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Find the longest subsequence of an array having LCM at most K

  • Difficulty Level : Hard
  • Last Updated : 08 Jun, 2021
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Given an array arr[] of N elements and a positive integer K. The task is to find the longest sub-sequence in the array having LCM (Least Common Multiple) at most K. Print the LCM and the length of the sub-sequence, following the indexes (starting from 0) of the elements of the obtained sub-sequence. Print -1 if it is not possible to do so.
Examples: 
 

Input: arr[] = {2, 3, 4, 5}, K = 14 
Output: 
LCM = 12, Length = 3 
Indexes = 0 1 2
Input: arr[] = {12, 33, 14, 52}, K = 4 
Output: -1 
 

 

Approach: Find all the unique elements of the array and their respective frequencies. Now the highest LCM that you are supposed to get is K. Suppose you have a number X such that 1 ≤ X ≤ K, obtain all the unique numbers from the array whom X is a multiple of and add their frequencies to numCount of X. The answer will be the number with highest numCount, let it be your LCM. Now, to obtain the indexes of the numbers of the sub-sequence, start traversing the array from the beginning and print the index i if LCM % arr[i] = 0.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the longest subsequence
// having LCM less than or equal to K
void findSubsequence(int* arr, int n, int k)
{
 
    // Map to store unique elements
    // and their frequencies
    map<int, int> M;
 
    // Update the frequencies
    for (int i = 0; i < n; ++i)
        ++M[arr[i]];
 
    // Array to store the count of numbers whom
    // 1 <= X <= K is a multiple of
    int* numCount = new int[k + 1];
 
    for (int i = 0; i <= k; ++i)
        numCount[i] = 0;
 
    // Check every unique element
    for (auto p : M) {
        if (p.first <= k) {
 
            // Find all its multiples <= K
            for (int i = 1;; ++i) {
                if (p.first * i > k)
                    break;
 
                // Store its frequency
                numCount[p.first * i] += p.second;
            }
        }
        else
            break;
    }
 
    int lcm = 0, length = 0;
 
    // Obtain the number having maximum count
    for (int i = 1; i <= k; ++i) {
        if (numCount[i] > length) {
            length = numCount[i];
            lcm = i;
        }
    }
 
    // Condition to check if answer
    // doesn't exist
    if (lcm == 0)
        cout << -1 << endl;
    else {
 
        // Print the answer
        cout << "LCM = " << lcm
             << ", Length = " << length << endl;
 
        cout << "Indexes = ";
        for (int i = 0; i < n; ++i)
            if (lcm % arr[i] == 0)
                cout << i << " ";
    }
}
 
// Driver code
int main()
{
 
    int k = 14;
    int arr[] = { 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    findSubsequence(arr, n, k);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
    // Function to find the longest subsequence
    // having LCM less than or equal to K
    static void findSubsequence(int []arr, int n, int k)
    {
     
        // Map to store unique elements
        // and their frequencies
        HashMap<Integer, Integer> M = new HashMap<Integer,Integer>();
     
        // Update the frequencies
        for (int i = 0; i < n; ++i)
        {
            if(M.containsKey(arr[i]))
                M.put(arr[i], M.get(arr[i])+1);
            else
                M.put(arr[i], 1);
        }
         
        // Array to store the count of numbers whom
        // 1 <= X <= K is a multiple of
        int [] numCount = new int[k + 1];
     
        for (int i = 0; i <= k; ++i)
            numCount[i] = 0;
     
        Iterator<HashMap.Entry<Integer, Integer>> itr = M.entrySet().iterator();
         
        // Check every unique element
        while(itr.hasNext())
        {
            HashMap.Entry<Integer, Integer> entry = itr.next();
            if (entry.getKey() <= k)
            {
     
                // Find all its multiples <= K
                for (int i = 1;; ++i)
                {
                    if (entry.getKey() * i > k)
                        break;
     
                    // Store its frequency
                    numCount[entry.getKey() * i] += entry.getValue();
                }
            }
            else
                break;
        }
     
        int lcm = 0, length = 0;
     
