Skip to content
Related Articles

Related Articles

Check if there exists a prime number which gives Y after being repeatedly subtracted from X

Improve Article
Save Article
  • Difficulty Level : Hard
  • Last Updated : 03 Mar, 2022
Improve Article
Save Article

Given two integers X and Y where X > Y, the task is to check if there exists a prime number P such that if P is repeatedly subtracted from X then it gives Y.

Examples: 

Input: X = 100, Y = 98 
Output: Yes 
(100 – (2 * 1) = 98)
Input: X = 45, Y = 31 
Output: Yes 
(45 – (7 * 2)) = 31 
 

Naive approach: Run a loop for every integer starting from 2 to x. If a current number is a prime number, and it meets the criteria given in the question, then it is the required number.
Efficient approach: Notice that for a valid prime p, x – k * p = y or x – y = k * p. Suppose, p = 2 then (x – y) = 2, 4, 6, … (all even numbers). This means if (x – y) is even then the answer is always true. If (x – y) is an odd number other than 1, it will always have a prime factor. Either it itself is prime or it is a product of a smaller prime and some other integers. So the answer is True for all odd numbers other than 1
What if (x – y) = 1, it is neither a prime nor composite. So this is the only case where the answer is false.
 

Below is the implementation of the approach:  

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if any
// prime number satisfies
// the given conditions
bool isPossible(int x, int y)
{
 
    // No such prime exists
    if ((x - y) == 1)
        return false;
 
    return true;
}
 
// Driver code
int main()
{
    int x = 100, y = 98;
 
    if (isPossible(x, y))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function that returns true if any
// prime number satisfies
// the given conditions
static boolean isPossible(int x, int y)
{
 
    // No such prime exists
    if ((x - y) == 1)
        return false;
 
    return true;
}
 
// Driver code
public static void main(String[] args)
{
    int x = 100, y = 98;
 
    if (isPossible(x, y))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 implementation of the approach
 
# Function that returns true if any
# prime number satisfies
# the given conditions
def isPossible(x, y):
 
    # No such prime exists
    if ((x - y) == 1):
        return False
 
    return True
 
# Driver code
x = 100
y = 98
 
if (isPossible(x, y)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Mohit Kumar

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function that returns true if any
// prime number satisfies
// the given conditions
static bool isPossible(int x, int y)
{
 
    // No such prime exists
    if ((x - y) == 1)
        return false;
 
    return true;
}
 
// Driver code
public static void Main(String[] args)
{
    int x = 100, y = 98;
 
    if (isPossible(x, y))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
// javascript implementation of the approach
 
    // Function that returns true if any
    // prime number satisfies
    // the given conditions
    function isPossible(x , y) {
 
        // No such prime exists
        if ((x - y) == 1)
            return false;
 
        return true;
    }
 
    // Driver code
     
        var x = 100, y = 98;
 
        if (isPossible(x, y))
            document.write("Yes");
        else
            document.write("No");
 
// This code contributed by umadevi9616
</script>

Output: 

Yes

 

Time Complexity: O(1)

Auxiliary Space: O(1)


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!