Find the minimum value of m that satisfies ax + by = m and all values after m also satisfy

Given two positive integers ‘a’ and ‘b’ that represent coefficients in equation ax + by = m. Find the minimum value of m that satisfies the equation for any positive integer values of x and y. And after this minimum value, the equation is satisfied by all (greater) values of m. If no such minimum value exists, return “-1”.
Examples:

Input: a = 4, b = 7
Output: 18
Explanation: 18 is the smallest value that
             can be satisfied by equation 4x + 7y.         
             4*1 + 7*2 = 18

             And after 18 all values are satisfied 
             4*3 + 7*1 = 19
             4*5 + 7*0 = 20
             ... and so on.

This is a variation of Frobenius coin problem. In Frobenius coin problem, we need to find the largest number that can not be represented using two coins. The largest amount for coins with denominations as ‘a’ and ‘b’ is a*b – (a+b). So the smallest number such that it can be represented using two coins and all numbers after it can also be represented is, a*b – (a+b) + 1.
One important case is when GCD of ‘a’ and ‘b’ is not 1. For example if ‘a’ = 4 and ‘b’ = 6, then all values that can be represented using two coins are even (or all values of m that can stratify the equation) are even. So all values that are NOT multiple of 2, cannot satisfy the equation. In this case there is no minimum value after which all values satisfy the equation.
Below is the implementation of above idea : 

C++

// C++ program to find the minimum value of m that satisfies
// ax + by = m and all values after m also satisfy
#include<bits/stdc++.h>
using namespace std;

int findMin(int a, int b)
{
    // If GCD is not 1, then there is no such value,
    // else value is obtained using "a*b-a-b+1'
    return (__gcd(a, b) == 1)? a*b-a-b+1 : -1;
}

// Driver code
int main()
{
    int a = 4, b = 7;
    cout << findMin(a, b) << endl;
    return 0;
}

Java

// Java program to find the
// minimum value of m that
// satisfies ax + by = m
// and all values after m
// also satisfy
import java.io.*;

class GFG
{
// Recursive function to
// return gcd of a and b
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);
}

static int findMin( int a, int b)
{

    // If GCD is not 1, then
    // there is no such value,
    // else value is obtained
    // using "a*b-a-b+1'
    return (__gcd(a, b) == 1)?
        a * b - a - b + 1 : -1;
}

// Driver code
public static void main (String[] args)
{
    int a = 4;
    int b = 7;
    System.out.println(findMin(a, b));
}
}

// This code is contributed
// by akt_mit

Python3

# Python3 program to find the minimum
# value of m that satisfies ax + by = m
# and all values after m also satisfy

# Recursive function to return
# gcd of a and b
def __gcd(a, b):

    # Everything divides 0
    if (a == 0 or 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);

def findMin( a, b):

    # If GCD is not 1, then
    # there is no such value,
    # else value is obtained
    # using "a*b-a-b+1'
    if(__gcd(a, b) == 1):
        return (a * b - a - b + 1)
    else:
        return -1

# Driver code
a = 4;
b = 7;
print(findMin(a, b));

# This code is contributed by mits

C#

// C# program to find the minimum
// value of m that satisfies
// ax + by = m and all values
// after m also satisfy
class GFG
{
// Recursive function to
// return gcd of a and b
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);
}

static int findMin( int a, int b)
{

    // If GCD is not 1, then
    // there is no such value,
    // else value is obtained
    // using "a*b-a-b+1'
    return (__gcd(a, b) == 1)?
           a * b - a - b + 1 : -1;
}

// Driver code
public static void Main()
{
    int a = 4;
    int b = 7;
    System.Console.WriteLine(findMin(a, b));
}
}

// This code is contributed
// by mits

PHP

<?php
// PHP program to find the
// minimum value of m that
// satisfies ax + by = m
// and all values after m
// also satisfy

// Recursive function to
// return gcd of a and b
function __gcd($a, $b)
{

    // Everything divides 0
    if ($a == 0 or $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 findMin( $a, $b)
{

    // If GCD is not 1, then
    // there is no such value,
    // else value is obtained
    // using "a*b-a-b+1'
    return (__gcd($a, $b) == 1)?
           $a * $b - $a - $b + 1 : -1;
}

    // Driver code
    $a = 4; $b = 7;
    echo findMin($a, $b) ;

// This code is contributed by anuj_67.
?>

Javascript

<script>

    // Javascript program
    // to find the minimum
    // value of m that satisfies
    // ax + by = m and all values
    // after m also satisfy

    // Recursive function to
    // return gcd of a and b
    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 findMin(a, b)
    {

        // If GCD is not 1, then
        // there is no such value,
        // else value is obtained
        // using "a*b-a-b+1'
        return (__gcd(a, b) == 1)?
                a * b - a - b + 1 : -1;
    }

    let a = 4;
    let b = 7;
    document.write(findMin(a, b));

</script>

Output:

18

Time Complexity: O(log(min(a, b)), where a and b are the two given integers.

Auxiliary Space: O(1), no extra space required so it is a constant.