Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Find the subset of Array with given LCM

  • Difficulty Level : Hard
  • Last Updated : 22 Nov, 2021

Given an array arr[] consisting of N positive integers and a positive integer X, the task is to find the subset of the given array whose Lowest Common Multiple(LCM) is X. If there doesn’t exists any subset then print “-1”.

Examples:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: arr[ ] = {2, 4, 3, 5}, X = 20
Output: {4, 5}
Explanation:
Consider the subset {4, 5}, the LCM of this subset is 20(= X). Therefore, print this subset.



Input: arr[ ] = {2, 3, 4, 5}, X = 18
Output: -1

 

Naive Approach: The simplest approach to solve the given problem is to find all possible subsets of the given array and if there exists any subset whose LCM is X, then print that subset. Otherwise, print “-1”.

Time Complexity: O(N*(log N)*2N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by using the fact that if an array element is not a divisor of X, then that element can’t be included in the subset as the LCM will never be X. Follow the steps below to solve the problem:

  • Initialize a variable, say LCM as 1 that stores the LCM of the resultant subset.
  • Initialize a vector, say subset[] to store the array element included in the resultant subset.
  • Traverse the given array arr[] and perform the following steps:
  • After completing the above steps, if the value LCM is X, then print the array subset[] as the resultant subset. Otherwise, print “-1”.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the LCM of two
// numbers P and Q
int LCM(int P, int Q)
{
    // Return the value of LCM
    return ((P * Q) / __gcd(P, Q));
}
 
// Function to find the subset with
// LCM as X
int subsetArrayWithLCM_X(int A[], int N,
                         int X)
{
    // Stores LCM of resultant subset
    int lcm = 1;
 
    // Stores elements of subset
    vector<int> subset;
 
    // Traverse the array A[]
    for (int i = 0; i < N; i++) {
 
        // Check if the current element
        // is a divisor of X
        if (X % A[i] == 0) {
 
            // Inserting it into subset
            subset.push_back(A[i]);
 
            // Update the lcm
            lcm = LCM(lcm, A[i]);
        }
    }
 
    // Check if the final LCM is
    // not equal to X
    if (X != lcm) {
        cout << "-1";
        return 0;
    }
 
    // Otherwise, print the subset
    for (int i = 0; i < subset.size(); i++) {
        cout << subset[i] << ' ';
    }
 
    return 0;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 4, 5 };
    int X = 20;
    int N = sizeof(arr) / sizeof(arr[0]);
    subsetArrayWithLCM_X(arr, N, X);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG {
 
    // Function to find the LCM of two
    // numbers P and Q
    static int LCM(int P, int Q)
    {
       
        // Return the value of LCM
        return ((P * Q) / __gcd(P, Q));
    }
 
    // Function to find the subset with
    // LCM as X
    static int subsetArrayWithLCM_X(int A[], int N, int X)
    {
       
        // Stores LCM of resultant subset
        int lcm = 1;
 
        // Stores elements of subset
        Vector<Integer> subset = new Vector<Integer>();
 
        // Traverse the array A[]
        for (int i = 0; i < N; i++) {
 
            // Check if the current element
            // is a divisor of X
            if (X % A[i] == 0) {
 
                // Inserting it into subset
                subset.add(A[i]);
 
                // Update the lcm
                lcm = LCM(lcm, A[i]);
            }
        }
 
        // Check if the final LCM is
        // not equal to X
        if (X != lcm) {
            System.out.print("-1");
            return 0;
        }
 
        // Otherwise, print the subset
        for (int i = 0; i < subset.size(); i++) {
            System.out.print(subset.get(i) + " ");
        }
 
        return 0;
 
    }
 
    static int __gcd(int a, int b) {
        return b == 0 ? a : __gcd(b, a % b);
    }
 
    // Driver Code
    public static void main(String[] args) {
        int arr[] = { 2, 3, 4, 5 };
        int X = 20;
        int N = arr.length;
        subsetArrayWithLCM_X(arr, N, X);
 
