Find a Co-Prime pair with maximum difference in a given range

Given two integers L and R, where  L < R, the task is to find Co-prime numbers X and Y such that L <= X < Y <= R and the difference between X and Y is maximum i.e., maximize (Y – X) over all pairs between  L and R where GCD(X, Y) = 1.

Note: Co-prime numbers are the numbers whose GCD (Greatest Common Divisor) is 1 i.e., A and B are Co-prime if and only if GCD(A, B) = 1.

Examples:

Input: L = 3, R = 9
Output: 5
Explanation: {3, 8}, and {4, 9} are both Co-prime number in the range [3, 9] with a maximum difference of 5. So the answer is 5.

Input: L = 4, R =5
Output: 1
Explanation: Only one pair {4, 5} is possible in the given range, so the maximum difference is 1.

Naive Approach: Follow the steps to solve the problem:

• Initialize a variable maxDifference = 0 to store the maximum difference as the answer.
• Run a loop on i from L to R – 1 and inside that loop perform the below steps
  • Run a nested loop on j from i + 1 to R
  • If GCD(i, j) == 1 update maxDifference = max(maxDifference, j – i ).
• Return maxDifference as the final answer.

Time complexity: O((R – L)^2)*log(R)), (R- L)^2 for nested loops and log(R) for calculating GCD(L, R) over all pairs.
Auxiliary Space: O(1)

Efficient Approach: To solve the problem use the following idea:

As we know that GCD(N, N + 1) is always 1 and we need to find the maximum difference of the pair with GCD 1 so GCD(L, R – 1) and GCD(L + 1, R) will be one as GCD(L, L + 1) and GCD(R – 1, R) will be 1 so there cannot be any common factors between (L and R – 1), and (L + 1, R).

Follow the below steps to solve the problem:

• If the GCD(L, R) = 1
  • Return (R – L) as the answer.
  • else return (R – L – 1) as the answer.

Below is the implementation of the above approach: 

C++

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

// Function to calculate the
// gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// function to find the maximum difference
int maxDifference(int l, int r)
{

    // if the GCD of L and R is 1
    // R - L is maximum difference
    if (gcd(l, r) == 1) {
        return (r - l);
    }

    // GCD(L + 1, R) or GCD(L, R - 1)
    // will always be 1 as N and N + 1
    // are co-primes
    return (r - l - 1);
}

// Driver code
int main()
{
    // Range for which we want to
    // calculate maximum difference
    int L = 3, R = 9;

    // Function Call
    cout << maxDifference(L, R);

    return 0;
}

Java

// Java code for the above approach
import java.io.*;

class GFG {
    // Driver Code
    public static void main(String[] args) {
        // Range for which we want to calculate maximum difference
        // input
        int L = 3, R = 9;

        // Function Call
        int answer = maxDifference(L, R);

        // Printing the answer
        System.out.println(answer);
    }

    static // function to find the maximum difference
    int maxDifference(int l, int r) {

        // if the GCD of L and R is 1
        // R - L is maximum difference
        if (gcd(l, r) == 1) {
            return (r - l);
        }

        // GCD(L + 1, R) or GCD(L, R - 1)
        // will always be 1 as N and N + 1
        // are co-primes
        return (r - l - 1);
    }

    // Function to calculate the gcd of a and b
    static int gcd(int a, int b) {
          if (a == 0)
            return b;
        return gcd(b % a, a);
    }
}

Python3

# Python code for the above approach

# function to find the maximum difference
def maxDifference(l, r):

    # if the GCD of L and R is 1
    # R - L is maximum difference
    if (gcd(l, r) == 1):
        return (r - l)

    # GCD(L + 1, R) or GCD(L, R - 1)
    # will always be 1 as N and N + 1
    # are co-primes
    return (r - l - 1)


# Function to calculate the gcd of a and b
def gcd(a, b):
    if (a == 0):
        return b
    return gcd(b % a, a)


# Driver Code

# Range for which we want to calculate maximum difference
# input
L = 3
R = 9

# Function Call
answer = maxDifference(L, R)

# Printing the answer
print(answer)

# This code is contributed by saurabh_jaiswal.

C#

// C# program for above approach
using System;
using System.Linq;

class GFG
{

    // function to find the maximum difference
    static int maxDifference(int l, int r) {

        // if the GCD of L and R is 1
        // R - L is maximum difference
        if (gcd(l, r) == 1) {
            return (r - l);
        }

        // GCD(L + 1, R) or GCD(L, R - 1)
        // will always be 1 as N and N + 1
        // are co-primes
        return (r - l - 1);
    }

    // Function to calculate the gcd of a and b
    static int gcd(int a, int b) {
          if (a == 0)
            return b;
        return gcd(b % a, a);
    }

// Driver Code
public static void Main()
{
    // Range for which we want to calculate maximum difference
        // input
        int L = 3, R = 9;

        // Function Call
        int answer = maxDifference(L, R);

        // Printing the answer
        Console.Write(answer);
}
}

// This code is contributed by code_hunt.

Javascript

<script>
// JavaScript code for the above approach

    // function to find the maximum difference
    function maxDifference(l, r) {

        // if the GCD of L and R is 1
        // R - L is maximum difference
        if (gcd(l, r) == 1) {
            return (r - l);
        }

        // GCD(L + 1, R) or GCD(L, R - 1)
        // will always be 1 as N and N + 1
        // are co-primes
        return (r - l - 1);
    }

    // Function to calculate the gcd of a and b
    function gcd(a, b) {
          if (a == 0)
            return b;
        return gcd(b % a, a);
    }

// Driver Code
    
        // Range for which we want to calculate maximum difference
        // input
        let L = 3, R = 9;

        // Function Call
        let answer = maxDifference(L, R);

        // Printing the answer
        document.write(answer);
    
</script>

Output

5

Time complexity: O(log(L)), for calculating GCD(L, R)
Auxiliary Space: O(1)

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