# 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); ` `} ` |

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## 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 ` |

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## Python 3

# 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 ` |

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## 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 ` `?> ` |

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**Output:**

1 10

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