# Count possible values of K such that X – 1 and Y – 1 modulo K is same

Given two numbers X and Y, the task is to find the count of integers K which satisfies the equation (X – 1) % K = (Y – 1) % K.

Examples:

Input: X = 2, Y = 6
Output: 3
Explanation:
K = {1, 2, 4} satisfies the given equation. Therefore, the count is 3.

Input: X = 4, Y = 9
Output: 2

Approach: Follow the steps below to solve the problem:

• The idea is to find the absolute difference of X and Y.
• Find the count of divisors of the calculated absolute difference and print the count as the required answer.

Below is the implementation of the above approach:

## C++14

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to count integers K` `// satisfying given equation` `int` `condition(``int` `a, ``int` `b)` `{` `    ``// Calculate the absoluter` `    ``// difference between a and b` `    ``int` `d = ``abs``(a - b), count = 0;`   `    ``// Iterate till sqrt of the diffeence` `    ``for` `(``int` `i = 1; i <= ``sqrt``(d); i++) {` `        ``if` `(d % i == 0) {` `            ``if` `(d / i == i)` `                ``count += 1;` `            ``else` `                ``count += 2;` `        ``}` `    ``}`   `    ``// Return the count` `    ``return` `count;` `}`   `// Driver Code` `int` `main()` `{`   `    ``int` `x = 2, y = 6;` `    ``cout << condition(x, y) << endl;`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.io.*;`   `class` `GFG{` ` `  `// Function to count integers K` `// satisfying given equation` `static` `int` `condition(``int` `a, ``int` `b)` `{` `    `  `    ``// Calculate the absoluter` `    ``// difference between a and b` `    ``int` `d = Math.abs(a - b), count = ``0``;` ` `  `    ``// Iterate till sqrt of the diffeence` `    ``for``(``int` `i = ``1``; i <= Math.sqrt(d); i++)` `    ``{` `        ``if` `(d % i == ``0``)` `        ``{` `            ``if` `(d / i == i)` `                ``count += ``1``;` `            ``else` `                ``count += ``2``;` `        ``}` `    ``}` ` `  `    ``// Return the count` `    ``return` `count;` `}` ` `  `// Driver Code` `public` `static` `void` `main (String[] args)` `{` `    ``int` `x = ``2``, y = ``6``;`   `    ``System.out.println(condition(x, y));` `}` `}`   `// This code is contributed by sanjoy_62`

## Python3

 `# Python3 program for the above approach`   `# Function to count integers K` `# satisfying given equation` `def` `condition(a, b):` `    `  `    ``# Calculate the absoluter` `    ``# difference between a and b` `    ``d ``=` `abs``(a ``-` `b)` `    ``count ``=` `0`   `    ``# Iterate till sqrt of the diffeence` `    ``for` `i ``in` `range``(``1``, d ``+` `1``):` `        ``if` `i ``*` `i > d:` `            ``break` `        ``if` `(d ``%` `i ``=``=` `0``):` `            ``if` `(d ``/``/` `i ``=``=` `i):` `                ``count ``+``=` `1` `            ``else``:` `                ``count ``+``=` `2`   `    ``# Return the count` `    ``return` `count`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:`   `    ``x ``=` `2` `    ``y ``=` `6` `    `  `    ``print``(condition(x, y))`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the ` `// above approach` `using` `System;` `class` `GFG{` ` `  `// Function to count ` `// integers K satisfying ` `// given equation` `static` `int` `condition(``int` `a, ` `                     ``int` `b)` `{    ` `  ``// Calculate the absoluter` `  ``// difference between a and b` `  ``int` `d = Math.Abs(a - b), count = 0;`   `  ``// Iterate till sqrt of ` `  ``// the diffeence` `  ``for``(``int` `i = 1; ` `          ``i <= Math.Sqrt(d); i++)` `  ``{` `    ``if` `(d % i == 0)` `    ``{` `      ``if` `(d / i == i)` `        ``count += 1;` `      ``else` `        ``count += 2;` `    ``}` `  ``}`   `  ``// Return the count` `  ``return` `count;` `}` ` `  `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `  ``int` `x = 2, y = 6;` `  ``Console.WriteLine(condition(x, y));` `}` `}`   `// This code is contributed by shikhasingrajput`

Output:

```3

```

Time Complexity: O(√abs(X – Y))
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.