# Smallest divisor D of N such that gcd(D, M) is greater than 1

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 ` `using` `namespace` `std;`   `// Function to find the minimum divisor` `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 = min(sec, mini);` `            ``}` `        ``}` `    ``}`   `    ``// If gcd is m itself` `    ``if` `(mini == m)` `        ``return` `-1;` `    ``else` `        ``return` `mini;` `}` `// Drivers code` `int` `main()` `{` `    ``int` `n = 8, m = 10;` `    ``cout << findMinimum(n, m);` `    ``return` `0;` `}`

## Java

 `// Java implementation of the above approach` `class` `GFG` `{`   `static` `int` `__gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(b == ``0``) ` `        ``return` `a; ` `    ``return` `__gcd(b, a % b); ` `    `  `} `   `// Function to find the minimum divisor` `static` `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 code` `public` `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 approach` `import` `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 approach` `using` `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 divisor` `static` `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 code` `static` `void` `Main()` `{` `    ``int` `n = 8, m = 10;` `    ``Console.WriteLine(findMinimum(n, m));` `}` `}`   `// This code is contributed by chandan_jnu`

## PHP

 ` 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

 ``

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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