closures - swift maximum consecutive positive numbers

Question

How to count maximum consecutive positive numbers using closures?

var numbers = [1,3,4,-1,-2,5,2,-2,-3,-4,5]
//in this case it should be 3

print(numbers.reduce(0, { $1 > 0 ? $0 + 1 : $0 } ))//this counts total positive numbers

Answer 1

Update: Simpler solution: Split the array into slices of positive elements, and determine the maximal slice length:

let  numbers = [1,3,4,-1,-2,5,2,-2,-3,-4,5]
let maxConsecutive = numbers.split(whereSeparator: { $0 <= 0 }).map { $0.count }.max()!
print(maxConsecutive) // 3

---

Old answer:) Using the ideas from Swift running sum:

let  numbers = [1,3,4,-1,-2,5,2,-2,-3,-4,5]

let maxConsecutive = numbers.map({
    () -> (Int) -> Int in var c = 0; return { c = $0 > 0 ? c + 1 : 0; return c }
}()).max()!

Here map() maps each array element to the count of consecutive positive numbers up to the elements position, in this case

[1, 2, 3, 0, 0, 1, 2, 0, 0, 0, 1]

The transformation is created as an "immediately evaluated closure" to capture a variable c which holds the current number of consecutive positive numbers. The transformation increments or resets c, and returns the updated value.

If the array is possibly large then change it to

let maxConsecutive = numbers.lazy.map( ... ).max()!

so that the maximum run length is determined without creating an intermediate array.