Minimum value that divides one number and divisible by other

Given two integer p and q, the task is to find the minimum possible number x such that q % x = 0 and x % p = 0. If the conditions aren’t true for any number then print -1.
Examples:

Input: p = 3, q = 99 
Output: 3 
99 % 3 = 0 
3 % 3 = 0
Input: p = 2, q = 7 
Output: -1 

Approach: If a number x satisfies the given condition then it’s obvious that q will be divided by p i.e. q % p = 0 because x is a multiple of p and q is a multiple of x. 
So the minimum possible value of x will be the GCD of p and q and when q is not divisible by p then no number will satisfy the given condition.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the minimum valid number
// that satisfies the given conditions
int minValidNumber(int p, int q)
{
    // If possible
    if (q % p == 0)
        return __gcd(p, q);
    else
        return -1;
}

// Driver Code
int main()
{
    int p = 2, q = 6;
    cout << minValidNumber(p, q);
    return 0;
}

Java

//Java  implementation of the approach

import java.io.*;

class GFG {

    // Function to calculate gcd
    static int __gcd(int a, int b)
    {

        // Everything divides 0
        if (a == 0 || b == 0)
            return 0;

        // 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 return the minimum valid number
// that satisfies the given conditions
static int minValidNumber(int p, int q)
{
    // If possible
    if (q % p == 0)
        return __gcd(p, q);
    else
        return -1;
}

// Driver Code
    public static void main (String[] args) {
        int p = 2, q = 6;
        System.out.print(minValidNumber(p, q));


    // THis code is contributed by  Sachin.
    }
}

Python3

# Python3 implementation of the approach
from math import gcd

# Function to return the minimum
# valid number that satisfies the
# given conditions
def minValidNumber(p, q) :

    # If possible
    if (q % p == 0) :
        return gcd(p, q)
    else :
        return -1

# Driver Code
if __name__ == "__main__" :

    p, q = 2, 6;
    print(minValidNumber(p, q))

# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;

class GFG
{
    // Function to calculate gcd
    static int __gcd(int a, int b)
    {

        // Everything divides 0
        if (a == 0 || b == 0)
            return 0;

        // 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 return the minimum valid number
// that satisfies the given conditions
static int minValidNumber(int p, int q)
{
    // If possible
    if (q % p == 0)
        return __gcd(p, q);
    else
        return -1;
}

// Driver Code
public static void Main()
{
    int p = 2, q = 6;
    Console.Write(minValidNumber(p, q));
}
}

// THis code is contributed
// by Mukul Singh

PHP

<?php
// Php implementation of the approach

function gcd($a, $b)
{

    // Everything divides 0
    if($b == 0)

        return $a ;

    return gcd($b , $a % $b);
}

// Function to return the minimum valid
// number that satisfies the given conditions
function minValidNumber($p, $q)
{
    // If possible
    if ($q % $p == 0)
        return gcd($p, $q);
    else
        return -1;
}

// Driver Code
$p = 2;
$q = 6;
echo minValidNumber($p, $q);

// This code is contributed by ita_c
?>

Javascript

<script>

    // Javascript implementation of the approach

    // Function to calculate gcd
    function __gcd(a, b)
    {

        // Everything divides 0
        if (a == 0 || b == 0)
            return 0;

        // 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 return the minimum valid number
    // that satisfies the given conditions
    function minValidNumber(p, q)
    {
        // If possible
        if (q % p == 0)
            return __gcd(p, q);
        else
            return -1;
    }

    let p = 2, q = 6;
    document.write(minValidNumber(p, q));

</script>

Output:

2

Time Complexity: O(logn)

Auxiliary Space: O(1)