how to find/filter the array element with the smallest positive difference

Question

Using Swift, I am trying to workout a complex array filter or sort and I am stuck. My array of integers can contain between 1 and 7 elements. Given a certain integer value (say X) which is not in the array, I would like to find the array element that gives the smallest difference between itself and X.

Answer 1

In Swift 2 you can do it as a "one-liner" with functional-style programming:

let numbers = [ 1, 3, 7, 11]
let x = 6

let closest = numbers.enumerate().minElement( { abs($0.1 - x) < abs($1.1 - x)} )!

print(closest.element) // 7 = the closest element
print(closest.index)   // 2 = index of the closest element

enumerate() iterates over all array elements together with the corresponding index, and minElement() returns the "smallest" (index, element) pair with respect to the closure. The closure compares the absolute values of the difference of two elements to x.

(It is assumed here that the array is not empty, so that minElement() does not return nil.)

Note that this is probably not the fastest solution for large arrays, because the absolute differences are computed twice for (almost) all array elements. But for small arrays this should not matter.

Swift 3:

let numbers = [ 1, 3, 7, 11]
let x = 6
let closest = numbers.enumerated().min( by: { abs($0.1 - x) < abs($1.1 - x) } )!
print(closest.element) // 7
print(closest.offset) // 2

The Swift 1.2 version can be found in the edit history.

Answer 2

Swift 4.2

Not exactly what the OP is asking, but you can use the first(where:) and firstIndex(where:) array methods on a sorted array to get the first value greater than x:

let numbers = [1, 3, 7, 11]

let x = 6
let index = numbers.firstIndex(where: { $0 >= x })!
let result = numbers.first(where: { $0 >= x })!

print(index, result) // 2 7