Maximum prime moves to convert X to Y

Given two integers X and Y, the task is to convert X to Y using the following operations: 

1. Add any prime number to X.
2. Subtract any prime number from Y.

Print the maximum number of such operations required or -1 if it is not possible to convert X to Y.
Examples:

Input: X = 2, Y = 4 
Output: 1 
2 -> 4
Input: X = 5, Y = 6 
Output: -1 
It is impossible to convert 5 to 6 
with the given operations. 

Approach: As the task is to maximize the operations, so the minimum possible value must be added to X in every operation. Since the value has to be prime, so the minimum two primes i.e. 2 and 3 can be used as they both are prime and can cover both even and odd parity. Now, there are three cases: 

• If X > Y then the answer will be -1 as X cannot be made equal to Y with the given operation.
• If X = Y then the answer will be 0.
• If X < Y then calculate P = Y – X and, 
  • If P = 1 then the answer will be -1 as 1 is not prime and it cannot be added or subtracted.
  • If P is even then 2 can be repeatedly added to X and the answer will be P / 2
  • If P is even then add 3 to X and then 2 can again be repeatedly added to the new X to make it equal to Y, the result, in this case, will be 1 + ((P – 3) / 2).

Below is the implementation of the above approach: 

C++

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

// Function to return the maximum operations
// required to convert X to Y
int maxOperations(int X, int Y)
{

    // X cannot be converted to Y
    if (X > Y)
        return -1;

    int diff = Y - X;

    // If the difference is 1
    if (diff == 1)
        return -1;

    // If the difference is even
    if (diff % 2 == 0)
        return (diff / 2);

    // Add 3 to X and the new
    // difference will be even
    return (1 + ((diff - 3) / 2));
}

// Driver code
int main()
{
    int X = 5, Y = 16;

    cout << maxOperations(X, Y);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the maximum operations
// required to convert X to Y
static int maxOperations(int X, int Y)
{

    // X cannot be converted to Y
    if (X > Y)
        return -1;

    int diff = Y - X;

    // If the difference is 1
    if (diff == 1)
        return -1;

    // If the difference is even
    if (diff % 2 == 0)
        return (diff / 2);

    // Add 3 to X and the new
    // difference will be even
    return (1 + ((diff - 3) / 2));
}

// Driver code
public static void main(String []args)
{
    int X = 5, Y = 16;

    System.out.println(maxOperations(X, Y));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach

# Function to return the maximum operations
# required to convert X to Y
def maxOperations(X, Y) :

    # X cannot be converted to Y
    if (X > Y) :
        return -1;

    diff = Y - X;

    # If the difference is 1
    if (diff == 1) :
        return -1;

    # If the difference is even
    if (diff % 2 == 0) :
        return (diff // 2);

    # Add 3 to X and the new
    # difference will be even
    return (1 + ((diff - 3) // 2));

# Driver code
if __name__ == "__main__" :

    X = 5; Y = 16;

    print(maxOperations(X, Y));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the maximum operations
// required to convert X to Y
static int maxOperations(int X, int Y)
{

    // X cannot be converted to Y
    if (X > Y)
        return -1;

    int diff = Y - X;

    // If the difference is 1
    if (diff == 1)
        return -1;

    // If the difference is even
    if (diff % 2 == 0)
        return (diff / 2);

    // Add 3 to X and the new
    // difference will be even
    return (1 + ((diff - 3) / 2));
}

// Driver code
public static void Main(String []args)
{
    int X = 5, Y = 16;

    Console.WriteLine(maxOperations(X, Y));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// JavaScript implementation of the approach

// Function to return the maximum operations
// required to convert X to Y
function maxOperations(X, Y)
{

    // X cannot be converted to Y
    if (X > Y)
        return -1;

    let diff = Y - X;

    // If the difference is 1
    if (diff == 1)
        return -1;

    // If the difference is even
    if (diff % 2 == 0)
        return (diff / 2);

    // Add 3 to X and the new
    // difference will be even
    return (1 + ((diff - 3) / 2));
}

// Driver code

    let X = 5, Y = 16;

    document.write(maxOperations(X, Y));

// This code is contributed by _saurabh_jaiswal

</script>

Output:

5

Time Complexity: O(1)

Auxiliary Space: O(1)