Find length of longest Fibonacci like subsequence
Last Updated :
07 Sep, 2022
Given a strictly increasing array A of positive integers where,
1 \leq A[i] \leq 10^{18}
The task is to find the length of the longest Fibonacci-like subsequence of A. If such subsequence does not exist, return 0.
Examples:
Input: A = [1, 3, 7, 11, 12, 14, 18]
Output: 3
Explanation:
The longest subsequence that is Fibonacci-like: [1, 11, 12]. Other possible subsequences are [3, 11, 14] or [7, 11, 18].
Input: A = [1, 2, 3, 4, 5, 6, 7, 8]
Output: 5
Explanation:
The longest subsequence that is Fibonacci-like: [1, 2, 3, 5, 8].
Naive Approach: A Fibonacci-like sequence is such that it has each two adjacent terms that determine the next expected term.
For example, with 1, 1, we expect that the sequence must continue 2, 3, 5, 8, 13, ... and so on.
- Use Set or Map to determine quickly whether the next term of Fibonacci sequence is present in the array A or not. Because of the exponential growth of these terms, there will be not more than log(M) searches to get next element on each iteration.
- For each starting pair A[i], A[j], we maintain the next expected value y = A[i] + A[j] and the previously seen largest value x = A[j]. If y is in the array, then we can then update these values (x, y) -> (y, x+y) otherwise we stop immediately.
Below is the implementation of above approach:
C++
// CPP implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the max Length of
// Fibonacci subsequence
int LongestFibSubseq(int A[], int n)
{
// Store all array elements in a hash
// table
unordered_set<int> S(A, A + n);
int maxLen = 0, x, y;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
x = A[j];
y = A[i] + A[j];
int length = 2;
// check until next fib element is found
while (S.find(y) != S.end()) {
// next element of fib subseq
int z = x + y;
x = y;
y = z;
maxLen = max(maxLen, ++length);
}
}
}
return maxLen >= 3 ? maxLen : 0;
}
// Driver program
int main()
{
int A[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
int n = sizeof(A) / sizeof(A[0]);
cout << LongestFibSubseq(A, n);
return 0;
}
// This code is written by Sanjit_Prasad
// Java implementation of above approach
import java.util.*;
public class GFG {
// Function to return the max Length of
// Fibonacci subsequence
static int LongestFibSubseq(int A[], int n) {
// Store all array elements in a hash
// table
TreeSet<Integer> S = new TreeSet<>();
for (int t : A) {
// Add each element into the set
S.add(t);
}
int maxLen = 0, x, y;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
x = A[j];
y = A[i] + A[j];
int length = 3;
// check until next fib element is found
while (S.contains(y) && (y != S.last())) {
// next element of fib subseq
int z = x + y;
x = y;
y = z;
maxLen = Math.max(maxLen, ++length);
}
}
}
return maxLen >= 3 ? maxLen : 0;
}
// Driver program
public static void main(String[] args) {
int A[] = {1, 2, 3, 4, 5, 6, 7, 8};
int n = A.length;
System.out.print(LongestFibSubseq(A, n));
}
}
// This code is contributed by 29AjayKumar
# Python3 implementation of the
# above approach
# Function to return the max Length
# of Fibonacci subsequence
def LongestFibSubseq(A, n):
# Store all array elements in
# a hash table
S = set(A)
maxLen = 0
for i in range(0, n):
for j in range(i + 1, n):
x = A[j]
y = A[i] + A[j]
length = 2
# check until next fib
# element is found
while y in S:
