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++ 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 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 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 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 ?> |
<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> |
LCM = 12, Length = 3 Indexes = 0 1 2
Time Complexity: O(nlogn)
Auxiliary Space: O(n)