Swift: Find closest value in array with binary search - arrays

Question

I I try to find the closest value in an array using binary search. Everything works fine as long as the value I am looking for is not smaller than the smallest value in the array.
Unfortunately, the debugger did not result in anything helpful. So I ask the community now. You can also try the code directly in the Xcode Playground. I tried to change an other searched value to a smaller value as in the array, but got the same error.
Error: error: Execution was interrupted, reason: EXC_BAD_INSTRUCTION (code=EXC_I386_INVOP, subcode=0x0).
func closestValue(_ arr: [Int],_ target: Int) -> Int {
var leftPointer = 0
var rightPointer = arr.count-1
while leftPointer < rightPointer {
let middleIndex = (leftPointer + rightPointer) / 2
let middleValue = arr[middleIndex]
if middleValue == target {
return middleValue
}
//Check for out of bounds error
let leftIndex = middleIndex-1
let leftValue = arr[leftIndex]
if leftValue <= target && middleValue >= target {
let leftDistance = abs(leftValue-target)
let rightDistance = abs(middleValue-target)
if leftDistance <= rightDistance {
return leftValue
} else {
return middleValue
}
}
if middleValue <= target {
leftPointer = middleIndex+1
} else {
rightPointer = middleIndex
}
}
guard let first = arr.first, let last = arr.last else {
fatalError()
}
if target <= first {
return first
} else if target >= last {
return last
} else {
fatalError()
}
}
let first = [1,2,3,5,5,5,7,9,19,11] // 6 --> 5
let second = [1,2,3] // 8 --> 3
let third = [9, 10, 22, 59, 67, 72, 100] // 70 --> 72
let fourth = [100, 101, 102] //5 --> 100 => Heres the error
print(closestValue(first, 6))
print(closestValue(second, 8))
print(closestValue(third, 110))
print(closestValue(fourth, 5))
I expected the fourth output 100. Because 100 is the closest value to 5 in the fourth array.

Answer 1

I can see you put in some boundary checks, but they should happen at the beginning of the function instead of the end. Allow me to rewrite the whole function:
func closestValue(_ arr: [Int],_ target: Int) -> Int {
// Array must not be empty
guard arr.count > 0 else { fatalError("Array must not be empty") }
// If array has only 1 element, that element is the closest
guard arr.count > 1 else { return arr[0] }
// To use binary search, your array must be ever-increasing or ever-decreasing
// Here, we require that the array must be ever-increasing
for index in 1..<arr.count {
if arr[index - 1] > arr[index] {
fatalError("Array must be monotonous increasing. Did you forget to sort it?")
}
}
// If the target is outside of the range of the array,
// return the edges of the array
guard arr.first! <= target else { return arr.first! }
guard target <= arr.last! else { return arr.last! }
// Now some actual searching
var left = 0
var right = arr.count - 1
while left < right {
if left == right - 1 {
return abs(arr[left] - target) <= abs(arr[right] - target) ? arr[left] : arr[right]
}
let middle = (left + right) / 2
switch arr[middle] {
case target:
return target
case ..<target:
left = middle
default:
right = middle
}
}
fatalError("It should never come here")
}
let first = [1,2,3,5,5,5,7,9,11,19] // 6 --> 5
let second = [1,2,3] // 8 --> 3
let third = [9, 10, 22, 59, 67, 72, 100] // 70 --> 72
let fourth = [100, 101, 102] //5 --> 100
print(closestValue(first, 6))
print(closestValue(second, 8))
print(closestValue(third, 70))
print(closestValue(fourth, 5))
Some notes:
if left == right - 1 { ... } allows the while loop to terminate. Otherwise, integer division will round middle down to `left, resulting in a infinite loop.
case ..<target is a short hand for "when arr[middle] < target"
The while should always find a solution and return from inside but I have not thoroughly tested that yet. If you find a case where it reaches the last fatalError, let me know.
The first example has two answers: both 5 and 7 are 1-away from target = 6.