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 pointsvoid 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 codeint main(){ int l1 = 5, r1 = 10, l2 = 1, r2 = 7; findPoints(l1, r1, l2, r2);} |
Java
// Java implementation of the approachclass GFG{ // Function to find the required pointsstatic 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 codepublic 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 pointsdef 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 codeif __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 approachusing 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 pointsfunction 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 pointsfunction 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 codevar 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.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
