Could someone explain to me the logic behind this hashMap algorithm? I'm getting confused about how the algorithm receives the total sum. I'm starting to learn about algorithms, so it's a little confusing for me. I made comments in my code to pinpoint each line code, but I'm not sure I'm grasping logic correctly. I'm just looking for an easier way to understand how the algorithm works to avoid confusing myself.
//**calculate Two Number Sum
func twoNumberSum(_ array: [Int], _ targetSum: Int) -> [Int] {
//1) initilize our Array to hold Integer Value: Boolean value to store value into hashTable
var numbersHashMap = [Int:Bool]()
//2) create placeHolder called number that iterates through our Array.
for number in array {
//3) variable = y - x
let match = targetSum - number
//4) ??
if let exists = numbersHashMap[match], exists {
//5) match = y / number = x
return [match, number] //
} else {
//6) Store number in HashTable and repeats
numbersHashMap[number] = true
}
}
return []
}
twoNumberSum([3,5,-4, 8, 11, 1, -1, -6], 10)
// x = Number
// y = Unknown *Solve for Y*
Sure, I can walk you through it. So we have a list of numbers, are we are trying to find two numbers that add together to make the specified target. To do this, for each number x, we check if (target - x) is in the list. If it is not, then we add x to the list. If it is, then we return x and (target - x).
Step 4 in your code is the part where we check if (target - x) is in the list. To see why this makes sense, let's walk through an example.
Say we have [2, 3, -1] and our target is 1. In this case, we first consider x = 2 and check our hashmap for (target - x) = (1 - 2) = -1. Since -1 is not in the hashmap, we add 2 to the hashmap. We then consider x = 3 and check for (1 - 3) = -2. Again, -2 is not in the hashmap, so we add it. Now we check x - -1. In this case, when we check (target - x) = (1 - (-1)) = 2, 2 is in the hashmap. Intuitively, we have already "seen" 2, and know that 2 and -1 can be added to get our value.
This is what provides the speed optimization over checking every two numbers in the list.
Related
Let's say I want to generate a random number between 1 and 100, but I don't want to include 42. How would I do this without repeating the random method until it is not 42.
Updated for Swift 5.1
Excluding 1 value
var nums = [Int](1...100)
nums.remove(at: 42)
let random = Int(arc4random_uniform(UInt32(nums.count)))
print(nums[random])
Excluding multiple values
This extension of Range does provide a solution when you want to exclude more than 1 value.
extension ClosedRange where Element: Hashable {
func random(without excluded:[Element]) -> Element {
let valid = Set(self).subtracting(Set(excluded))
let random = Int(arc4random_uniform(UInt32(valid.count)))
return Array(valid)[random]
}
}
Example
(1...100).random(without: [40,50,60])
I believe the computation complexity of this second solution is O(n) where n is the number of elements included in the range.
The assumption here is the no more than n excluded values are provided by the caller.
appzYourLife has some great general purpose solutions, but I want to tackle the specific problem in a lightweight way.
Both of these approaches work roughly the same way: Narrow the range to the random number generator to remove the impossible answer (99 answers instead of 100), then map the result so it isn't the illegal value.
Neither approach increases the probability of an outcome relative to another outcome. That is, assuming your random number function is perfectly random the result will still be random (and no 2x chance of 43 relative to 5, for instance).
Approach 1: Addition.
Get a random number from 1 to 99. If it's greater than or equal to the number you want to avoid, add one to it.
func approach1()->Int {
var number = Int(arc4random_uniform(99)+1)
if number >= 42 {
number = number + 1
}
return number
}
As an example, trying to generate a random number from 1-5 that's not 3, take a random number from 1 to 4 and add one if it's greater than or equal to 3.
rand(1..4) produces 1, +0, = 1
rand(1..4) produces 2, +0, = 2
rand(1..4) produces 3, +1, = 4
rand(1..4) produces 4, +1, = 5
Approach 2: Avoidance.
