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

Last Updated : 09 Jun, 2022

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# 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

Time Complexity: O(nlogn)

Auxiliary Space: O(n)

C++ Program for Longest Increasing Subsequence

NaimishSingh

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

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