Given two integers. We need to find if the first number x is divisible by all prime divisors of y.
Examples :
Input : x = 120, y = 75
Output : Yes
Explanation :
120 = (2^3)*3*5
75 = 3*(5^2)
120 is divisible by both 3 and 5 which
are the prime divisors of 75. Hence,
answer is "Yes".
Input : x = 15, y = 6
Output : No
Explanation :
15 = 3*5.
6 = 2*3,
15 is not divisible by 2 which is a
prime divisor of 6. Hence, answer
is "No".
A simple solution is to find all prime factors of y. For every prime factor, check if it divides x or not.
An efficient solution is based on the below facts.
1) if y == 1, then it no prime divisors. Hence answer is “Yes”
2) We find GCD of x and y.
a) If GCD == 1, then clearly there are no common divisors of x and y, hence answer is “No”.
b) If GCD > 1, the GCD contains prime divisors which divide x also. Now, we have all unique prime divisors if and only if y/GCD has such unique prime divisor. So we have to find uniqueness for pair (x, y/GCD) using recursion.
Below is the implementation of the above approach:
C++
// CPP program to find if all prime factors// of y divide x.#include <bits/stdc++.h>using namespace std;// Returns true if all prime factors of y// divide x.bool isDivisible(int x, int y){ if (y == 1) return true; int z = __gcd(x, y); if (z == 1) return false; return isDivisible(x, y / z);}// Driver Codeint main(){ int x = 18, y = 12; if (isDivisible(x, y)) cout << "Yes" << endl; else cout << "No" << endl; return 0;} |
Java
// Java program to find if all // prime factors of y divide x.public class Divisible { public static int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } // Returns true if all prime factors // of y divide x. static boolean isDivisible(int x, int y) { if (y == 1) return true; int z = gcd(x, y); if (z == 1) return false; return isDivisible(x, y / z); } // Driver program to test above functions public static void main(String[] args) { int x = 18, y = 12; if (isDivisible(x, y)) System.out.println("Yes"); else System.out.println("No"); }};// This code is contributed by Prerna Saini |
Python3
# python program to find if all# prime factors of y divide x.def gcd(a, b): if(b == 0): return a else: return gcd(b, a % b) # Returns true if all prime # factors of y divide x.def isDivisible(x,y): if (y == 1): return 1 z = gcd(x, y); if (z == 1): return false; return isDivisible(x, y / z);# Driver Codex = 18y = 12if (isDivisible(x, y)): print("Yes")else: print("No")# This code is contributed by Sam007 |
C#
// C# program to find if all // prime factors of y divide x.using System;class GFG { public static int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } // Returns true if all prime factors // of y divide x. static bool isDivisible(int x, int y) { if (y == 1) return true; int z = gcd(x, y); if (z == 1) return false; return isDivisible(x, y / z); } // Driver program to test above functions public static void Main() { int x = 18, y = 12; if (isDivisible(x, y)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}// This code is contributed by vt_m. |
Javascript
<script>// Javascript program to find if all // prime factors of y divide x.function gcd(a , b) { return b == 0 ? a : gcd(b, a % b); }// Returns true if all prime factors// of y divide x.function isDivisible(x , y){ if (y == 1) return true; var z = gcd(x, y); if (z == 1) return false; return isDivisible(x, y / z);}// Driver program to test above functionsvar x = 18, y = 12;if (isDivisible(x, y)) document.write("Yes");else document.write("No");// This code is contributed by Amit Katiyar </script> |
PHP
<?php// PHP program to find if all // prime factors of y divide x.function gcd ($a, $b) { return $b == 0 ? $a : gcd($b, $a % $b);} // Returns true if all prime // factors of y divide x.function isDivisible($x, $y){ if ($y == 1) return true; $z = gcd($x, $y); if ($z == 1) return false; return isDivisible($x, $y / $z);}// Driver Code$x = 18;$y = 12;if (isDivisible($x, $y)) echo "Yes";else echo "No";// This code is contributed by Sam007?> |
Output
Yes
Time Complexity: The time complexity for calculating GCD is O(log min(x, y)), and recursion will terminate after log y steps because we are reducing it by a factor greater than one. Overall Time complexity: O(log2y)
Auxiliary Space: O(log min(x, y))
Approach 2:Factorization:
Here’s another approach to check if all prime factors of y divide x:
- Find all the prime factors of y.
