Count possible values of K such that A%K = B%K

Given two integers A and B, the task is to count the possible values of K such that A%K = B%K. If the count is infinite, print -1.

Examples:

Input: A = 2, B = 4
Output: 2
Explanation: The set of all possible values of K is {1, 2}. 
As 2%1 = 4%1 = 0 and 2%2 = 4%2 = 0.

Input: A = 5, B = 5
Output: -1
Explanation: There are infinite values of K as all possible integer value of K satisfies the given condition.

Approach: The given problem can be solved using the following observation that all values of A and B can be divided into the following two cases:

• The case where A = B. In such cases, all possible integer value of K is valid answer. Hence, the value of the required count is infinite.
• The case where A > B. In such cases, it can be observed that a value of K is valid if K divides (A – B). For cases with B > A, simply swap the values of A and B.

Therefore, calculate all the possible values of K such that it divides (A – B) completely which is the required value.

Below is the implementation of the above approach:  

C++14

// C++ Program of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the count of the
// values of K such that A%K = B%K
int countInt(int A, int B)
{
    // If the count is Infinite
    if (A == B)
        return -1;

    int diff = abs(A - B);

    // Stores the required count
    int count = 0;

    // Loop to calculate all the
    // divisors of diff
    for (int i = 1; i * i <= diff; i++) {
        if (diff % i == 0) {

            // Increment count for i
            if (diff == i * i)
                count++;

            // Increment count for i
            // and diff / i both
            else
                count += 2;
        }
    }

    // Return Answer
    return count;
}

// Driver code
int main()
{
    int A = 2, B = 4;
    cout << countInt(A, B);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {

  // Function to find the count of the
  // values of K such that A%K = B%K
  static int countInt(int A, int B)
  {

    // If the count is Infinite
    if (A == B)
      return -1;

    int diff = Math.abs(A - B);

    // Stores the required count
    int count = 0;

    // Loop to calculate all the
    // divisors of diff
    for (int i = 1; i * i <= diff; i++) {
      if (diff % i == 0) {

        // Increment count for i
        if (diff == i * i)
          count++;

        // Increment count for i
        // and diff / i both
        else
          count += 2;
      }
    }

    // Return Answer
    return count;
  }

  // Driver code
  public static void main (String[] args) {

    int A = 2, B = 4;
    System.out.print(countInt(A, B));
  }
}

// This code is contributed by hrithikgarg03188.

Python3

# Python code for the above approach

# Function to find the count of the
# values of K such that A%K = B%K
def countInt(A, B):

    # If the count is Infinite
    if (A == B):
        return -1;

    diff = abs(A - B);

    # Stores the required count
    count = 0;

    # Loop to calculate all the
    # divisors of diff
    i = 1;
    while((i * i) <= diff):
        if (diff % i == 0):

            # Increment count for i
            if (diff == i * i):
                count += 1

            # Increment count for i
            # and diff / i both
            else:
                count += 2;
        i += 1
    # Return Answer
    return count;

# Driver code

A = 2
B = 4
print(countInt(A, B));

# This code is contributed by Saurabh Jaiswal

C#

// C# Program of the above approach
using System;
class GFG {
    // Function to find the count of the
    // values of K such that A%K = B%K
    static int countInt(int A, int B)
    {
        // If the count is Infinite
        if (A == B)
            return -1;

        int diff = Math.Abs(A - B);

        // Stores the required count
        int count = 0;

        // Loop to calculate all the
        // divisors of diff
        for (int i = 1; i * i <= diff; i++) {
            if (diff % i == 0) {

                // Increment count for i
                if (diff == i * i)
                    count++;

                // Increment count for i
                // and diff / i both
                else
                    count += 2;
            }
        }

        // Return Answer
        return count;
    }

    // Driver code
    public static int Main()
    {
        int A = 2, B = 4;
        Console.Write(countInt(A, B));
        return 0;
    }
}

// This code is contributed by Taranpreet

Javascript

<script>
      // JavaScript code for the above approach

      // Function to find the count of the
      // values of K such that A%K = B%K
      function countInt(A, B) {
          // If the count is Infinite
          if (A == B)
              return -1;

          let diff = Math.abs(A - B);

          // Stores the required count
          let count = 0;

          // Loop to calculate all the
          // divisors of diff
          for (let i = 1; i * i <= diff; i++) {
              if (diff % i == 0) {

                  // Increment count for i
                  if (diff == i * i)
                      count++;

                  // Increment count for i
                  // and diff / i both
                  else
                      count += 2;
              }
          }

          // Return Answer
          return count;
      }

      // Driver code

      let A = 2, B = 4;
      document.write(countInt(A, B));

       // This code is contributed by Potta Lokesh
  </script>

Output

2

Time Complexity: O(√(A – B))
Auxiliary Space: O(1)