Find a distinct pair (x, y) in given range such that x divides y


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 divides 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 occurence 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 solution.

C++

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// 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);
}

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Java

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// 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

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C#

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// 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

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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 that 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 minimum pair to fall in given range.

Below is the implementation of the above approach:

C++

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// 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);
}

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Java

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// 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

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C#

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// 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

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Output:

1, 2


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