, pos + nextStr.length) !== nextStr) break;
fibSeq.push(next);
pos += nextStr.length;
}
if (pos === length && fibSeq.length >= 3) return fibSeq;
}
}
return [];
}
const str1 = "1101111";
twoPointerFibonacciSequence(str1); //output: [11, 0, 11, 11]
const str2 = "112358130";
twoPointerFibonacciSequence(str2); //output: []
const str3 = "0123";
twoPointerFibonacciSequence(str3); //output: []
Solution 3: Recursive Depth-First Search
function dfsFibonacciSequence(num) { // Recursive Depth-First Search
function isValidSegment(segment) {
return segment.length === 1 || segment[0] !== '0';
}
function dfs(startIndex, currentSeq) {
if (startIndex === num.length && currentSeq.length >= 3) return currentSeq;
for (let length = 1; length <= 10 && startIndex + length <= num.length; length++) {
const segment = num.slice(startIndex, startIndex + length);
if (!isValidSegment(segment)) continue;
const segmentValue = parseInt(segment, 10);
if (segmentValue >= 2**31) continue;
if (currentSeq.length < 2) {
// Start the sequence with the first two numbers
currentSeq.push(segmentValue);
const result = dfs(startIndex + length, currentSeq);
if (result) return result;
currentSeq.pop();
} else {
// Check if the current segment can continue the sequence
const [secondLast, last] = [currentSeq[currentSeq.length - 2], currentSeq[currentSeq.length - 1]];
if (secondLast + last === segmentValue) {
currentSeq.push(segmentValue);
const result = dfs(startIndex + length, currentSeq);
if (result) return result;
currentSeq.pop();
}
}
}
return null;
}
return dfs(0, []) || [];
}
const str1 = "1101111";
dfsFibonacciSequence(str1); //output: [11, 0, 11, 11]
const str2 = "112358130";
dfsFibonacciSequence(str2); //output: []
const str3 = "0123";
dfsFibonacciSequence(str3); //output: []
Solution 4: Dynamic Programming Approach
function dpFibonacciSequence(num) { // Dynamic Programming Approach
const len = num.length;
const dp = Array(len).fill(false);
dp[0] = true;
function isValidNumber(segment) {
return segment.length === 1 || segment[0] !== '0';
}
for (let firstSplit = 1; firstSplit < len; firstSplit++) {
for (let secondSplit = firstSplit + 1; secondSplit < len; secondSplit++) {
const first = num.slice(0, firstSplit);
const second = num.slice(firstSplit, secondSplit);
if (isValidNumber(first) && isValidNumber(second)) {
const firstNum = parseInt(first, 10);
const secondNum = parseInt(second, 10);
let rest = num.slice(secondSplit);
const sequence = [firstNum, secondNum];
while (rest.length > 0) {
const nextNum = sequence[sequence.length - 1] + sequence[sequence.length - 2];
const nextStr = nextNum.toString();
if (rest.startsWith(nextStr)) {
sequence.push(nextNum);
rest = rest.slice(nextStr.length);
} else {
break;
}
}
if (rest.length === 0 && sequence.length >= 3) return sequence;
}
}
}
return [];
}
const str1 = "1101111";
dpFibonacciSequence(str1); //output: [11, 0, 11, 11]
const str2 = "112358130";
dpFibonacciSequence(str2); //output: []
const str3 = "0123";
dpFibonacciSequence(str3); //output: []
Solution 5: Greedy Approach with Early Termination
function greedyFibonacciSequence(num) { // Greedy Approach with Early Termination
function isValidPart(part) {
return part.length === 1 || part[0] !== '0';
}
for (let i = 1; i < num.length; i++) {
for (let j = i + 1; j < num.length; j++) {
const first = num.slice(0, i);
const second = num.slice(i, j);
if (!isValidPart(first) || !isValidPart(second)) continue;
const sequence = [parseInt(first, 10), parseInt(second, 10)];
let k = j;
while (k < num.length) {
const nextNumber = sequence[sequence.length - 1] + sequence[sequence.length - 2];
const nextStr = nextNumber.toString();
if (num.slice(k, k + nextStr.length) !== nextStr) break;
sequence.push(nextNumber);
k += nextStr.length;
}
if (k === num.length && sequence.length >= 3) return sequence;
}
}
return [];
}
const str1 = "1101111";
greedyFibonacciSequence(str1); //output: [11, 0, 11, 11]
const str2 = "112358130";
greedyFibonacciSequence(str2); //output: []
const str3 = "0123";
greedyFibonacciSequence(str3); //output: []
Solution 6: String Manipulation with Validation
function stringManipulationFibonacci(num) { // String Manipulation with Validation
function isValid(part) {
return part.length === 1 || part[0] !== '0';
}
for (let i = 1; i < num.length / 2; i++) {
for (let j = i + 1; j < num.length; j++) {
const first = num.slice(0, i);
const second = num.slice(i, j);
if (!isValid(first) || !isValid(second)) continue;
const fibSequence = [parseInt(first, 10), parseInt(second, 10)];
let k = j;
while (k < num.length) {
const nextValue = fibSequence[fibSequence.length - 1] + fibSequence[fibSequence.length - 2];
const nextStr = nextValue.toString();
if (num.slice(k, k + nextStr.length) !== nextStr) break;
fibSequence.push(nextValue);
k += nextStr.length;
}
if (k === num.length && fibSequence.length >= 3) return fibSequence;
}
}
return [];
}
const str1 = "1101111";
stringManipulationFibonacci(str1); //output: [11, 0, 11, 11]
const str2 = "112358130";
stringManipulationFibonacci(str2); //output: []
const str3 = "0123";
stringManipulationFibonacci(str3); //output: []
Solution 7: Iterative Approach with Early Exit
function iterativeFibonacciSequence(num) { // Iterative Approach with Early Exit
const length = num.length;
for (let firstEnd = 1; firstEnd < length; firstEnd++) {
for (let secondEnd = firstEnd + 1; secondEnd < length; secondEnd++) {
const firstPart = num.slice(0, firstEnd);
const secondPart = num.slice(firstEnd, secondEnd);
if (firstPart.length > 1 && firstPart[0] === '0' || secondPart.length > 1 && secondPart[0] === '0') continue;
const sequence = [parseInt(firstPart, 10), parseInt(secondPart, 10)];
let index = secondEnd;
while (index < length) {
const nextNumber = sequence[sequence.length - 1] + sequence[sequence.length - 2];
const nextStr = nextNumber.toString();
if (num.slice(index, index + nextStr.length) !== nextStr) break;
sequence.push(nextNumber);
index += nextStr.length;
}
if (index === length && sequence.length >= 3) return sequence;
}
}
return [];
}
const str1 = "1101111";
iterativeFibonacciSequence(str1); //output: [11, 0, 11, 11]
const str2 = "112358130";
iterativeFibonacciSequence(str2); //output: []
const str3 = "0123";
iterativeFibonacciSequence(str3); //output: []