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
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-sequenceint 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 codeint 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 approachimport 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-sequencedef 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 codeif __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-sequencefunction 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 codevar 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.
