# 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 gretest 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 maximimum 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 ` `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 gretest 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 gretest 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 gretest 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 gretest 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

 ` `

Output:

```3
```

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