[java] How to find GCD, LCM on a set of numbers

What would be the easiest way to calculate Greatest Common Divisor and Least Common Multiple on a set of numbers? What math functions can be used to find this information?

There is an Euclid's algorithm for GCD,

public int GCF(int a, int b) {
    if (b == 0) return a;
    else return (GCF (b, a % b));
}

By the way, a and b should be greater or equal 0, and LCM = |ab| / GCF(a, b)

With Java 8, there are more elegant and functional ways to solve this.

LCM:

private static int lcm(int numberOne, int numberTwo) {
    final int bigger = Math.max(numberOne, numberTwo);
    final int smaller = Math.min(numberOne, numberTwo);

    return IntStream.rangeClosed(1,smaller)
                    .filter(factor -> (factor * bigger) % smaller == 0)
                    .map(factor -> Math.abs(factor * bigger))
                    .findFirst()
                    .getAsInt();
}

GCD:

private static int gcd(int numberOne, int numberTwo) {
    return (numberTwo == 0) ? numberOne : gcd(numberTwo, numberOne % numberTwo);
}

Of course if one argument is 0, both methods will not work.

import java.util.Scanner; public class Lcmhcf {

/**
 * @param args the command line arguments
 */
public static void main(String[] args) {
    // TODO code application logic here
    Scanner scan = new Scanner(System.in);
    int n1,n2,x,y,lcm,hcf;
    System.out.println("Enter any 2 numbers....");
    n1=scan.nextInt();
    n2=scan.nextInt();
    x=n1;
    y=n2;

    do{
       if(n1>n2){
         n1=n1-n2;
       }
       else{
         n2=n2-n1;
       }
     } while(n1!=n2);
     hcf=n1;
     lcm=x*y/hcf;
     System.out.println("HCF IS = "+hcf);
     System.out.println("LCM IS = "+lcm);

     }
 }
//## Heading ##By Rajeev Lochan Sen

Basically to find gcd and lcm on a set of numbers you can use below formula,

LCM(a, b) X HCF(a, b) = a * b

Meanwhile in java you can use euclid's algorithm to find gcd and lcm, like this

public static int GCF(int a, int b)
{
    if (b == 0)
    {
       return a;
    }
    else
    {
       return (GCF(b, a % b));
    }
}

You can refer this resource to find examples on euclid's algorithm.

If you can use Java 8 (and actually want to) you can use lambda expressions to solve this functionally:

private static int gcd(int x, int y) {
    return (y == 0) ? x : gcd(y, x % y);
}

public static int gcd(int... numbers) {
    return Arrays.stream(numbers).reduce(0, (x, y) -> gcd(x, y));
}

public static int lcm(int... numbers) {
    return Arrays.stream(numbers).reduce(1, (x, y) -> x * (y / gcd(x, y)));
}

I oriented myself on Jeffrey Hantin's answer, but

• calculated the gcd functionally
• used the varargs-Syntax for an easier API (I was not sure if the overload would work correctly, but it does on my machine)
• transformed the gcd of the numbers-Array into functional syntax, which is more compact and IMO easier to read (at least if you are used to functional programming)

This approach is probably slightly slower due to additional function calls, but that probably won't matter at all for the most use cases.

int lcmcal(int i,int y)
{
    int n,x,s=1,t=1;
    for(n=1;;n++)
    {
        s=i*n;
        for(x=1;t<s;x++)
        {
            t=y*x;
        }
        if(s==t)
            break;
    }
    return(s);
}

import java.util.*;
public class lcm {
    public static void main(String args[])
    {
        int lcmresult=1;
        System.out.println("Enter the number1: ");
        Scanner s=new Scanner(System.in);
        int a=s.nextInt();
        System.out.println("Enter the number2: ");
        int b=s.nextInt();
        int max=a>b?a:b;
        for(int i=2;i<=max;i++)
        {
            while(a%i==0||b%i==0)
            {
                lcmresult=lcmresult*i;
                if(a%i==0)
                    a=a/i;
                if(b%i==0)
                    b=b/i;
                if(a==1&&b==1)
                    break;
            }
        }
    System.out.println("lcm: "+lcmresult);
}
}

There are no build in function for it. You can find the GCD of two numbers using Euclid's algorithm.