    }
}
 
// This code is contributed by 29AjayKumar

Python3




# Python 3 program for the above approach
from math import gcd
 
# Function to find the LCM of two
# numbers P and Q
def LCM(P, Q):
   
    # Return the value of LCM
    return ((P * Q) // gcd(P, Q))
 
# Function to find the subset with
# LCM as X
def subsetArrayWithLCM_X(A, N, X):
   
    # Stores LCM of resultant subset
    lcm = 1
 
    # Stores elements of subset
    subset = []
 
    # Traverse the array A[]
    for i in range(N):
       
        # Check if the current element
        # is a divisor of X
        if (X % A[i] == 0):
           
            # Inserting it into subset
            subset.append(A[i])
 
            # Update the lcm
            lcm = LCM(lcm, A[i])
 
    # Check if the final LCM is
    # not equal to X
    if (X != lcm):
        print("-1")
        return 0
 
    # Otherwise, print the subset
    for i in range(len(subset)):
        print(subset[i],end = ' ')
 
    return 0
 
# Driver Code
if __name__ == '__main__':
    arr = [2, 3, 4, 5]
    X = 20
    N = len(arr)
    subsetArrayWithLCM_X(arr, N, X)
     
    # This code is contributed by SURENDRA_GANGWAR.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG {
 
    // Function to find the LCM of two
    // numbers P and Q
    static int LCM(int P, int Q)
    {
       
        // Return the value of LCM
        return ((P * Q) / __gcd(P, Q));
    }
 
    // Function to find the subset with
    // LCM as X
    static int subsetArrayWithLCM_X(int []A, int N, int X)
    {
       
        // Stores LCM of resultant subset
        int lcm = 1;
 
        // Stores elements of subset
        List<int> subset = new List<int>();
 
        // Traverse the array []A
        for (int i = 0; i < N; i++) {
 
            // Check if the current element
            // is a divisor of X
            if (X % A[i] == 0) {
 
                // Inserting it into subset
                subset.Add(A[i]);
 
                // Update the lcm
                lcm = LCM(lcm, A[i]);
            }
        }
 
        // Check if the readonly LCM is
        // not equal to X
        if (X != lcm) {
            Console.Write("-1");
            return 0;
        }
 
        // Otherwise, print the subset
        for (int i = 0; i < subset.Count; i++) {
            Console.Write(subset[i] + " ");
        }
 
        return 0;
 
    }
 
    static int __gcd(int a, int b) {
        return b == 0 ? a : __gcd(b, a % b);
    }
 
    // Driver Code
    public static void Main(String[] args) {
        int []arr = { 2, 3, 4, 5 };
        int X = 20;
        int N = arr.Length;
        subsetArrayWithLCM_X(arr, N, X);
 
    }
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to find the LCM of two
// numbers P and Q
function gcd(a, b)
{
     
    // Everything divides 0
    if (a == 0)
        return b;
    if (b == 0)
        return a;
 
    // Base case
    if (a == b)
        return a;
 
    // a is greater
    if (a > b)
        return gcd(a - b, b);
         
    return gcd(a, b - a);
}
 
// Function to find the LCM of two
// numbers P and Q
function LCM(P, Q)
{
     
    // Return the value of LCM
    return ((P * Q) / gcd(P, Q));
}
 
// Function to find the subset with
// LCM as X
function subsetArrayWithLCM_X(A, N, X)
{
     
    // Stores LCM of resultant subset
    let lcm = 1;
 
    // Stores elements of subset
    let subset = [];
 
    // Traverse the array A[]
    for(let i = 0; i < N; i++)
    {
         
        // Check if the current element
        // is a divisor of X
        if (X % A[i] == 0)
        {
             
            // Inserting it into subset
            subset.push(A[i]);
 
            // Update the lcm
            lcm = LCM(lcm, A[i]);
        }
    }
 
    // Check if the final LCM is
    // not equal to X
    if (X != lcm)
    {
        document.write("-1");
        return 0;
    }
 
    // Otherwise, print the subset
    for(let i = 0; i < subset.length; i++)
    {
        document.write(subset[i] + ' ');
    }
    return 0;
}
 
// Driver Code
let arr = [ 2, 3, 4, 5 ];
let X = 20;
let N = arr.length;
 
subsetArrayWithLCM_X(arr, N, X);
 
// This code is contributed by Potta Lokesh
 
</script>
Output: 
2 4 5

 

Time Complexity: O(N*log M), where M is the maximum element of the array.
Auxiliary Space: O(N)




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!