# 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].

**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:**

## 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 ` |

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## Java

`// 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 ` |

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## Python3

`# 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 ` |

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**Output:**

5

**Time Complexity:** O(N^{2}*log(M)), where N is the length of array and M is max(A).

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