Given three numbers x, y and k, find the k’th common factor of x and y. Print -1 if there are less than k common factors of x and y.
Examples :
Input : x = 20, y = 24
k = 3
Output : 4
Common factors are 1, 2, 4, ...
Input : x = 4, y = 24
k = 2
Output : 2
Input : x = 22, y = 2
k = 3
Output : -1
We find the smaller of two numbers as common factor cannot be greater than the smaller number. Then we run a loop from 1 to the smaller number. For every number i, we check if it is a common factor. If yes, we increment count of common factors.
Below is the Implementation :
C++
// C++ program to find kth common factor// of two numbers#include<iostream>using namespace std;// Returns k'th common factor of x and y.int findKHCF(int x, int y, int k){ // Find smaller of two numbers int small = min(x, y); // Count common factors until we either // reach small or count becomes k. int count = 1; for (int i=2; i<=small; i++) { if (x % i==0 && y % i == 0) count++; if (count == k) return i; } // If we reached small return -1;}// Driver codeint main(){ int x = 4, y = 24, k = 3; cout << findKHCF(x, y, k); return 0;} |
Java
// Java program to find kth // common factor of two numbersimport java.lang.*;class GFG { // Returns k'th common factor of x and y.static int findKHCF(int x, int y, int k) { // Find smaller of two numbers int small = Math.min(x, y); // Count common factors until we either // reach small or count becomes k. int count = 1; for (int i = 2; i <= small; i++) { if (x % i == 0 && y % i == 0) count++; if (count == k) return i; } // If we reached small return -1;}// Driver codepublic static void main(String[] args) { int x = 4, y = 24, k = 3; System.out.print(findKHCF(x, y, k));}}// This code is contributed by Anant Agarwal. |
Python3
# Python program to find# kth common factor# of two numbers# Returns k'th common# factor of x and y.def findKHCF(x,y,k): # Find smaller of two numbers small = min(x, y) # Count common factors # until we either # reach small or count # becomes k. count = 1 for i in range(2,small+1): if (x % i==0 and y % i == 0): count=count + 1 if (count == k): return i # If we reached small return -1# Driver codex = 4y = 24k = 3print(findKHCF(x, y, k))# This code is contributed# by Anant Agarwal. |
C#
// C# program to find kth // common factor of two numbersusing System;class GFG { // Returns k'th common factor of x and y.static int findKHCF(int x, int y, int k){ // Find smaller of two numbers int small = Math.Min(x, y); // Count common factors until we either // reach small or count becomes k. int count = 1; for (int i = 2; i <= small; i++) { if (x % i == 0 && y % i == 0) count++; if (count == k) return i; } // If we reached small return -1;}// Driver codepublic static void Main() { int x = 4, y = 24, k = 3; Console.Write(findKHCF(x, y, k));}}// This code is contributed by Nitin Mittal. |
PHP
<?php// PHP program to find kth // common factor of two numbers// Returns k'th common // factor of x and y.function findKCF($x, $y, $k){// Find smaller of two numbers$small = min($x, $y);// Count common factors until we either// reach small or count becomes k.$count = 1;for ($i = 2; $i <= $small; $i++){ if ($x % $i == 0 && $y % $i == 0) $count++; if ($count == $k) return $i;}// If we reached smallreturn -1;}// Driver code$x = 4; $y = 24; $k = 3;echo findKCF($x, $y, $k);// This code is contributed by nitin mittal. ?> |
Javascript
<script>// Javascript program to find kth // common factor of two numbers// Returns k'th common factor of x and y.function findKHCF(x, y, k) { // Find smaller of two numbers let small = Math.min(x, y); // Count common factors until we either // reach small or count becomes k. let count = 1; for (let i = 2; i <= small; i++) { if (x % i == 0 && y % i == 0) count++; if (count == k) return i; } // If we reached small return -1;} // Driver code let x = 4, y = 24, k = 3; document.write(findKHCF(x, y, k)); </script> |
Time complexity: O(n) where n is smallest among (x,y)
Auxiliary space: O(1)
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