Find the GCD that lies in given range

Given two positive integer a and b and a range [low, high]. The task is to find the greatest common divisor of a and b which lie in the given range. If no divisor exist in the range, print -1.
Examples:

Input : a = 9, b = 27, low = 1, high = 5
Output : 3
3 is the highest number that lies in range 
[1, 5] and is common divisor of 9 and 27.

Input : a = 9, b = 27, low = 10, high = 11
Output : -1

The idea is to find the Greatest Common Divisor GCD(a, b) of a and b. Now observe, divisor of GCD(a, b) is also the divisor of a and b. So, we will iterate a loop i from 1 to sqrt(GCD(a, b)) and check if i divides GCD(a, b). Also, observe if i is divisor of GCD(a, b) then GCD(a, b)/i will also be divisor. So, for each iteration, if i divides GCD(a, b), we will find maximum of i and GCD(a, b)/i if they lie in the range. 
Below is the implementation of this approach: 

C++

// CPP Program to find the Greatest Common divisor
// of two number which is in given range
#include <bits/stdc++.h>
using namespace std;

// Return the greatest common divisor
// of two numbers
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

// Return the greatest common divisor of a
// and b which lie in the given range.
int maxDivisorRange(int a, int b, int l, int h)
{
    int g = gcd(a, b);
    int res = -1;

    // Loop from 1 to sqrt(GCD(a, b).
    for (int i = l; i * i <= g && i <= h; i++)

        // if i divides the GCD(a, b), then
        // find maximum of three numbers res,
        // i and g/i
        if (g % i == 0)
            res = max({res, i, g / i});

    return res;
}

// Driven Program
int main()
{
    int a = 3, b = 27, l = 1, h = 5;
    cout << maxDivisorRange(a, b, l, h) << endl;
    return 0;
}

Java

// Java Program to find the Greatest Common
// divisor of two number which is in given
// range
import java.io.*;

class GFG {

    // Return the greatest common divisor
    // of two numbers
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    // Return the greatest common divisor of a
    // and b which lie in the given range.
    static int maxDivisorRange(int a, int b,
                                   int l, int h)
    {
        int g = gcd(a, b);
        int res = -1;

        // Loop from 1 to sqrt(GCD(a, b).
        for (int i = l; i * i <= g && i <= h; i++)

            // if i divides the GCD(a, b), then
            // find maximum of three numbers res,
            // i and g/i
            if (g % i == 0)
                res = Math.max(res,
                             Math.max(i, g / i));

        return res;
    }

    // Driven Program
    public static void main (String[] args)
    {
        int a = 3, b = 27, l = 1, h = 5;
        System.out.println(
             maxDivisorRange(a, b, l, h));
    }
}

// This code is contributed by anuj_67.

Python3

# Python3 Program to find the
# Greatest Common divisor
# of two number which is
# in given range


# Return the greatest common
# divisor of two numbers
def gcd(a, b):
    if(b == 0):
        return a;
    return gcd(b, a % b);

# Return the greatest common
# divisor of a and b which
# lie in the given range.
def maxDivisorRange(a, b, l, h):
    g = gcd(a, b);
    res = -1;
    # Loop from 1 to
    # sqrt(GCD(a, b).
    i = l;
    while(i * i <= g and i <= h):
        # if i divides the GCD(a, b),
        # then find maximum of three
        # numbers res, i and g/i
        if(g % i == 0):
            res = max(res,max(i, g/i));
        i+=1;
    return int(res);

# Driver Code
if __name__ == "__main__":
    a = 3;
    b = 27;
    l = 1;
    h = 5;

    print(maxDivisorRange(a, b, l, h));

# This code is contributed by mits

C#

// C# Program to find the Greatest Common
// divisor of two number which is in given
// range
using System;

class GFG {

    // Return the greatest common divisor
    // of two numbers
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    // Return the greatest common divisor of a
    // and b which lie in the given range.
    static int maxDivisorRange(int a, int b,
                                int l, int h)
    {
        int g = gcd(a, b);
        int res = -1;

        // Loop from 1 to sqrt(GCD(a, b).
        for (int i = l; i * i <= g && i <= h; i++)

            // if i divides the GCD(a, b), then
            // find maximum of three numbers res,
            // i and g/i
            if (g % i == 0)
                res = Math.Max(res,
                            Math.Max(i, g / i));

        return res;
    }

    // Driven Program
    public static void Main ()
    {
        int a = 3, b = 27, l = 1, h = 5;
        Console.WriteLine(
            maxDivisorRange(a, b, l, h));
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// PHP Program to find the
// Greatest Common divisor
// of two number which is
// in given range


// Return the greatest common
// divisor of two numbers
function gcd($a, $b)
{
    if ($b == 0)
        return $a;
    return gcd($b, $a % $b);
}

// Return the greatest common
// divisor of a and b which
// lie in the given range.
function maxDivisorRange($a, $b,
                         $l, $h)
{
    $g = gcd($a, $b);
    $res = -1;

    // Loop from 1 to
    // sqrt(GCD(a, b).
    for ($i = $l; $i * $i <= $g and
                  $i <= $h; $i++)

        // if i divides the GCD(a, b),
        // then find maximum of three
        // numbers res, i and g/i
        if ($g % $i == 0)
            $res = max($res,
                   max($i, $g / $i));

    return $res;
}

// Driver Code
$a = 3; $b = 27;
$l = 1; $h = 5;

echo maxDivisorRange($a, $b, $l, $h);

// This code is contributed by anuj_67.
?>

Javascript

<script>

// JavaScript Program to find the Greatest Common
// divisor of two number which is in given
// range

    // Return the greatest common divisor
    // of two numbers
    function gcd(a , b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    // Return the greatest common divisor of a
    // and b which lie in the given range.
    function maxDivisorRange(a , b , l , h) {
        var g = gcd(a, b);
        var res = -1;

        // Loop from 1 to sqrt(GCD(a, b).
        for (i = l; i * i <= g && i <= h; i++)

            // if i divides the GCD(a, b), then
            // find maximum of three numbers res,
            // i and g/i
            if (g % i == 0)
                res = Math.max(res, Math.max(i, g / i));

        return res;
    }

    // Driven Program

        var a = 3, b = 27, l = 1, h = 5;
        document.write(maxDivisorRange(a, b, l, h));

// This code is contributed by todaysgaurav

</script>

Output:

3

Time complexity: O(log(min(a,b)) + min(?m,h)). where m is the gcd of inputs a and b.
Auxiliary Space:  O(1)