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Find a distinct pair (x, y) in given range such that x divides y
  • Difficulty Level : Easy
  • Last Updated : 02 Jun, 2021

Given a range of positive integers from l to r. Find such a pair of integers (x, y) that l <= x, y <= r, x != y and x divide y. 
If there are multiple pairs, you need to find any one of them.

Examples: 

Input : 1 10 
Output : 1 2

Input : 2 4
Output : 2 4

The brute force solution is to traverse through the given range of (l, r) and find the first occurrence where x divides y and x!=y.This solution is feasible if the difference between l and r is small. 
Time Complexity of this solution is O((r-l)*(r-l)).
Following are codes based on brute force solutions. 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
void findpair(int l, int r)
{
    int c = 0;
    for (int i = l; i <= r; i++) {
        for (int j = i + 1; j <= r; j++) {
            if (j % i == 0 && j != i) {
                cout << i << ", " << j;
                c = 1;
                break;
            }
        }
        if (c == 1)
            break;
    }
}
 
int main()
{
    int l = 1, r = 10;
    findpair(l, r);
}

Java




// Java implementation of the approach
class GFG
{
 
static void findpair(int l, int r)
{
    int c = 0;
    for (int i = l; i <= r; i++)
    {
        for (int j = i + 1; j <= r; j++)
        {
            if (j % i == 0 && j != i)
            {
                System.out.println( i +", " + j);
                c = 1;
                break;
            }
        }
        if (c == 1)
            break;
    }
}
 
// Driver code
public static void main(String args[])
{
    int l = 1, r = 10;
    findpair(l, r);
}
}
 
// This code is contributed by Arnab Kundu

Python 3




# Python 3 implementation of the approach
def findpair(l, r):
    c = 0
    for i in range(l, r + 1):
        for j in range(i + 1, r + 1):
            if (j % i == 0 and j != i):
                print( i, ", ", j)
                c = 1
                break
        if (c == 1):
            break
     
# Driver Code
if __name__ == "__main__":
 
    l = 1
    r = 10
    findpair(l, r)
 
# This code is contributed by ita_c

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
static void findpair(int l, int r)
{
    int c = 0;
    for (int i = l; i <= r; i++)
    {
        for (int j = i + 1; j <= r; j++)
        {
            if (j % i == 0 && j != i)
            {
                Console.Write( i + ", " + j);
                c = 1;
                break;
            }
        }
        if (c == 1)
            break;
    }
}
 
// Driver code
static void Main()
{
    int l = 1, r = 10;
    findpair(l, r);
}
}
 
// This code is contributed by mits

PHP




<?php
// PHP implementation of the approach
function findpair($l, $r)
{
    $c = 0;
    for ($i = $l; $i <= $r; $i++)
    {
        for ($j = $i + 1; $j <= $r; $j++)
        {
            if ($j % $i == 0 && $j != $i)
            {
                echo($i . ", " . $j);
                $c = 1;
                break;
            }
        }
        if ($c == 1)
            break;
    }
}
 
// Driver code
$l = 1; $r = 10;
findpair($l, $r);
 
// This code is contributed
// by Code_Mech.
?>

Javascript




<script>
 
// JavaScript implementation of the approach
 
function findpair(l,r)
{
    let c = 0;
    for (let i = l; i <= r; i++)
    {
        for (let j = i + 1; j <= r; j++)
        {
            if (j % i == 0 && j != i)
            {
                document.write( i +", " + j+"<br>");
                c = 1;
                break;
            }
        }
        if (c == 1)
            break;
    }
}
 
// Driver code
let l = 1, r = 10;
findpair(l, r);
 
 
// This code is contributed by avanitrachhadiya2155
 
</script>
Output: 
1, 2

 

Efficient Solution: 
The problem can be solved in O(1) time complexity if you find the value of l and 2l.
Explanation: 
1) As we know, the smallest value of y/x you can have is 2 and if any greater value is in the range, then 2 is also in the given range.
2) The difference between x and 2x also increases when you increase the value of x. So l and 2l will be the minimum pair to fall in the given range.
Below is the implementation of the above approach:  



C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the possible pair
void findpair(int l, int r)
{
    // ans1, ans2 store value of x
    // and y respectively
    int ans1 = l;
    int ans2 = 2 * l;
 
    cout << ans1 << ", " << ans2 << endl;
}
 
// Driver Code
int main()
{
    int l = 1, r = 10;
    findpair(l, r);
}

Java




// Java implementation of the approach
class GFG
{
 
// Function to return the possible pair
static void findpair(int l, int r)
{
    // ans1, ans2 store value of x
    // and y respectively
    int ans1 = l;
    int ans2 = 2 * l;
 
    System.out.println( ans1 + ", " + ans2 );
}
 
// Driver Code
public static void main(String args[])
{
    int l = 1, r = 10;
    findpair(l, r);
}
}
 
// This code is contruibuted by Arnab Kundu

Python3




# Python3 implementation of the approach
 
# Function to return the possible pair
def findpair(l, r):
     
    # ans1, ans2 store value of x
    # and y respectively
    ans1 = l
    ans2 = 2 * l
 
    print(ans1, ", ", ans2)
 
# Driver Code
if __name__ == '__main__':
    l, r = 1, 10
    findpair(l, r)
 
# This code is contributed
# by 29AjayKumar

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the possible pair
    static void findpair(int l, int r)
    {
        // ans1, ans2 store value of x
        // and y respectively
        int ans1 = l;
        int ans2 = 2 * l;
     
        Console.WriteLine( ans1 + ", " + ans2 );
    }
     
    // Driver Code
    public static void Main()
    {
        int l = 1, r = 10;
        findpair(l, r);
    }
}
 
// This code is contruibuted by Ryuga

PHP




<?php
// PHP implementation of the approach
 
// Function to return the possible pair
function findpair($l, $r)
{
    // ans1, ans2 store value of x
    // and y respectively
    $ans1 = $l;
    $ans2 = 2 * $l;
 
    echo ($ans1 . ", " . $ans2);
}
 
// Driver Code
$l = 1; $r = 10;
findpair($l, $r);
 
// This code contributed by Rajput-Ji
?>

Javascript




<script>
// Javascript implementation of the approach
 
// Function to return the possible pair
function findpair(l,r)
{
    // ans1, ans2 store value of x
    // and y respectively
    let ans1 = l;
    let ans2 = 2 * l;
  
    document.write( ans1 + ", " + ans2 );
}
 
// Driver Code
let l = 1, r = 10;
findpair(l, r);
 
 
// This code is contributed by rag2127
</script>
Output: 
1, 2

 

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