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++

// 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);
}

Java

// 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

Python3

# 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#

// 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

PHP

<?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
?>

Javascript

<script>

// Javascript implementation of the approach

// Function to find the required points
function findPoints(l1 , r1 , l2 , r2)
{
    var x = (l1 != l2) ? Math.min(l1, l2) : -1;
    var y = (r1 != r2) ? Math.max(r1, r2) : -1;
    document.write(x + " " + y);
}

// Driver code
var l1 = 5, r1 = 10, l2 = 1, r2 = 7;

findPoints(l1, r1, l2, r2);

// This code is contributed by Rajput-Ji

</script>

Output:

1 10

Time Complexity : O(1), since there is only a basic arithmetic operation that takes constant time.
Auxiliary Space : O(1), since no extra space has been taken.