# Find length of longest Fibonacci like subsequence

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].

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
A Fibonacci-like sequence is such that it has each two adjacent terms that determines the next expected term. For example, with 1, 1, we expect that the sequence must continue 2, 3, 5, 8, 13, … and so on.

We will 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:

 // CPP implementation of above approach  #include  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);      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 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

Output:
5


Time Complexity: O(N2*log(M)), where N is the length of array and M is max(A).

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Improved By : rituraj_jain, 29AjayKumar

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