LINQ to find the closest number that is greater / less than an input

Question

Suppose I have this number list:

List<int> = new List<int>(){3,5,8,11,12,13,14,21}

Suppose that I want to get the closest number that is less than 11, it would be 8 Suppose that I want to get the closest number that is greater than 13 that would be 14.

The numbers in list can't be duplicated and are always ordered. How can I write Linq for this?

Answer 1

with Linq assuming that the list is ordered I would do it like this:

var l = new List<int>() { 3, 5, 8, 11, 12, 13, 14, 21 };
var lessThan11 = l.TakeWhile(p => p < 11).Last();
var greaterThan13 = l.SkipWhile(p => p <= 13).First();

EDIT:

As I have received negative feedback about this answer and for the sake of people that may see this answer and while it's accepted don't go further, I explored the other comments regarding BinarySearch and decided to add the second option in here (with some minor change).

This is the not sufficient way presented somewhere else:

var l = new List<int>() { 3, 5, 8, 11, 12, 13, 14, 21 };
var indexLessThan11 = ~l.BinarySearch(10) -1;
var value = l[indexLessThan11];

Now the code above doesn't cope with the fact that the value 10 might actually be in the list (in which case one shouldn't invert the index)! so the good way is to do it:

var l = new List<int>() { 3, 5, 8, 11, 12, 13, 14, 21 };
var indexLessThan11 = l.BinarySearch(10);
if (indexLessThan11 < 0) // the value 10 wasn't found
{    
    indexLessThan11 = ~indexLessThan11;
    indexLessThan11 -= 1;
}
var value = l[indexLessThan11];

I simply want to note that:

l.BinarySearch(11) == 3
//and
l.BinarySearch(10) == -4;

Answer 2

Use Array.BinarySearch - no need for LINQ or visiting on average half the elements to find your target.

There are also a variety of SortedXXX classes that may be suitable for what you're doing [that will have such efficient O(log N) searches built-in]