Find numbers that divide X and Y to produce the same remainder

Given two integers X and Y, the task is to find and print the numbers that divide X and Y to produce the same remainder.
Examples:

Input: X = 1, Y = 5 
Output: 1, 2, 4 
Explanation:
Let the number be M. It can be any value in the range [1, 5]: 
If M = 1, 1 % 1 = 0 and 5 % 1 = 0 
If M = 2, 1 % 2 = 1 and 5 % 2 = 1 
If M = 3, 1 % 3 = 1 and 5 % 3 = 2 
If M = 4, 1 % 4 = 1 and 5 % 4 = 1 
If M = 5, 1 % 5 = 1 and 5 % 5 = 0 
Therefore, the possible M values are 1, 2, 4
Input: X = 8, Y = 10 
Output: 1, 2 

Naive Approach: The naive approach for this problem is to check the modulo value for all the possible values of M in the range [1, max(X, Y)] and print the value of M if the condition satisfies. 
Below is the implementation of the above approach: 

C++

// C++ program to find numbers
// that divide X and Y
// to produce the same remainder

#include <iostream>
using namespace std;

// Function to find
// the required number as M
void printModulus(int X, int Y)
{
    // Finding the maximum
    // value among X and Y
    int n = max(X, Y);

    // Loop to iterate through
    // maximum value among X and Y.
    for (int i = 1; i <= n; i++) {

        // If the condition satisfies, then
        // print the value of M
        if (X % i == Y % i)
            cout << i << " ";
    }
}

// Driver code
int main()
{

    int X, Y;
    X = 10;
    Y = 20;
    printModulus(X, Y);
    return 0;
}

Java

// Java program to find numbers
// that divide X and Y
// to produce the same remainder
import java.util.*;
import java.io.*;
class GFG{

// Function to find
// the required number as M
static void printModulus(int X, int Y)
{
    // Finding the maximum
    // value among X and Y
    int n = Math.max(X, Y);

    // Loop to iterate through
    // maximum value among X and Y.
    for (int i = 1; i <= n; i++) {

        // If the condition satisfies, then
        // print the value of M
        if (X % i == Y % i)
            System.out.print(i + " ");
    }
}

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

    int X, Y;
    X = 10;
    Y = 20;
    printModulus(X, Y);
}
}

// This code is contributed by Princi Singh

Python3

# Python program to find numbers
# that divide X and Y
# to produce the same remainder

# Function to find
# the required number as M
def printModulus( X, Y):

    # Finding the maximum
    # value among X and Y
    n = max(X, Y)

    # Loop to iterate through
    # maximum value among X and Y.
    for i in range(1, n + 1):

        # If the condition satisfies, then
        # print the value of M
        if (X % i == Y % i):
            print(i,end=" ")

# Driver code
X = 10
Y = 20
printModulus(X, Y)

# This code is contributed by Atul_kumar_Shrivastava

C#

// C# program to find numbers
// that divide X and Y
// to produce the same remainder
using System;

class GFG{

// Function to find
// the required number as M
static void printModulus(int X, int Y)
{
    // Finding the maximum
    // value among X and Y
    int n = Math.Max(X, Y);

    // Loop to iterate through
    // maximum value among X and Y.
    for (int i = 1; i <= n; i++) {

        // If the condition satisfies, then
        // print the value of M
        if (X % i == Y % i)
            Console.Write(i + " ");
    }
}

// Driver code
public static void Main()
{
    int X, Y;
    X = 10;
    Y = 20;
    printModulus(X, Y);
}
}

// This code is contributed by AbhiThakur

Javascript

<script>

// Javascript program to find numbers
// that divide X and Y
// to produce the same remainder

// Function to find
// the required number as M
function printModulus(X, Y)
{
    // Finding the maximum
    // value among X and Y
    var n = Math.max(X, Y);

    // Loop to iterate through
    // maximum value among X and Y.
    for (var i = 1; i <= n; i++) {

        // If the condition satisfies, then
        // print the value of M
        if (X % i == Y % i)
            document.write(i+" ");
    }
}

// Driver code
X = 10;
Y = 20;
printModulus(X, Y);

// This code is contributed by noob2000.
</script>

Output:

1 2 5 10

Time Complexity: O(max(X, Y))
Auxiliary Space: O(1)

Efficient Approach: Let’s assume that Y is greater than X by a difference of D. 

