Minimum value that divides one number and divisible by other

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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:

Time Complexity: O(logn)

Auxiliary Space: O(1)

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