Choose points from two ranges such that no point lies in both the ranges

Given two segments [L1, R1] and [L2, R2], the task is to choose two elements x and y from both the ranges (one from range one and other from range two) such that no element belongs to both the ranges i.e. x belongs to first range and y belongs to second range. If no such element exists then print -1 instead.

Examples:

Input: L1 = 1, R1 = 6, L2 = 3, R2 = 11
Output: 1 11
1 lies only in range [1, 6] and 11 lies only in [3, 11]

Input: L1 = 5, R1 = 10, L2 = 1, R2 = 7
Output: 1 10



Approach:

  • If L1 != L2 and R1 != R2 then the points will be min(L1, L2) and max(R1, R2).
  • Else only one point can be chosen from one of the ranges as one of the range is completely inside the other so we print -1 for that point.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the required points
void findPoints(int l1, int r1, int l2, int r2)
{
  
    int x = (l1 != l2) ? min(l1, l2) : -1;
    int y = (r1 != r2) ? max(r1, r2) : -1;
    cout << x << " " << y;
}
  
// Driver code
int main()
{
    int l1 = 5, r1 = 10, l2 = 1, r2 = 7;
    findPoints(l1, r1, l2, r2);
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
class GFG
{
      
// Function to find the required points
static void findPoints(int l1, int r1, 
                       int l2, int r2)
{
  
    int x = (l1 != l2) ? Math.min(l1, l2) : -1;
    int y = (r1 != r2) ? Math.max(r1, r2) : -1;
    System.out.println(x + " " + y);
}
  
// Driver code
public static void main(String[] args)
{
    int l1 = 5, r1 = 10, l2 = 1, r2 = 7;
    findPoints(l1, r1, l2, r2);
}
}
  
// This code is contributed by Code_Mech

chevron_right


Python 3

# Python3 implementation of the approach

# Function to find the required points
def findPoints(l1, r1, l2, r2):

x = min(l1, l2) if(l1 != l2) else -1
y = max(r1, r2) if(r1 != r2) else -1
print(x, y)

# Driver code
if __name__ == “__main__”:

l1 = 5
r1 = 10
l2 = 1
r2 = 7
findPoints(l1, r1, l2, r2)

# This code is contributed by ita_c

C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG
{
    // Function to find the required points
    static void findPoints(int l1, int r1,
                            int l2, int r2)
    {
        int x = (l1 != l2) ? Math.Min(l1, l2) : -1;
        int y = (r1 != r2) ? Math.Max(r1, r2) : -1;
        Console.WriteLine(x + " " + y);
    }
      
    // Driver code
    public static void Main()
    {
        int l1 = 5, r1 = 10, l2 = 1, r2 = 7;
        findPoints(l1, r1, l2, r2);
    }
}
  
// This code is contributed by Ryuga

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the approach
  
// Function to find the required points
function findPoints($l1, $r1, $l2, $r2)
{
  
    $x = ($l1 != $l2) ? min($l1, $l2) : -1;
    $y = ($r1 != $r2) ? max($r1, $r2) : -1;
    echo $x , " " , $y;
}
  
// Driver code
$l1 = 5;
$r1 = 10;
$l2 = 1;
$r2 = 7;
findPoints($l1, $r1, $l2, $r2);
  
// This code is contributed by ajit
?>

chevron_right


Output:

1 10


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : jit_t, Code_Mech, Ryuga, Ita_c