Minimum Difference between multiples of three integers

Given three integers X, Y, and Z. the task is to find the minimum distance between any two multiples of the given integers.

Note: Common multiples are not considered. 

Examples:

Input: X = 5, Y = 2, Z = 3
Output: 1
Explanation: The multiples if arranged in sorted order are 0, 2, 3, 4, 5, 6, 8. . . 
Out of which the minimum possible difference is 1 between 2 and 3.

Input: X = 3, Y = 6, Z = 12
Output: 3

Approach: To solve the problem follow the below idea:

• Difference between the multiples of two numbers A and B is actually the multiples of GCD(A, B).
• To calculate the minimum possible difference between the multiple of any two numbers, calculate the GCD of those two numbers.

Follow the given steps to solve the problem:

• Calculate the greatest common divisor(GCD) between every pair of numbers.
• Return the minimum value by comparing the values calculated in the above step.

Below is the implementation for the above approach:

C++

// C++ program for above approach

#include <bits/stdc++.h>
using namespace std;

// Function to calculate the minimum
// possible difference between any two numbers
int minimumdifference(int H1, int H2, int H3)
{
    int ans = 0;
    int d1, d2, d3;

    // Calculating GCD between different
    // pairs of numbers
    d3 = __gcd(H1, H3);
    d1 = __gcd(H1, H2);
    d2 = __gcd(H2, H3);

    // Taking minimum of all GCD's
    ans = min(d1, d2);
    ans = min(d3, ans);
    return ans;
}

// Driver code
int main()
{
    int X = 5, Y = 2, Z = 3;

    // Function call
    cout << minimumdifference(X, Y, Z);
    return 0;
}

Java

// JAVA program for the above approach
import java.io.*;

class GFG {

  // Function to calculate gcd of two numbers
  public static int gcd(int a, int b)
  {
    return b==0 ? a : gcd(b, a%b);
  }

  // Function to calculate the minimum
  // possible difference between any two numbers
  static int minimumdifference(int H1,int H2,int H3)
  {
    int ans=0;
    int d1, d2, d3;

    // Calculating GCD between different
    // pairs of numbers
    d3 = gcd(H1, H3);
    d1 = gcd(H1, H2);
    d2 = gcd(H2, H3);

    // Taking minimum of all GCD's
    ans = Math.min(d1, d2);
    ans = Math.min(d3, ans);
    return ans;
  }

  // Driver code
  public static void main (String[] args)
  {
    int X = 5, Y = 2, Z = 3;

    // function call
    System.out.println(minimumdifference(X, Y, Z));
  }
}

// This code is written by Ujjwal Kumar Bhardwaj

Python3

# python 3 code to implement the above approach

def __gcd(x, y):

    while(y):
       x, y = y, x % y
    return abs(x)

# Function to calculate the minimum
# possible difference between any two numbers
def minimumdifference(H1, H2, H3) :

    ans = 0
    d1,d2,d3 = 0,0,0

    # Calculating GCD between different
    # pairs of numbers
    d3 = __gcd(H1, H3)
    d1 = __gcd(H1, H2)
    d2 = __gcd(H2, H3)

    # Taking minimum of all GCD's
    ans = min(d1, d2)
    ans = min(d3, ans)
    return ans

# Driver Code
if __name__ == "__main__" :

    X,Y,Z = 5,2,3

    # Function call
    print(minimumdifference(X, Y, Z))

#this code is contributed by aditya942003patil

C#

// C# program for above approach:
using System;
class GFG {

  // Function to calculate gcd of two numbers
  public static int gcd(int a, int b)
  {
    return b==0 ? a : gcd(b, a%b);
  }

  // Function to calculate the minimum
  // possible difference between any two numbers
  static int minimumdifference(int H1,int H2,int H3)
  {
    int ans=0;
    int d1, d2, d3;

    // Calculating GCD between different
    // pairs of numbers
    d3 = gcd(H1, H3);
    d1 = gcd(H1, H2);
    d2 = gcd(H2, H3);

    // Taking minimum of all GCD's
    ans = Math.Min(d1, d2);
    ans = Math.Min(d3, ans);
    return ans;
  }

// Driver Code
public static void Main()
{
    int X = 5, Y = 2, Z = 3;

    // function call
    Console.Write(minimumdifference(X, Y, Z));

}
}

// This code is contributed by code_hunt.

Javascript

<script>
    // JavaScript program for above approach

    // Function for __gcd
    const __gcd = (a, b) => {
        if (!b) return a;
        return __gcd(b, a % b);
    }

    // Function to calculate the minimum
    // possible difference between any two numbers
    const minimumdifference = (H1, H2, H3) => {
        let ans = 0;
        let d1, d2, d3;

        // Calculating GCD between different
        // pairs of numbers
        d3 = __gcd(H1, H3);
        d1 = __gcd(H1, H2);
        d2 = __gcd(H2, H3);

        // Taking minimum of all GCD's
        ans = Math.min(d1, d2);
        ans = Math.min(d3, ans);
        return ans;
    }

    // Driver code
    let X = 5, Y = 2, Z = 3;

    // Function call
    document.write(minimumdifference(X, Y, Z));

// This code is contributed by rakeshsahni

</script>

Output

1

Time Complexity: O(max(logX, logY, logZ))
Auxiliary Space: O(1)