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

• Last Updated : 07 May, 2021

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:

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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 ``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

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## Javascript

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Output:
`1 10`

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