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Longest sub-sequence of array containing Lucas numbers

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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:
1, 2, 3 and 18 are the only elements from the Lucas sequence.

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

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 <bits/stdc++.h>
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[0]);
 
    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<Integer> 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


Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the length of
// the longest required sub-sequence
function LucasSequence(arr, n)
{
    // Find the maximum element from
    // the array
    var max = arr.reduce((a,b)=> Math.max(a,b));
 
    // push all lucas numbers
    // below max to the set
    // a and b are first two elements
    // of the Lucas sequence
    var s = [];
    var a = 2, b = 1, c;
    s.push(a);
    s.push(b);
    while (b < max) {
        var c = a + b;
        a = b;
        b = c;
        s.push(b);
    }
 
    s.sort((a,b) => a-b)
    var count = 0;
    for (var i = 0; i < n; i++) {
 
        // If current element is a Lucas
        // number, increment count
        if(s.includes(arr[i]))
        {
            s.pop(arr[i]);
            count++;
        }
    }
 
    // Return the count
    return count;
}
 
// Driver code
var arr = [7, 11, 22, 4, 2, 1, 8, 9 ];
var n = arr.length;
document.write( LucasSequence(arr, n));
 
</script>


Output: 

5

 

Time complexity:  O(n), where n is the number of elements in the input array. This is because the function iterates through the array once and performs a constant-time lookup for each element in the unordered_set.
Auxiliary Space:  O(n), where n is the number of elements in the input array. This is because the function creates an unordered_set that stores all the Lucas numbers up to the maximum element in the input array. The size of this set is determined by the number of Lucas numbers that are less than or equal to the maximum element in the input array, which is directly proportional to the number of elements in the input array.



Last Updated : 31 Jan, 2023
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