# next element of fib subseq
z = x + y
x = y
y = z
length += 1
maxLen = max(maxLen, length)
return maxLen if maxLen >= 3 else 0
# Driver Code
if __name__ == "__main__":
A = [1, 2, 3, 4, 5, 6, 7, 8]
n = len(A)
print(LongestFibSubseq(A, n))
# This code is contributed
# by Rituraj Jain
// C# implementation of above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the max Length of
// Fibonacci subsequence
static int LongestFibSubseq(int []A, int n)
{
// Store all array elements in a hash
// table
SortedSet<int> S = new SortedSet<int>();
foreach (int t in A)
{
// Add each element into the set
S.Add(t);
}
int maxLen = 0, x, y;
for (int i = 0; i < n; ++i)
{
for (int j = i + 1; j < n; ++j)
{
x = A[j];
y = A[i] + A[j];
int length = 3;
// check until next fib element is found
while (S.Contains(y) && y != last(S))
{
// next element of fib subseq
int z = x + y;
x = y;
y = z;
maxLen = Math.Max(maxLen, ++length);
}
}
}
return maxLen >= 3 ? maxLen : 0;
}
static int last(SortedSet<int> S)
{
int ans = 0;
foreach(int a in S)
ans = a;
return ans;
}
// Driver Code
public static void Main(String[] args)
{
int []A = {1, 2, 3, 4, 5, 6, 7, 8};
int n = A.Length;
Console.Write(LongestFibSubseq(A, n));
}
}
// This code is contributed by 29AjayKumar
<script>
// Javascript implementation of above approach
// Function to return the max Length of
// Fibonacci subsequence
function LongestFibSubseq(A, n)
{
// Store all array elements in a hash
// table
var S = new Set(A);
var maxLen = 0, x, y;
for (var i = 0; i < n; ++i) {
for (var j = i + 1; j < n; ++j) {
x = A[j];
y = A[i] + A[j];
var length = 2;
// check until next fib element is found
while (S.has(y)) {
// next element of fib subseq
var z = x + y;
x = y;
y = z;
maxLen = Math.max(maxLen, ++length);
}
}
}
return maxLen >= 3 ? maxLen : 0;
}
// Driver program
var A = [1, 2, 3, 4, 5, 6, 7, 8];
var n = A.length;
document.write( LongestFibSubseq(A, n));
// This code is contributed by famously.
</script>
Complexity Analysis:
- Time Complexity: O(N2 * log(M)), where N is the length of array and M is max(A).
- Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach the idea is to implement Dynamic Programming. Initialize a dp table, dp[a, b] that represents the length of Fibonacci sequence ends up with (a, b). Then update the table as dp[a, b] = (dp[b - a, a] + 1 ) or 2
Below is the implementation of the above approach:
C++
// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the max Length of
// Fibonacci subsequence
int LongestFibSubseq(int A[], int n)
{
// Initialize the unordered map
unordered_map<int, int> m;
int N = n, res = 0;
// Initialize dp table
int dp[N][N];
// Iterate till N
for (int j = 0; j < N; ++j) {
m[A[j]] = j;
for (int i = 0; i < j; ++i) {
// Check if the current integer
// forms a fibonacci sequence
int k = m.find(A[j] - A[i]) == m.end()
? -1
: m[A[j] - A[i]];
// Update the dp table
dp[i][j] = (A[j] - A[i] < A[i] && k >= 0)
? dp[k][i] + 1
: 2;
res = max(res, dp[i][j]);
}
}
// Return the answer
return res > 2 ? res : 0;
}
// Driver program
int main()
{
int A[] = { 1, 3, 7, 11, 12, 14, 18 };
int n = sizeof(A) / sizeof(A[0]);
cout << LongestFibSubseq(A, n);
return 0;
}
/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
class GFG
{
// Function to return the max Length of
// Fibonacci subsequence
static int LongestFibSubseq(int[] A, int n)
{
// Initialize the unordered map
HashMap<Integer,Integer>m = new HashMap<>();
