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++ 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 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);
}
} |
# 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# 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. |
<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> |
5
Time complexity: O(log(L)), for calculating GCD(L, R)
Auxiliary Space: O(1)