LCM of digits of a given number

Given a number n, find the LCM of its digits.

Examples: 

Input : 397
Output : 63
LCM of 3, 9 and 7 is 63.

Input : 244
Output : 4
LCM of 2, 4 and 4 is 4.

Method 1: 

Follow the steps to solve this problem:

• Initialise a variable l = 1
• While n is greater than 0. Do the following:
  • Find the LCM of n % 10 and lcm
  • Check if lcm == 0 , return 0
  • Initialise n = n / 10;
• Finally, return l.

Follow the steps below to implement the above approach:

// CPP program to find LCM of digits of a number
#include <bits/stdc++.h>
using namespace std;

// Recursive function to return gcd of a and b
long long gcd(long long int a, long long int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

// Function to return LCM of two numbers
long long lcm(int a, int b) { return (a / gcd(a, b)) * b; }

int digitLCM(int n)
{
    int l = 1;
    while (n > 0) {
        l = lcm(n % 10, l);

        // If at any point LCM become 0.
        // return it
        if (lcm == 0)
            return 0;

        n = n / 10;
    }
    return l;
}

// driver code
int main()
{
    long n = 397;
    cout << digitLCM(n);
    return 0;
}

// Java program to find LCM of digits of a number
import java.io.*;

class GFG
{
// define lcm function
static int lcm_fun(int a, int b)
{
    if (b == 0)
        return a;
    return lcm_fun(b, a % b);
}

static int digitLCM(int n)
{
    int lcm = 1;
    while (n > 0)
    {
        lcm = (n % 10 * lcm) / lcm_fun(n % 10, lcm);

        // If at any point LCM become 0.
        // return it
        if (lcm == 0)
            return 0;

        n = n/10;
    }
    return lcm;
}

// driver code
public static void main(String[] args)
{
    int n = 397;
    System.out.println(digitLCM(n));
}
}
// This code is contributed by mits

# Python3 program to find
# LCM of digits of a number

# define lcm function
def lcm_fun(a, b):

    if (b == 0):
        return a;
    return lcm_fun(b, a % b);

def digitLCM(n):

    lcm = 1;
    while (n > 0):
        lcm = int((n % 10 * lcm) / 
              lcm_fun(n % 10, lcm));

        # If at any point LCM 
        # become 0. return it
        if (lcm == 0):
            return 0;

        n = int(n / 10);
    
    return lcm;

# Driver code
n = 397;
print(digitLCM(n));

# This code is contributed by mits

// C# program to find LCM of digits
// of a number
class GFG
{
    
// define lcm function
static int lcm_fun(int a, int b)
{
    if (b == 0)
        return a;
    return lcm_fun(b, a % b);
}

static int digitLCM(int n)
{
    int lcm = 1;
    while (n > 0)
    {
        lcm = (n % 10 * lcm) / lcm_fun(n % 10, lcm);

        // If at any point LCM become 0.
        // return it
        if (lcm == 0)
            return 0;

        n = n/10;
    }
    return lcm;
}

// Driver Code
public static void Main()
{
    int n = 397;
    System.Console.WriteLine(digitLCM(n));
}
}

// This code is contributed by mits

<?php
// PHP program to find
// LCM of digits of a number

// define lcm function
function lcm_fun($a, $b)
{
    if ($b == 0)
        return $a;
    return lcm_fun($b, $a % $b);
}

function digitLCM($n)
{
    $lcm = 1;
    while ($n > 0)
    {
        $lcm = (int)(($n % 10 * $lcm) / 
              lcm_fun($n % 10, $lcm));

        // If at any point LCM 
        // become 0. return it
        if ($lcm == 0)
            return 0;

        $n = (int)($n / 10);
    }
    return $lcm;
}

// Driver code
$n = 397;
echo digitLCM($n);

// This code is contributed by mits
?>

<script>
// Javascript program to find LCM of digits of a number

    // define lcm function
    function lcm_fun( a, b) 
    {
        if (b == 0)
            return a;
        return lcm_fun(b, a % b);
    }

    function digitLCM( n)
    {
        let lcm = 1;
        while (n > 0) 
        {
            lcm = (n % 10 * lcm) / lcm_fun(n % 10, lcm);

            // If at any point LCM become 0.
            // return it
            if (lcm == 0)
                return 0;

            n = parseInt(n / 10);
        }
        return lcm;
    }

    // Driver code     
        let n = 397;
        document.write(digitLCM(n));

// This code is contributed by gauravrajput1 
</script>

63

Time Complexity: O(log n), the time complexity of this algorithm is O(log n) as we are making a single iteration and each iteration is taking O(1) time for computation.
Auxiliary Space: O(1), the space complexity of this algorithm is O(1) as we are using a single variable l to store the lcm of the digits.