# Longest sub-sequence of array containing Lucas numbers

Given an array arr[] of N elements, the task is to find the length of the longest sub-sequence in arr[] such that all the elements of the sequence are Lucas Numbers.

Examples:

Input: arr[] = {2, 3, 55, 6, 1, 18}
Output: 4
1, 2, 3 and 18 are the only elements from the Lucas sequence.

Input: arr[] = {22, 33, 2, 123}
Output: 2

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

Approach:

• Find the maximum element in the array.
• Generate Lucas numbers upto to the max and store them in a set.
• Traverse the array arr[] and check if the current element is present in the set.
• If it is present in the set, and increment the count.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the length of ` `// the longest required sub-sequence ` `int` `LucasSequence(``int` `arr[], ``int` `n) ` `{ ` `    ``// Find the maximum element from  ` `    ``// the array ` `    ``int` `max = *max_element(arr, arr+n); ` ` `  `    ``// Insert all lucas numbers ` `    ``// below max to the set ` `    ``// a and b are first two elements ` `    ``// of the Lucas sequence ` `    ``unordered_set<``int``> s; ` `    ``int` `a = 2, b = 1, c; ` `    ``s.insert(a); ` `    ``s.insert(b); ` `    ``while` `(b < max) { ` `        ``int` `c = a + b; ` `        ``a = b; ` `        ``b = c; ` `        ``s.insert(b); ` `    ``} ` ` `  `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// If current element is a Lucas  ` `        ``// number, increment count ` `        ``auto` `it = s.find(arr[i]); ` `        ``if` `(it != s.end())  ` `            ``count++; ` `    ``} ` ` `  `    ``// Return the count ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 7, 11, 22, 4, 2, 1, 8, 9 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << LucasSequence(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to return the length of  ` `    ``// the longest required sub-sequence  ` `    ``static` `int` `LucasSequence(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``// Find the maximum element from  ` `        ``// the array  ` `        ``int` `max = Arrays.stream(arr).max().getAsInt(); ` `        ``int` `counter = ``0``; ` ` `  `        ``// Insert all lucas numbers  ` `        ``// below max to the set  ` `        ``// a and b are first two elements  ` `        ``// of the Lucas sequence  ` `        ``HashSet s = ``new` `HashSet<>(); ` ` `  `        ``int` `a = ``2``, b = ``1``; ` `        ``s.add(a); ` `        ``s.add(b); ` ` `  `        ``while` `(b < max) ` `        ``{ ` `            ``int` `c = a + b; ` `            ``a = b; ` `            ``b = c; ` `            ``s.add(b); ` `        ``} ` ` `  `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` ` `  `            ``// If current element is a Lucas  ` `            ``// number, increment count  ` `            ``if` `(s.contains(arr[i]))  ` `            ``{ ` `                ``counter++; ` `            ``} ` `        ``} ` ` `  `        ``// Return the count  ` `        ``return` `counter; ` `    ``} ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int``[] arr = {``7``, ``11``, ``22``, ``4``, ``2``, ``1``, ``8``, ``9``}; ` `        ``int` `n = arr.length; ` ` `  `        ``System.out.println(LucasSequence(arr, n)); ` `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## Python3

 `# Python 3 implementation of the approach ` ` `  `# Function to return the length of ` `# the longest required sub-sequence ` `def` `LucasSequence(arr, n): ` `     `  `    ``# Find the maximum element from  ` `    ``# the array ` `    ``max` `=` `arr[``0``] ` `    ``for` `i ``in` `range``(``len``(arr)): ` `        ``if``(arr[i] > ``max``): ` `            ``max` `=` `arr[i] ` ` `  `    ``# Insert all lucas numbers below max  ` `    ``# to the set a and b are first two  ` `    ``# elements of the Lucas sequence ` `    ``s ``=` `set``() ` `    ``a ``=` `2` `    ``b ``=` `1` `    ``s.add(a) ` `    ``s.add(b) ` `    ``while` `(b < ``max``): ` `        ``c ``=` `a ``+` `b ` `        ``a ``=` `b ` `        ``b ``=` `c ` `        ``s.add(b) ` ` `  `    ``count ``=` `0` `    ``for` `i ``in` `range``(n): ` `         `  `        ``# If current element is a Lucas  ` `        ``# number, increment count ` `        ``if``(arr[i] ``in` `s): ` `            ``count ``+``=` `1` ` `  `    ``# Return the count ` `    ``return` `count ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``7``, ``11``, ``22``, ``4``, ``2``, ``1``, ``8``, ``9``] ` `    ``n ``=` `len``(arr) ` ` `  `    ``print``(LucasSequence(arr, n)) ` ` `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# implementation of the approach  ` `using` `System; ` `using` `System.Collections.Generic; ` `using` `System.Linq; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to return the length of  ` `    ``// the longest required sub-sequence  ` `    ``static` `int` `LucasSequence(``int` `[]arr, ``int` `n)  ` `    ``{  ` `        ``// Find the maximum element from  ` `        ``// the array  ` `        ``int` `max = arr.Max();  ` `        ``int` `counter = 0;  ` ` `  `        ``// Insert all lucas numbers  ` `        ``// below max to the set  ` `        ``// a and b are first two elements  ` `        ``// of the Lucas sequence  ` `        ``HashSet<``int``> s = ``new` `HashSet<``int``>() ; ` `         `  `        ``int` `a = 2, b = 1 ; ` `        ``s.Add(a);  ` `        ``s.Add(b);  ` `         `  `        ``while` `(b < max)  ` `        ``{  ` `            ``int` `c = a + b;  ` `            ``a = b;  ` `            ``b = c;  ` `            ``s.Add(b);  ` `        ``}  ` `     `  `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{  ` `     `  `            ``// If current element is a Lucas  ` `            ``// number, increment count  ` `            ``if` `(s.Contains(arr[i])) ` `                ``counter++;  ` `        ``}  ` `     `  `        ``// Return the count  ` `        ``return` `counter;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``static` `public` `void` `Main()  ` `    ``{  ` `        ``int` `[]arr = { 7, 11, 22, 4, 2, 1, 8, 9 };  ` `        ``int` `n = arr.Length ; ` `     `  `        ``Console.WriteLine(LucasSequence(arr, n)) ; ` `    ``}  ` `} ` ` `  `// This code is contributed by Ryuga `

Output:

```5
```

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