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 pairsvoid 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 codeint main(){ int L, R; L = 1, R = 8; CountPair(L, R); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class solution{// Function to count possible pairsstatic 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 codepublic 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 # pairsdef 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 codeif __name__=='__main__': L,R=1,8 CountPair(L,R) # This code is contributed by # Indrajit Sinha. |
C#
// C# implementation of the approachusing System;class GFG{// Function to count possible pairsstatic 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 codepublic 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 pairsfunction 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 pairsfunction 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
Time Complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.
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