I got a simple code which works and which I am programming in and old fashioned way and I am sure there is a more elegant way of doing this in swift. Here is the code:
var cardsInCompartment1:Int = 0
var cardsInCompartment2:Int = 0
for card in cards{
if card.compartment == 1{
cardsInCompartment1 += 1
print(cardsInCompartment1)
}
if card.compartment == 2{
cardsInCompartment2 += 1
print(cardsInCompartment2)
}
}
I basically got cards in different compartments and now I want to count how many cards are in each compartment.
How about using filter to select the cards you want? Then you can just count them:
let cardsInCompartment1 = cards.filter { $0.compartment == 1 }.count
let cardsInCompartment2 = cards.filter { $0.compartment == 2 }.count
If you have a bunch of compartments, you could store the counts in a dictionary:
var compartmentCounts = [Int:Int]()
cards.forEach {
compartmentCounts[$0.compartment] = (compartmentCounts[$0.compartment] ?? 0) + 1
}
In this case, the key would be the compartment#, and the value would be the card count. Something like [1: 32, 2: 42] if there are 32 and 42 cards in each respective compartment.
Try this:
var cardsInCompartment1:Int = 0
var cardsInCompartment2:Int = 0
for card in cards {
(card.compartment == 1) ? (cardsInCompartment1 += 1) : (cardsInCompartment2 += 1)
}
I think you should store the cardsInCompartment as arrays:
var cardsInCompartment = [0, 0] // you can add more to this array
Then you can just loop through the cards and add the values to the array elements:
for card in cards {
cardsInCompartment[card.compartment - 1] += 1
print(cardsInCompartment[card.compartment - 1])
}
What about a switch statement? Something like this?
var card:Int = 1
var CardsInCompartment:Int = 0
switch (card) {
case 1:
CardsInCompartment += 1
print("CardsInCompartment \(CardsInCompartment)")
case 2:
CardsInCompartment += 2
print("CardsInCompartment \(CardsInCompartment)")
default:
}
Or, use an Array to keep your counts:
var counts = [ 0, 0, 0 ] // create an array of integers, where the Ints in the array represent the count of cards in each compartment
cards.forEach { counts[ $0.compartment ] += 1 } // for each card, increment the count in array corresponding to the compartment of the card. (if card.compartment == 1, increment counts[1], and so on
print("cards in compartment 1 \(counts[1])")
print("cards in compartment 2 \(counts[2])")
(This assumes your only compartments are integers 1 and 2)
I like Aaron Brager's idea which counts values into dictionary. I am using reduce to eliminate mutable dictionary outside the 'loop' (more functional)
let d = cards.reduce([:]) { (d, card) -> [Int:Int] in
var d = d
let s = d[card.compartment] ?? 0
d[card.compartment] = s + 1
return d
}
Related
I want to add the numbers together and print every 4 elements, however i cannot wrap my head around using the stride function, if i am using the wrong approach please explain a better method
var numbers = [1,2,3,4,5,6,7,8,9,10,11,12,13]
func addNumbersByStride(){
var output = Stride...
//first output = 1+2+3+4 = 10
//second output = 5+6+7+8 = 26 and so on
print(output)
}
It seems you would like to use stride ...
let arr = [1,2,3,4,5,6,7,8,9,10,11,12,13]
let by = 4
let i = stride(from: arr.startIndex, to: arr.endIndex, by: by)
var j = i.makeIterator()
while let n = j.next() {
let e = min(n.advanced(by: by), arr.endIndex)
let sum = arr[n..<e].reduce(0, +)
print("summ of arr[\(n)..<\(e)]", sum)
}
prints
summ of arr[0..<4] 10
summ of arr[4..<8] 26
summ of arr[8..<12] 42
summ of arr[12..<13] 13
You can first split the array into chunks, and then add the chunks up:
extension Array {
// split array into chunks of n
func chunked(into size: Int) -> [[Element]] {
return stride(from: 0, to: count, by: size).map {
Array(self[$0 ..< Swift.min($0 + size, count)])
}
}
}
// add each chunk up:
let results = numbers.chunked(into: 4).map { $0.reduce(0, +) }
If you would like to discard the last sum if the length of the original array is not divisible by 4, you can add an if statement like this:
let results: [Int]
if numbers.count % 4 != 0 {
results = Array(numbers.chunked(into: 4).map { $0.reduce(0, +) }.dropLast())
} else {
results = numbers.chunked(into: 4).map { $0.reduce(0, +) }
}
This is quite a basic solution and maybe not so elegant. First calculate and print sum of every group of 4 elements
var sum = 0
var count = 0
for n in stride(from: 4, to: numbers.count, by: 4) {
sum = 0
for i in n-4..<n {
sum += numbers[i]
}
count = n
print(sum)
}
Then calculate the sum of the remaining elements
sum = 0
for n in count..<numbers.count {
sum += numbers[n]
}
print(sum)
My question is as on the title. I'm trying to print even numbers from 1 to 500 suing a while loop and break keyword. Below is my best possible answer I can think of, but this only print number 2. I've been spending hours but I wasn't able to solve it.