Another simple way would be to get a number from 1 to 99. If it's exactly equal to the number you're trying to avoid, make it 100 instead.
func approach2()->Int {
var number = Int(arc4random_uniform(99)+1)
if number == 42 {
number = 100
}
return number
}
Using this algorithm and narrowing the range to 1-5 (while avoiding 3) again, we get these possible outcomes:
rand(1..4) produces 1; allowed, so Result = 1
rand(1..4) produces 2, allowed, so Result = 2
rand(1..4) produces 3; not allowed, so Result = 5
rand(1..4) produces 4, allowed, so Result = 4
I'm making an app where I have to put a lot of if/else statements. I know you can do as in the title in some other coding language, but I'm not sure if you can do it in Swift.
How do you shorten this:
if x == y || x == z {
//do something
}
To something like this:
if x == y || z {
//do something
}
Perhaps you could consider using an array and checking to see if x is in the array, like in the following example:
let (x, y, z) = (3, 8, 3)
if [y, z].contains(x) {
//True
}
If you're comparing objects (like UIImage), use containsObject instead of contains:
if [x, y, z].containsObject(y) {
//True
}
I conject that there is no sensible language (swift included), that distributes comparison == across logical or ||.
The way you've written it - x == y || x == z - is the most compact form.
Jack Greenhill's answer does indeed go into the right direction. However with more and more values, his method will get very inefficient, since it has to check every element of the array against equality, therefore complexity O(n).
A very underrated data structure, which can do this kind of operation in O(1) should be used instead: The Set. It uses hash values to check quickly whether a value is present or not. You can use it like this:
let x = 3
let values : Set = [1, 3, 6, 1, 7] // {6, 7, 3, 1}
if values.contains(x) {
// ...
}
This takes the same amount of time, whether values contains just one or 1000 elements. An array would be 1000 times slower.
Oftentimes the decision to use an array is made before even considering a set. If your elements don't have any order and can only occur once (which is actually more often the case than you'd think), you probably want a set. A set gives you useful methods, such as union, isSubset, interception and more, for free by just putting your elements in it. The only additional requirement for the element type is to conform to Hashable.
Perhaps this:
switch x {
case y, z : // do then
default : // do else
}
I have an array of coordinates (965 in total). I want to use these coordinates in Google Roads API however the limit is 100.
I have a function that determines how many items in the array and then gets the value to use.
round(Double(userCoordinatesHardCoded.count / 100))
produces 9.
I would like to remove ALL items that are not at indexes that are multiples of, in this case, 9. So in theory I will only ever have no more than 100 items in the array.
If possible, I would like to keep the first and last array item.
I know this has already been answered, but this is a great use case for using the filter function built into Swift:
let newCoordinates = oldCoordinates.filter { coord in
return (coord % 9 != 0)
}
If space is not a problem, you can create another array with the desired multiples.
var oldCoordinates // Imagine it having all those elements
var newCoordinates : GoogleCoordinates = [] // This one will have new ones
newCoordinates.append(oldCoordinates[0])
var x = 1
for (x; x < oldCoordinates.count ; x++ ) {
if (x % 5 == 0) {
newCoordinates.append(oldCoordinates[x])
}
}
if (x != (oldCoordinates.count - 1)) {
newCoordinates.append(oldCoordinates[oldCoordinates.count - 1])
}
x is an object that holds an array called point.
x implements the subscript operator so you can do things, like x[i] to get the array's ith element (of type T, which is usually an Int or Double).
This is what I want to do:
x[0...2] = [0...2]
But I get an error that says ClosedInterval<T> is not convertible to Int/Double.
Edit1:
Here is my object x:
let x = Point<Double>(dimensions:3)
For kicks and giggles: define x as [1.0,2.0,0.0]
I can get the first n elements via x[0...2].
What I want to know is how to update x[0...2] to hold [0.0, 0.0.0.0] in one fell swoop. Intuitively, I would want to do x[0...2] = [0...2]. This does not work as can be seen in the answers. I want to update x without iteration (on my end) and by hiding the fact that x is not an array (even though it is not).