        // Obtain the number having maximum count
        for (int i = 1; i <= k; ++i)
        {
            if (numCount[i] > length)
            {
                length = numCount[i];
                lcm = i;
            }
        }
     
        // Condition to check if answer
        // doesn't exist
        if (lcm == 0)
            System.out.println(-1);
        else
        {
     
            // Print the answer
            System.out.println("LCM = " + lcm
                + " Length = " + length );
     
            System.out.print( "Indexes = ");
            for (int i = 0; i < n; ++i)
                if (lcm % arr[i] == 0)
                    System.out.print(i + " ");
        }
    }
     
    // Driver code
    public static void main (String[] args)
    {
     
        int k = 14;
        int arr[] = { 2, 3, 4, 5 };
        int n = arr.length;
     
        findSubsequence(arr, n, k);
    }
}
 
// This code is contributed by ihritik

Python3




# Python3 implementation of the approach
from collections import defaultdict
 
# Function to find the longest subsequence
# having LCM less than or equal to K
def findSubsequence(arr, n, k):
 
    # Map to store unique elements
    # and their frequencies
    M = defaultdict(lambda:0)
 
    # Update the frequencies
    for i in range(0, n):
        M[arr[i]] += 1
 
    # Array to store the count of numbers
    # whom 1 <= X <= K is a multiple of
    numCount = [0] * (k + 1)
 
    # Check every unique element
    for p in M:
        if p <= k:
 
            # Find all its multiples <= K
            i = 1
            while p * i <= k:
                 
                # Store its frequency
                numCount[p * i] += M[p]
                i += 1
         
        else:
            break
     
    lcm, length = 0, 0
 
    # Obtain the number having maximum count
    for i in range(1, k + 1):
        if numCount[i] > length:
            length = numCount[i]
            lcm = i
 
    # Condition to check if answer doesn't exist
    if lcm == 0:
        print(-1)
    else:
 
        # Print the answer
        print("LCM = {0}, Length = {1}".format(lcm, length))
 
        print("Indexes = ", end = "")
        for i in range(0, n):
            if lcm % arr[i] == 0:
                print(i, end = " ")
     
# Driver code
if __name__ == "__main__":
 
    k = 14
    arr = [2, 3, 4, 5]
    n = len(arr)
 
    findSubsequence(arr, n, k)
 
# This code is contributed by Rituraj Jain

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
    // Function to find the longest subsequence
    // having LCM less than or equal to K
    static void findSubsequence(int []arr, int n, int k)
    {
     
        // Map to store unique elements
        // and their frequencies
        Dictionary<int, int> M = new Dictionary<int, int>();
     
        // Update the frequencies
        for (int i = 0; i < n; ++i)
        {
            if(M.ContainsKey(arr[i]))
                M[arr[i]]++;
            else
                M[arr[i]] = 1;
        }
         
        // Array to store the count of numbers whom
        // 1 <= X <= K is a multiple of
        int [] numCount = new int[k + 1];
     
        for (int i = 0; i <= k; ++i)
            numCount[i] = 0;
     
        Dictionary<int, int>.KeyCollection keyColl = M.Keys;
         
        // Check every unique element
        foreach(int key in keyColl)
        {
            if ( key <= k)
            {
     
                // Find all its multiples <= K
                for (int i = 1;; ++i)
                {
                    if (key * i > k)
                        break;
     
                    // Store its frequency
                    numCount[key * i] += M[key];
                }
            }
            else
                break;
        }
     
        int lcm = 0, length = 0;
     
        // Obtain the number having maximum count
        for (int i = 1; i <= k; ++i)
        {
            if (numCount[i] > length)
            {
                length = numCount[i];
                lcm = i;
            }
        }
     
        // Condition to check if answer
        // doesn't exist
        if (lcm == 0)
            Console.WriteLine(-1);
        else
        {
     
            // Print the answer
            Console.WriteLine("LCM = " + lcm
                + " Length = " + length );
     
            Console.Write( "Indexes = ");
            for (int i = 0; i < n; ++i)
                if (lcm % arr[i] == 0)
                    Console.Write(i + " ");
        }
    }
     
