Given two positive integers N and M., The task is to find the smallest divisor D of N such that gcd(D, M) > 1. If there are no such divisors, then print -1.
Examples:
Input: N = 8, M = 10
Output: 2
Input: N = 8, M = 1
Output: -1
A naive approach is to iterate for every factor and calculate the gcd of the factor and M. If it exceeds M, then we have the answer.
Time Complexity: O(N * log max(N, M))
An efficient approach is to iterate till sqrt(n) and check for gcd(i, m). If gcd(i, m) > 1, then we print and break it, else we check for gcd(n/i, m) and store the minimum of them.
Below is the implementation of the above approach.
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum divisorint findMinimum(int n, int m){ int mini = m; // Iterate for all factors of N for (int i = 1; i * i <= n; i++) { if (n % i == 0) { int sec = n / i; // Check for gcd > 1 if (__gcd(m, i) > 1) { return i; } // Check for gcd > 1 else if (__gcd(sec, m) > 1) { mini = min(sec, mini); } } } // If gcd is m itself if (mini == m) return -1; else return mini;}// Drivers codeint main(){ int n = 8, m = 10; cout << findMinimum(n, m); return 0;} |
Java
// Java implementation of the above approachclass GFG{static int __gcd(int a, int b) { if (b == 0) return a; return __gcd(b, a % b); } // Function to find the minimum divisorstatic int findMinimum(int n, int m){ int mini = m; // Iterate for all factors of N for (int i = 1; i * i <= n; i++) { if (n % i == 0) { int sec = n / i; // Check for gcd > 1 if (__gcd(m, i) > 1) { return i; } // Check for gcd > 1 else if (__gcd(sec, m) > 1) { mini = Math.min(sec, mini); } } } // If gcd is m itself if (mini == m) return -1; else return mini;}// Driver codepublic static void main (String[] args) { int n = 8, m = 10; System.out.println(findMinimum(n, m));}}// This code is contributed by chandan_jnu |
Python3
# Python3 implementation of the above approachimport math# Function to find the minimum divisor def findMinimum(n, m): mini, i = m, 1 # Iterate for all factors of N while i * i <= n: if n % i == 0: sec = n // i # Check for gcd > 1 if math.gcd(m, i) > 1: return i # Check for gcd > 1 elif math.gcd(sec, m) > 1: mini = min(sec, mini) i += 1 # If gcd is m itself if mini == m: return -1 else: return mini # Drivers code if __name__ == "__main__": n, m = 8, 10 print(findMinimum(n, m)) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the above approachusing System;class GFG{static int __gcd(int a, int b) { if (b == 0) return a; return __gcd(b, a % b); } // Function to find the minimum divisorstatic int findMinimum(int n, int m){ int mini = m; // Iterate for all factors of N for (int i = 1; i * i <= n; i++) { if (n % i == 0) { int sec = n / i; // Check for gcd > 1 if (__gcd(m, i) > 1) { return i; } // Check for gcd > 1 else if (__gcd(sec, m) > 1) { mini = Math.Min(sec, mini); } } } // If gcd is m itself if (mini == m) return -1; else return mini;}// Driver codestatic void Main(){ int n = 8, m = 10; Console.WriteLine(findMinimum(n, m));}}// This code is contributed by chandan_jnu |
PHP
<?php// PHP implementation of the above approachfunction __gcd($a, $b) { if ($b == 0) return $a; return __gcd($b, $a % $b); } // Function to find the minimum divisorfunction findMinimum($n, $m){ $mini = $m; // Iterate for all factors of N for ($i = 1; $i * $i <= $n; $i++) { if ($n % $i == 0) { $sec = $n / $i; // Check for gcd > 1 if (__gcd($m, $i) > 1) { return $i; } // Check for gcd > 1 else if (__gcd($sec, $m) > 1) { $mini = min($sec, $mini); } } } // If gcd is m itself if ($mini == $m) return -1; else return $mini;}// Driver code$n = 8; $m = 10;echo(findMinimum($n, $m));// This code is contributed by Code_Mech. |
Javascript
<script>// javascript implementation of the above approach function __gcd(a , b) { if (b == 0) return a; return __gcd(b, a % b); } // Function to find the minimum divisor function findMinimum(n , m) { var mini = m; // Iterate for all factors of N for (var i = 1; i * i <= n; i++) { if (n % i == 0) { var sec = n / i; // Check for gcd > 1 if (__gcd(m, i) > 1) { return i; } // Check for gcd > 1 else if (__gcd(sec, m) > 1) { mini = Math.min(sec, mini); } } } // If gcd is m itself if (mini == m) return -1; else return mini; } // Driver code var n = 8, m = 10; document.write(findMinimum(n, m));// This code is contributed by todaysgaurav</script> |
Output:
2
Time Complexity: O(sqrt(N) * log max(N, M)), as we are using a loop to traverse sqrt(N) times and we are using the inbuilt GCD function in each traversal which costs logN time. Where N and M are the two integers provided.
Auxiliary Space: O(1), as we are not using any extra space.
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