• Then Y can be expressed as 
  

Y = X + D
and
Y % M = (X + D) % M
      = (X % M) + (D % M)

• Now, the condition becomes whether X % M and X % M + D % M are equal or not.
• Here, since X % M is common on both the sides, the value of M is true if for some M, D % M = 0.
• Therefore, the required values of M will be the factors of D.

Below is the implementation of the above approach: 

CPP

// C++ program to find numbers
// that divide X and Y to
// produce the same remainder

#include <iostream>
using namespace std;

// Function to print all the possible values
// of M such that X % M = Y % M
void printModulus(int X, int Y)
{
    // Finding the absolute difference
    // of X and Y
    int d = abs(X - Y);

    // Iterating from 1
    int i = 1;

    // Loop to print all the factors of D
    while (i * i <= d) {

        // If i is a factor of d, then print i
        if (d % i == 0) {
            cout << i << " ";

            // If d / i is a factor of d,
            // then print d / i
            if (d / i != i)
                cout << d / i << " ";
        }
        i++;
    }
}

// Driver code
int main()
{

    int X = 10;
    int Y = 26;

    printModulus(X, Y);
    return 0;
}

Java

// Java program to find numbers
// that divide X and Y to
// produce the same remainder
import java.util.*;
import java.io.*;
class GFG{

// Function to print all the possible values
// of M such that X % M = Y % M
static void printModulus(int X, int Y)
{
    // Finding the absolute difference
    // of X and Y
    int d = Math.abs(X - Y);

    // Iterating from 1
    int i = 1;

    // Loop to print all the factors of D
    while (i * i <= d) {

        // If i is a factor of d, then print i
        if (d % i == 0) {
            System.out.print(i+ " ");

            // If d / i is a factor of d,
            // then print d / i
            if (d / i != i)
                System.out.print(d / i+ " ");
        }
        i++;
    }
}

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

    int X = 10;
    int Y = 26;

    printModulus(X, Y);
}
}

// This code is contributed by Princi Singh

Python3

# Python program to find numbers
# that divide X and Y to
# produce the same remainder

# Function to print all the possible values
# of M such that X % M = Y % M
def printModulus(X, Y):
    # Finding the absolute difference
    # of X and Y
    d = abs(X - Y);

    # Iterating from 1
    i = 1;

    # Loop to print all the factors of D
    while (i * i <= d):

        # If i is a factor of d, then pri
        if (d % i == 0):
            print(i, end="");

            # If d / i is a factor of d,
            # then prd / i
            if (d // i != i):
                print(d // i, end=" ");

        i+=1;



# Driver code
if __name__ == '__main__':

    X = 10;
    Y = 26;

    printModulus(X, Y);

# This code contributed by Princi Singh

C#

// C# program to find numbers
// that divide X and Y to
// produce the same remainder
using System;

public class GFG{

// Function to print all the possible values
// of M such that X % M = Y % M
static void printModulus(int X, int Y)
{
    // Finding the absolute difference
    // of X and Y
    int d = Math.Abs(X - Y);

    // Iterating from 1
    int i = 1;

    // Loop to print all the factors of D
    while (i * i <= d) {

        // If i is a factor of d, then print i
        if (d % i == 0) {
            Console.Write(i+ " ");

            // If d / i is a factor of d,
            // then print d / i
            if (d / i != i)
                Console.Write(d / i+ " ");
        }
        i++;
    }
}

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

    int X = 10;
    int Y = 26;

    printModulus(X, Y);
}
}

// This code contributed by Princi Singh

Javascript

  <script>
    // Javascript program to find numbers
    // that divide X and Y to
    // produce the same remainder

    // Function to print all the possible values
    // of M such that X % M = Y % M
    function printModulus(X, Y)
    {

      // Finding the absolute difference
      // of X and Y
      var d = Math.abs(X - Y);

      // Iterating from 1
      var i = 1;

      // Loop to print all the factors of D
      while (i * i <= d) {

        // If i is a factor of d, then print i
        if (d % i == 0) {
          document.write(i + " ");

          // If d / i is a factor of d,
          // then print d / i
          if (d / i != i)
            document.write(parseInt(d / i) + " ");
        }
        i++;
      }
    }

    // Driver code
    var X = 10;
    var Y = 26;
    printModulus(X, Y);

// This code is contributed by rrrtnx.
  </script>

Output:

1 16 2 8 4

Time Complexity Analysis O(sqrt(D)), where D is the difference between the values X and Y.
Auxiliary Space: O(1)