For a set of number

GCD(a_1,a_2,a_3,...,a_n) = GCD( GCD(a_1, a_2), a_3, a_4,..., a_n )

Apply it recursively.

Same for LCM:

LCM(a,b) = a * b / GCD(a,b)
LCM(a_1,a_2,a_3,...,a_n) = LCM( LCM(a_1, a_2), a_3, a_4,..., a_n )

for gcd you cad do as below:

    String[] ss = new Scanner(System.in).nextLine().split("\\s+");
    BigInteger bi,bi2 = null;
    bi2 = new BigInteger(ss[1]);
    for(int i = 0 ; i<ss.length-1 ; i+=2 )
    {
        bi = new BigInteger(ss[i]);
        bi2 = bi.gcd(bi2);
    }
    System.out.println(bi2.toString());

import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int n0 = input.nextInt(); // number of intended input.
        int [] MyList = new int [n0];

        for (int i = 0; i < n0; i++)
            MyList[i] = input.nextInt();
            //input values stored in an array
        int i = 0;
        int count = 0;
            int gcd = 1; // Initial gcd is 1
            int k = 2; // Possible gcd
            while (k <= MyList[i] && k <= MyList[i]) {
                if (MyList[i] % k == 0 && MyList[i] % k == 0)
                    gcd = k; // Update gcd
                k++;
                count++; //checking array for gcd
            }
           // int i = 0;
            MyList [i] = gcd;
            for (int e: MyList) {
                System.out.println(e);

            }

            }

        }

int lcm = 1;
int y = 0;
boolean flag = false;
for(int i=2;i<=n;i++){
            if(lcm%i!=0){
                for(int j=i-1;j>1;j--){
                    if(i%j==0){
                        flag =true;
                        y = j;
                        break;
                    }
                }
                if(flag){
                    lcm = lcm*i/y;
                }
                else{
                    lcm = lcm*i;
                }
            }
            flag = false;
        }

here, first for loop is for getting every numbers starting from '2'. then if statement check whether the number(i) divides lcm if it does then it skip that no. and if it doesn't then next for loop is for finding a no. which can divides the number(i) if this happens we don't need that no. we only wants its extra factor. so here if the flag is true this means there already had some factors of no. 'i' in lcm. so we divide that factors and multiply the extra factor to lcm. If the number isn't divisible by any of its previous no. then when simply multiply it to the lcm.

public class HcfLcm {
    public static void main(String[] args) {
        System.out.println("HCF: "+ getHcf(20, 15)); //5
        System.out.println("LCM: "+ getLcm2(20, 15)); //60
    }

    private static Integer getLcm2(int n1, int n2) {
        int lcm = Math.max(n1, n2);
        // Always true
        while (true) {
            if (lcm % n1 == 0 && lcm % n2 == 0) {
                break;
            }
            ++lcm;
        }
        return lcm;
    }

    private static Integer getLcm(int i, int j) {
        int hcf = getHcf(i, j);
        return hcf * i/hcf * j/hcf; // i*j*hcf
    }

    private static Integer getHcf(int i, int j) {
        while(i%j != 0) {
            int temp = i%j;
            i = j;
            j = temp;
        }
        return j;
    }
}

int gcf(int a, int b)
{
    while (a != b) // while the two numbers are not equal...
    { 
        // ...subtract the smaller one from the larger one

        if (a > b) a -= b; // if a is larger than b, subtract b from a
        else b -= a; // if b is larger than a, subtract a from b
    }

    return a; // or return b, a will be equal to b either way
}

int lcm(int a, int b)
{
    // the lcm is simply (a * b) divided by the gcf of the two

    return (a * b) / gcf(a, b);
}