Given an array of numbers, reset the array to put all the non-zero numbers in front of all the zeros in the array, then return the count of non-zero numbers. e.g., for an input array of [3,0,2,0,0,1,0,4], you’ll end up with a return value of 4 and an array of [3,2,1,4,0,0,0,0]

resetZeros.swift

import Foundation

/*Given an array of numbers, reset the array to put all the non-zero numbers in front of all the zeros in the array, then return the count of non-zero numbers.
 
 e.g., for an input array of [3,0,2,0,0,1,0,4], you’ll end up with a return value of 4 and an array of [3,2,1,4,0,0,0,0]
 */
//idea

//get first zero
//return func(array.removeZero) + [0]
//do until array.firstZero() is nil -> there, return array

func get<T: Equatable>(value: T, toBottomOf array: [T]) -> [T] {
    
    func removeFirst(value: T, from array: [T]) -> [T] {
        
        guard let firstItem = array.first else {
            
            return array
        }
        
        if firstItem == value {
            
            return Array(array.dropFirst())
            
        } else {
            
            return [firstItem] + removeFirst(value: value, from: Array(array.dropFirst()))
        }
    }
    
    guard array.filter({ $0 == value}).count > 0 else {
        
        return array
    }
    
    return get(value: value, toBottomOf: removeFirst(value: value, from: array)) + [value]
}

let array = [3,0,2,0,0,1,0,4]

let resetArray = get(value: 0, toBottomOf: array)

You could always filter the zeros out and then create a new array with the zeros.