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
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 <bits/stdc++.h>using namespace std;// Return the greatest common divisor// of two numbersint 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 Programint 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// rangeimport 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 numbersdef 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 Codeif __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// rangeusing 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
<?php// PHP Program to find the// Greatest Common divisor// of two number which is// in given range// Return the greatest common// divisor of two numbersfunction gcd($a, $b){ if ($b == 0) return $a; return gcd($b, $a % $b);}// Return the gretest common// divisor of a and b which// lie in the given range.function maxDivisorRange($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 gretest 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
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