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

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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.

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