Find length of longest Fibonacci like subsequence - GeeksforGeeks

05 Apr, 2022
Given a strictly increasing array A of positive integers where, 
 
. 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++
Java
Python3
C#
Javascript
// CPP implementation of above approach#include <bits/stdc++.h>using namespace std; // Function to return the max Length of// Fibonacci subsequenceint 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 programint 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 approachimport 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 subsequencedef 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 Codeif __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 approachusing 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 subsequencefunction 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 programvar 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>
5
Time Complexity: O(N2 * log(M)), where N is the length of array and M is max(A).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++
Python3
Javascript
// CPP program for the above approach#include <bits/stdc++.h>using namespace std; // Function to return the max Length of// Fibonacci subsequenceint 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 programint main(){    int A[] = { 1, 3, 7, 11, 12, 14, 18 };    int n = sizeof(A) / sizeof(A[0]);    cout << LongestFibSubseq(A, n);    return 0;}
# Python program for the above approach # Function to return the max Length of# Fibonacci subsequencedef 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 programA = [ 1, 3, 7, 11, 12, 14, 18 ]n = len(A)print(LongestFibSubseq(A, n)) # This code is contributed by shinjanpatra
<script> // JavaScript program for the above approach  // Function to return the max Length of// Fibonacci subsequencefunction 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>
3
Time Complexity: O(N2), where N is the length of the array. 
rituraj_jain
29AjayKumar
rishabhgaurav007
famously
sagar0719kumar
shinjanpatra
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