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# Find the GCD that lies in given range

• Difficulty Level : Easy
• Last Updated : 06 Apr, 2021

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```

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

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## Javascript

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Output:
`3`

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