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:Â
Input: L1 = 1, R1 = 6, L2 = 3, R2 = 11Â
Output: 1 11Â
1 lies only in range [1, 6] and 11 lies only in [3, 11]Input: L1 = 5, R1 = 10, L2 = 1, R2 = 7Â
Output: 1 10Â
Â
Approach:Â Â
- 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:Â
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;Â
// Function to find the required pointsvoid findPoints(int l1, int r1, int l2, int r2){Â
    int x = (l1 != l2) ? min(l1, l2) : -1;    int y = (r1 != r2) ? max(r1, r2) : -1;    cout << x << " " << y;}Â
// Driver codeint main(){Â Â Â Â int l1 = 5, r1 = 10, l2 = 1, r2 = 7;Â Â Â Â findPoints(l1, r1, l2, r2);} |
Java
// Java implementation of the approachclass GFG{Â Â Â Â Â // Function to find the required pointsstatic void findPoints(int l1, int r1, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â int l2, int r2){Â
    int x = (l1 != l2) ? Math.min(l1, l2) : -1;    int y = (r1 != r2) ? Math.max(r1, r2) : -1;    System.out.println(x + " " + y);}Â
// Driver codepublic static void main(String[] args){Â Â Â Â int l1 = 5, r1 = 10, l2 = 1, r2 = 7;Â Â Â Â findPoints(l1, r1, l2, r2);}}Â
// This code is contributed by Code_Mech |
Python3
# Python3 implementation of the approachÂ
# Function to find the required pointsdef findPoints(l1, r1, l2, r2):Â
    x = min(l1, l2) if(l1 != l2) else -1    y = max(r1, r2) if(r1 != r2) else -1    print(x, y)Â
# Driver codeif __name__ == "__main__":Â Â Â Â Â Â Â Â Â l1 = 5Â Â Â Â r1 = 10Â Â Â Â l2 = 1Â Â Â Â r2 = 7Â Â Â Â findPoints(l1, r1, l2, r2)Â
# This code is contributed by ita_c |
C#
// C# implementation of the approachusing System;Â
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);    }}Â
// This code is contributed by Ryuga |
PHP
<?php// PHP implementation of the approachÂ
// Function to find the required pointsfunction findPoints($l1, $r1, $l2, $r2){Â
    $x = ($l1 != $l2) ? min($l1, $l2) : -1;    $y = ($r1 != $r2) ? max($r1, $r2) : -1;    echo $x , " " , $y;}Â
// Driver code$l1 = 5;$r1 = 10;$l2 = 1;$r2 = 7;findPoints($l1, $r1, $l2, $r2);Â
// This code is contributed by ajit?> |
Javascript
<script>Â
// Javascript implementation of the approach   Â
// Function to find the required pointsfunction findPoints(l1 , r1 , l2 , r2){Â Â Â Â var x = (l1 != l2) ? Math.min(l1, l2) : -1;Â Â Â Â var y = (r1 != r2) ? Math.max(r1, r2) : -1;Â Â Â Â document.write(x + " " + y);}Â
// Driver codevar l1 = 5, r1 = 10, l2 = 1, r2 = 7;Â
findPoints(l1, r1, l2, r2);Â
// This code is contributed by Rajput-Ji Â
</script> |
Output:Â
1 10
Â
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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