For each prime factor p, check if it divides x or not. If it doesn’t, return false.
If all prime factors of y divide x, return true. - First, the function isDivisible() finds all the prime factors of y using trial division. It starts with 2 and goes up to the square root of y, checking if each number i divides y or not. If i divides y, it is added to a vector factors. The loop continues until i*i becomes greater than y. If y is still greater than 1 after this loop, it means that y itself is a prime factor and it is added to the vector.
- Next, the function checks if each prime factor in the vector factors divides x or not. If any prime factor does not divide x, the function returns false. Otherwise, if all prime factors of y divide x, the function returns true.
Here is the code of this approach:
C++
#include <bits/stdc++.h>using namespace std;// Function to check if all prime factors of y divide xbool isDivisible(int x, int y){ // Find all prime factors of y vector<int> factors; for (int i = 2; i * i <= y; i++) { while (y % i == 0) { factors.push_back(i); y /= i; } } if (y > 1) factors.push_back(y); // Check if all prime factors divide x for (int p : factors) if (x % p != 0) return false; return true;}// Driver codeint main(){ int x = 18, y = 12; if (isDivisible(x, y)) cout << "Yes" << endl; else cout << "No" << endl; return 0;} |
Java
import java.util.ArrayList;import java.util.List;class Main { // Function to check if all prime factors of y divide x static boolean isDivisible(int x, int y) { // Find all prime factors of y List<Integer> factors = new ArrayList<>(); for (int i = 2; i * i <= y; i++) { while (y % i == 0) { factors.add(i); y /= i; } } if (y > 1) factors.add(y); // Check if all prime factors divide x for (int p : factors) if (x % p != 0) return false; return true; } // Driver code public static void main(String[] args) { int x = 18, y = 12; if (isDivisible(x, y)) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by Prajwal Kandekar |
Python
import math# Function to check if all prime factors of y divide xdef is_divisible(x, y): # Find all prime factors of y factors = [] for i in range(2, int(math.sqrt(y)) + 1): while y % i == 0: factors.append(i) y //= i if y > 1: factors.append(y) # Check if all prime factors divide x for p in factors: if x % p != 0: return False return True# Driver codeif __name__ == '__main__': x = 18 y = 12 if is_divisible(x, y): print("Yes") else: print("No") |
C#
using System;using System.Collections.Generic;class Program { // Function to check if all prime factors of y divide x static bool IsDivisible(int x, int y) { // Find all prime factors of y List<int> factors = new List<int>(); for (int i = 2; i * i <= y; i++) { while (y % i == 0) { factors.Add(i); y /= i; } } if (y > 1) factors.Add(y); // Check if all prime factors divide x foreach(int p in factors) if (x % p != 0) return false; return true; } // Driver code static void Main(string[] args) { int x = 18, y = 12; if (IsDivisible(x, y)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }} |
Javascript
// Function to check if all prime factors of y divide xfunction isDivisible(x, y) { // Find all prime factors of y let factors = []; for (let i = 2; i * i <= y; i++) { while (y % i === 0) { factors.push(i); y /= i; } } if (y > 1) factors.push(y); // Check if all prime factors divide x for (let p of factors) if (x % p !== 0) return false; return true;}// Driver codelet x = 18, y = 12;if (isDivisible(x, y)) console.log("Yes");else console.log("No"); |
Output
Yes
Time Complexity: O(sqrt(y) + log(x)), where sqrt(y) is the time required to find all prime factors of y, and log(x) is the time required to check if each prime factor
Auxiliary Space: O(sqrt(y)), as we are storing the prime factors in a vector.
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