var number = 0
while true{
number += 2
print(number)
if number % 2 == 0 && number <= 500 {
break
}
}
You can use Stride
for evenNumber in stride(from: 0, through: 500, by: 2) {
print(evenNumber)
}
To specifically do this with while and break:
var i = 0
while true {
print(i)
i += 2
if i > 500 {
break
}
}
for i in 0...500 {
if i % 2 == 0 {
print(i)
}
}
Use below code
var numbers = 0...500
for number in numbers {
if number % 2 == 0 {
print(number)
}
}
I think it's easier to use build-in stride
let arr = Array(stride(from: 0, to: 502, by: 2))
print(arr)
//
For manually
var counter = 0
var arr = [Int]()
while counter <= 500 {
if counter % 2 == 0 {
print(counter)
arr.append(counter)
}
counter += 1
}
var number = 0
while true {
number += 2
print(number)
// ↓ Your code goes here ↓
if number > 499 {
break
}
}
Per Codefighters:
Note: Write a solution with O(n) time complexity and O(1) additional space complexity, since this is what you would be asked to do during a real interview.
Given an array a that contains only numbers in the range from 1 to a.length, find the first duplicate number for which the second occurrence has the minimal index. In other words, if there are more than 1 duplicated numbers, return the number for which the second occurrence has a smaller index than the second occurrence of the other number does. If there are no such elements, return -1.
Example
For a = [2, 3, 3, 1, 5, 2], the output should be firstDuplicate(a) = 3.
There are 2 duplicates: numbers 2 and 3. The second occurrence of 3 has a smaller index than than second occurrence of 2 does, so the answer is 3.
For a = [2, 4, 3, 5, 1], the output should be firstDuplicate(a) = -1.
So here is what I came up with. It works but fails on the final test because it ran over 4000ms. I am stuck to what else I can do. Any Ideas to improve speed?
func firstDuplicate(a : [Int]) -> Int {
var duplicateIndexArray = [Int]()
for firstIndex in 0..<a.count {
for secondIndex in 0..<a.count {
if a[firstIndex] == a[secondIndex] && firstIndex != secondIndex {
print(firstIndex, secondIndex)
if !(duplicateIndexArray.contains(firstIndex)){
duplicateIndexArray.append(secondIndex)
break
}
}
}
}
// Check for duplicacy
if duplicateIndexArray.count > 0 {
print(duplicateIndexArray)
return a[duplicateIndexArray.min()!]
}
return -1
}
The O(n) time part is easy, but the O(1) additional space is a bit tricky. Usually, a hash set (or bit array in your case) can be used to check if a number occurred more than once, but that requires O(n) additional space. For O(1) additional space, we can use the source array itself as a bit array by making some of the numbers in it negative.
For example if the first number in the array is 3, then we make the number at position 3-1 negative. If one of the other numbers in the array is also 3, we can check if the number at position 3-1 is negative.
I don't have any experience with Swift, so I'll try to write a function in pseudocode:
function firstDuplicate(a)
result = -1
for i = 0 to a.count - 1
if a[abs(a[i])-1] < 0 then
result = a[i]
exit for loop
else
a[abs(a[i])-1] = -a[abs(a[i])-1]
// optional restore the negative numbers back to positive
for i = 0 to a.count - 1
if a[i] < 0 then
a[i] = -a[i]
return result
Replace this line
for secondIndex in 0..<a.count
with
for secondIndex in firstIndex..<a.count
There is no requirement of double checking
So Your Final code is
func firstDuplicate(a : [Int]) -> Int {
var duplicateIndexArray = [Int]()
for firstIndex in 0..<a.count {
for secondIndex in firstIndex..<a.count {
if a[firstIndex] == a[secondIndex] && firstIndex != secondIndex {
print(firstIndex, secondIndex)
if !(duplicateIndexArray.contains(firstIndex))
{
duplicateIndexArray.append(secondIndex)
break
}
}
}
}
// Check for duplicacy
if duplicateIndexArray.count > 0
{
print(duplicateIndexArray)
return a[duplicateIndexArray.min()!]
