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All possible co-prime distinct element pairs within a range [L, R]

  • Difficulty Level : Basic
  • Last Updated : 21 May, 2021

Given a range [L, R], the task is to find all possible co-prime pairs from the range such that an element doesn’t appear in more than a single pair.
Examples: 
 

Input : L=1 ; R=6
Output : 3
The answer is 3 [(1, 2) (3, 4) (5, 6)], 
all these pairs have GCD 1.

Input : L=2 ; R=4
Output : 1
The answer is 1 [(2, 3) or (3, 4)] 
as '3' can only be chosen for a single pair.

 

Approach: The key observation of the problem is that the numbers with the difference of ‘1’ are always relatively prime to each other i.e. co-primes. 
GCD of this pair is always ‘1’. So, the answer will be (R-L+1)/2 [ (total count of numbers in range) / 2 ] 
 

  • If R-L+1 is odd then there will be one element left which can not form a pair.
  • If R-L+1 is even then all elements can form pairs.

Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count possible pairs
void CountPair(int L, int R)
{
 
    // total count of numbers in range
    int x = (R - L + 1);
 
    // Note that if 'x' is odd then
    // there will be '1' element left
    // which can't form a pair
 
    // printing count of pairs
    cout << x / 2 << "\n";
}
 
// Driver code
int main()
{
 
    int L, R;
 
    L = 1, R = 8;
    CountPair(L, R);
 
    return 0;
}

Java




   
// Java implementation of the approach
import java.util.*;
class solution
{
 
// Function to count possible pairs
static void CountPair(int L, int R)
{
 
    // total count of numbers in range
    int x = (R - L + 1);
 
    // Note that if 'x' is odd then
    // there will be '1' element left
    // which can't form a pair
 
    // printing count of pairs
    System.out.println(x / 2 + "\n");
}
 
// Driver code
public static void main(String args[])
{
 
    int L, R;
 
    L = 1; R = 8;
    CountPair(L, R);
 
}
}
//contributed by Arnab Kundu

Python3




# Python3 implementation of
# the approach
 
# Function to count possible
# pairs
def CountPair(L,R):
 
    # total count of numbers
    # in range
    x=(R-L+1)
 
    # Note that if 'x' is odd then
    # there will be '1' element left
    # which can't form a pair
    # printing count of pairs
    print(x//2)
 
# Driver code
if __name__=='__main__':
    L,R=1,8
    CountPair(L,R)
     
# This code is contributed by
# Indrajit Sinha.

C#




// C# implementation of the approach
using System;
class GFG
{
 
// Function to count possible pairs
static void CountPair(int L, int R)
{
 
    // total count of numbers in range
    int x = (R - L + 1);
 
    // Note that if 'x' is odd then
    // there will be '1' element left
    // which can't form a pair
 
    // printing count of pairs
    Console.WriteLine(x / 2 + "\n");
}
 
// Driver code
public static void Main()
{
    int L, R;
 
    L = 1; R = 8;
    CountPair(L, R);
}
}
 
// This code is contributed
// by inder_verma..

PHP




<?php
// PHP implementation of the above approach
 
// Function to count possible pairs
function CountPair($L, $R)
{
 
    // total count of numbers in range
    $x = ($R - $L + 1);
 
    // Note that if 'x' is odd then
    // there will be '1' element left
    // which can't form a pair
 
    // printing count of pairs
    echo $x / 2, "\n";
}
 
// Driver code
$L = 1;
$R = 8;
CountPair($L, $R);
 
// This code is contributed by ANKITRAI1
?>

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to count possible pairs
function CountPair(L, R)
{
   
    // total count of numbers in range
    let x = (R - L + 1);
   
    // Note that if 'x' is odd then
    // there will be '1' element left
    // which can't form a pair
   
    // printing count of pairs
   document.write(x / 2 + "<br/>");
}
 
 
// driver code
 
    let L, R;
   
    L = 1; R = 8;
    CountPair(L, R);
   
</script>
Output: 



4

 

Complexity: O(1)
 

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