[0...2] is an array with one element which, at best, will be a Range<Int> from 0 through 2. You can't assign that to a slice containing, say, Ints.
x[0...2] on the other hand is (probably) a slice, and Sliceable only defines a get subscript, not a setter. So even if the types were more compatible - that is, if you tried x[0...2] = 0...2, which at least is attempting to replace a range within x with the values of a similarly-sized collection - it still wouldn't work.
edit: as #rintaro points out, Array does support a setter subscript for ranges – so if x were a range you could do x[0...2] = Slice(0...2) – but it has to be a slice you assign, so I'd still go with replaceRange.
If what you mean is you want to replace entries 0 through 2 with some values, what you want is replaceRange, as long as your collection conforms to RangeReplaceableCollection (which, for example, Array does):
var x = [0,1,2,3,4,5]
var y = [200,300,400]
x.replaceRange(2..<5, with: y)
// x is now [0,1,200,300,400,5]
Note, the replaced range and y don't have to be the same size, the collection will expand/contract as necessary.
Also, y doesn't have to an array, it can be any kind of collection (has to be a collection though, not a sequence). So the above code could have been written as:
var x = [0,1,2,3,4,5]
var y = lazy(2...4).map { $0 * 100 }
x.replaceRange(2..<5, with: y)
edit: so, per your edit, to in-place zero out an array of any size in one go, you can do:
var x = [1.0,2.0,0.0]
// range to replace is the whole array's range,
// Repeat just generates any given value n times
x.replaceRange(indices(x), with: Repeat(count: x.count, repeatedValue: 0.0))
Adjust the range (and count of replacing entries) accordingly if you want to just zero out a subrange.
Given your example Point class, here is how you could implement this behavior assuming it's backed by an array under the hood:
struct Point<T: FloatLiteralConvertible> {
private var _vals: [T]
init(dimensions: Int) {
_vals = Array(count: dimensions, repeatedValue: 0.0)
}
mutating func replaceRange
<C : CollectionType where C.Generator.Element == T>
(subRange: Range<Array<T>.Index>, with newElements: C) {
// just forwarding on the request - you could perhaps
// do some additional validation first to ensure dimensions
// aren't being altered...
_vals.replaceRange(subRange, with: newElements)
}
}
var x = Point<Double>(dimensions:3)
x.replaceRange(0...2, with: [1.1,2.2,3.3])
You need to implement subscript(InvervalType) to handle the case of multiple assignments like this. That isn't done for you automatically.
I'm in the process of getting comfortable passing unnamed functions as arguments and I am using this to practice with, based off of the examples in the Swift Programming Guide.
So we have an array of Ints:
var numbers: Int[] = [1, 2, 3, 4, 5, 6, 7]
And I apply a transform like so: (7)
func transformNumber(number: Int) -> Int {
let result = number * 3
return result
}
numbers = numbers.map(transformNumber)
Which is equal to: (7)
numbers = numbers.map({(number: Int) -> Int in
let result = number * 3
return result;
})
Which is equal to: (8)
numbers = numbers.map({number in number * 3})
Which is equal to: (8)
numbers = numbers.map({$0 * 3})
Which is equal to: (8)
numbers = numbers.map() {$0 * 3}
As you can see in the following graphic, the iteration count in the playground sidebar shows that in the furthest abstraction of a function declaration, it has an 8 count.
Question
Why is it showing as 8 iterations for the last two examples?
It's not showing 8 iterations, really. It's showing that 8 things executed on that line. There were 7 executions as part of the map function, and an 8th to do the assignment back into the numbers variable.
It looks like this could probably provide more helpful diagnostics. I would highly encourage you to provide feedback via https://bugreport.apple.com.
Slightly rewriting your experiment to use only closures, the call counts still differ by one:
Case 1: Explicitly specifying argument types (visit count is 7)
var f1 = {(number: Int) -> Int in
let result = number * 3
return result
}
numbers.map(f1)
Case 2: Implicit argument types (visit count is 8)
var f2 = {$0 * 3}
numbers.map(f2)
If the (x times) count reported by the REPL does indeed represent a count of visits to that code location, and noting that the count is greater by one in cases where the closure type arguments are not explicitly specified (e.g. f2), my guess is that at least in the playground REPL, the extra visit is to establish actual parameter types and fill that gap in the underlying AST.