    // Driver code
    public static void Main ()
    {
     
        int k = 14;
        int []arr = { 2, 3, 4, 5 };
        int n = arr.Length;
     
        findSubsequence(arr, n, k);
    }
}
 
// This code is contributed by ihritik

PHP




<?php
// PHP implementation of the approach
 
// Function to find the longest subsequence
// having LCM less than or equal to K
function findSubsequence($arr, $n, $k)
{
 
    // Map to store unique elements
    // and their frequencies
    $M = array();
     
    for($i = 0; $i < $n; $i++)
        $M[$arr[$i]] = 0 ;
 
    // Update the frequencies
    for ($i = 0; $i < $n; ++$i)
        ++$M[$arr[$i]];
 
    // Array to store the count of numbers
    // whom 1 <= X <= K is a multiple of
    $numCount = array();
 
    for ($i = 0; $i <= $k; ++$i)
        $numCount[$i] = 0;
 
    // Check every unique element
    foreach($M as $key => $value)
    {
        if ($key <= $k)
        {
 
            // Find all its multiples <= K
            for ($i = 1;; ++$i)
            {
                if ($key * $i > $k)
                    break;
 
                // Store its frequency
                $numCount[$key * $i] += $value;
            }
        }
        else
            break;
    }
 
    $lcm = 0; $length = 0;
 
    // Obtain the number having
    // maximum count
    for ($i = 1; $i <= $k; ++$i)
    {
        if ($numCount[$i] > $length)
        {
            $length = $numCount[$i];
            $lcm = $i;
        }
    }
 
    // Condition to check if answer
    // doesn't exist
    if ($lcm == 0)
        echo -1 << "\n";
    else
    {
 
        // Print the answer
        echo "LCM = ", $lcm,
             ", Length = ", $length, "\n";
 
        echo "Indexes = ";
        for ($i = 0; $i < $n; ++$i)
            if ($lcm % $arr[$i] == 0)
                echo $i, " ";
    }
}
 
// Driver code
$k = 14;
$arr = array( 2, 3, 4, 5 );
$n = count($arr);
 
findSubsequence($arr, $n, $k);
 
// This code is contributed by Ryuga
?>

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function to find the longest subsequence
// having LCM less than or equal to K
function findSubsequence(arr, n, k) {
 
    // Map to store unique elements
    // and their frequencies
    let M = new Map();
 
    // Update the frequencies
    for (let i = 0; i < n; ++i) {
        if (M.has(arr[i])) {
            M.set(arr[i], M.get(arr[i]) + 1)
        } else {
            M.set(arr[i], 1)
        }
    }
 
    // Array to store the count of numbers whom
    // 1 <= X <= K is a multiple of
    let numCount = new Array(k + 1);
 
    for (let i = 0; i <= k; ++i)
        numCount[i] = 0;
 
    // Check every unique element
    for (let p of M) {
        if (p[0] <= k) {
 
            // Find all its multiples <= K
            for (let i = 1; ; ++i) {
                if (p[0] * i > k)
                    break;
 
                // Store its frequency
                numCount[p[0] * i] += p[1];
            }
        }
        else
            break;
    }
 
    let lcm = 0, length = 0;
 
    // Obtain the number having maximum count
    for (let i = 1; i <= k; ++i) {
        if (numCount[i] > length) {
            length = numCount[i];
            lcm = i;
        }
    }
 
    // Condition to check if answer
    // doesn't exist
    if (lcm == 0)
        document.write(-1 + "<br>");
    else {
 
        // Print the answer
        document.write(
        "LCM = " + lcm + ", Length = " + length + "<br>"
        );
 
        document.write("Indexes = ");
        for (let i = 0; i < n; ++i)
            if (lcm % arr[i] == 0)
                document.write(i + " ");
    }
}
 
// Driver code
 
 
let k = 14;
let arr = [2, 3, 4, 5];
let n = arr.length;
 
findSubsequence(arr, n, k);
 
 
// This code is contributed by _saurabh_jaiswal
 
</script>
Output: 
LCM = 12, Length = 3
Indexes = 0 1 2

 

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