int N = n, res = 0;
// Initialize dp table
int[][] dp = new int[N][N];
// Iterate till N
for (int j = 0; j < N; ++j)
{
m.put(A[j], j);
for (int i = 0; i < j; ++i)
{
// Check if the current integer
// forms a fibonacci sequence
int k = m.containsKey(A[j] - A[i])? m.get(A[j] - A[i]):-1;
// Update the dp table
dp[i][j] = (A[j] - A[i] < A[i] && k >= 0)
? dp[k][i] + 1
: 2;
res = Math.max(res, dp[i][j]);
}
}
// Return the answer
return res > 2 ? res : 0;
}
// Drivers code
public static void main(String args[]){
int[] A = { 1, 3, 7, 11, 12, 14, 18 };
int n = A.length;
System.out.println(LongestFibSubseq(A, n));
}
}
// This code is contributed by shinjanpatra
# Python program for the above approach
# Function to return the max Length of
# Fibonacci subsequence
def LongestFibSubseq(A, n):
# Initialize the unordered map
m = {}
N, res = n, 0
# Initialize dp table
dp = [ [0 for i in range(N) ] for J in range(N) ]
# Iterate till N
for j in range(N):
m[A[j]] = j
for i in range(j):
# Check if the current integer
# forms a fibonacci sequence
k = -1 if ((A[j] - A[i]) not in m) else m[A[j] - A[i]]
# Update the dp table
dp[i][j] = dp[k][i] + 1 if (A[j] - A[i] < A[i] and k >= 0) else 2
res = max(res, dp[i][j])
# Return the answer
return res if res > 2 else 0
# Driver program
A = [ 1, 3, 7, 11, 12, 14, 18 ]
n = len(A)
print(LongestFibSubseq(A, n))
# This code is contributed by shinjanpatra
// C# implementation of above approach
using System;
using System.Collections.Generic;
class HelloWorld {
// Function to return the max Length of
// Fibonacci subsequence
static int LongestFibSubseq(int[] A, int n)
{
// Initialize the unordered map
var m = new Dictionary<int, int>();
int N = n;
int res = 0;
// Initialize dp table
int[, ] dp = new int[N, N];
// Iterate till N
for (int j = 0; j < N; ++j) {
m[A[j]] = j;
for (int i = 0; i < j; ++i) {
// Check if the current integer
// forms a fibonacci sequence
int k = m.ContainsKey(A[j] - A[i])
? m[A[j] - A[i]]
: -1;
// Update the dp table
dp[i, j] = (A[j] - A[i] < A[i] && k >= 0)
? dp[k, i] + 1
: 2;
res = Math.Max(res, dp[i, j]);
}
}
// Return the answer
return res > 2 ? res : 0;
}
// Driver program
static void Main()
{
int[] A = { 1, 3, 7, 11, 12, 14, 18 };
int n = A.Length;
Console.WriteLine(LongestFibSubseq(A, n));
}
}
// The code is contributed by Gautam goel (gautamgoel962)
<script>
// JavaScript program for the above approach
// Function to return the max Length of
// Fibonacci subsequence
function LongestFibSubseq(A,n)
{
// Initialize the unordered map
let m = new Map();
let N = n, res = 0;
// Initialize dp table
let dp = new Array(N);
for(let i=0;i<N;i++){
dp[i] = new Array(N);
}
// Iterate till N
for (let j = 0; j < N; ++j) {
m.set(A[j],j);
for (let i = 0; i < j; ++i) {
// Check if the current integer
// forms a fibonacci sequence
let k = m.has(A[j] - A[i]) == false
? -1
: m.get(A[j] - A[i]);
// Update the dp table
dp[i][j] = (A[j] - A[i] < A[i] && k >= 0)
? dp[k][i] + 1
: 2;
res = Math.max(res, dp[i][j]);
}
}
// Return the answer
return res > 2 ? res : 0;
}
// Driver program
let A = [ 1, 3, 7, 11, 12, 14, 18 ];
let n = A.length;
document.write(LongestFibSubseq(A, n));
// code is contributed by shinjanpatra
</script>
Complexity Analysis:
- Time Complexity: O(N2), where N is the length of the array.
- Auxiliary Space: O(N2)
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