}
return -1
}
func firstDuplicate(input: [Int]) -> Int{
var map : [String : Int] = [:]
var result = -1
for i in 0 ..< input.count {
if map["\(input[i])"] != nil {
result = i
break
}
else {
map["\(input[i])"] = i
}
}
return result
}
I am trying to solve the task
Using a standard for-in loop add all odd numbers less than or equal to 100 to the oddNumbers array
I tried the following:
var oddNumbers = [Int]()
var numbt = 0
for newNumt in 0..<100 {
var newNumt = numbt + 1; numbt += 2; oddNumbers.append(newNumt)
}
print(oddNumbers)
This results in:
1,3,5,7,9,...199
My question is: Why does it print numbers above 100 although I specify the range between 0 and <100?
You're doing a mistake:
for newNumt in 0..<100 {
var newNumt = numbt + 1; numbt += 2; oddNumbers.append(newNumt)
}
The variable newNumt defined inside the loop does not affect the variable newNumt declared in the for statement. So the for loop prints out the first 100 odd numbers, not the odd numbers between 0 and 100.
If you need to use a for loop:
var odds = [Int]()
for number in 0...100 where number % 2 == 1 {
odds.append(number)
}
Alternatively:
let odds = (0...100).filter { $0 % 2 == 1 }
will filter the odd numbers from an array with items from 0 to 100. For an even better implementation use the stride operator:
let odds = Array(stride(from: 1, to: 100, by: 2))
If you want all the odd numbers between 0 and 100 you can write
let oddNums = (0...100).filter { $0 % 2 == 1 }
or
let oddNums = Array(stride(from: 1, to: 100, by: 2))
Why does it print numbers above 100 although I specify the range between 0 and <100?
Look again at your code:
for newNumt in 0..<100 {
var newNumt = numbt + 1; numbt += 2; oddNumbers.append(newNumt)
}
The newNumt used inside the loop is different from the loop variable; the var newNumt declares a new variable whose scope is the body of the loop, so it gets created and destroyed each time through the loop. Meanwhile, numbt is declared outside the loop, so it keeps being incremented by 2 each time through the loop.
I see that this is an old question, but none of the answers specifically address looping over odd numbers, so I'll add another. The stride() function that Luca Angeletti pointed to is the right way to go, but you can use it directly in a for loop like this:
for oddNumber in stride(from:1, to:100, by:2) {
// your code here
}
stride(from:,to:,by:) creates a list of any strideable type up to but not including the from: parameter, in increments of the by: parameter, so in this case oddNumber starts at 1 and includes 3, 5, 7, 9...99. If you want to include the upper limit, there's a stride(from:,through:,by:) form where the through: parameter is included.
If you want all the odd numbers between 0 and 100 you can write
for i in 1...100 {
if i % 2 == 1 {
continue
}
print(i - 1)
}
For Swift 4.2
extension Collection {
func everyOther(_ body: (Element) -> Void) {
let start = self.startIndex
let end = self.endIndex
var iter = start
while iter != end {
body(self[iter])
let next = index(after: iter)
if next == end { break }
iter = index(after: next)
}
}
}
And then you can use it like this:
class OddsEvent: UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
(1...900000).everyOther{ print($0) } //Even
(0...100000).everyOther{ print($0) } //Odds
}
}
This is more efficient than:
let oddNums = (0...100).filter { $0 % 2 == 1 } or
let oddNums = Array(stride(from: 1, to: 100, by: 2))
because supports larger Collections
Source: https://developer.apple.com/videos/play/wwdc2018/229/
For a project, I'm trying to find the sum of the multiples of both 3 and 5 under 10,000 using Swift. Insert NoobJokes.
Printing the multiples of both 3 and 5 was fairly easy using a ForLoop, but I'm wondering how I can..."sum" all of the items that I printed.
for i in 0...10000 {
if i % 3 == 0 || i % 5 == 0 {
print(i)
}
}
(468 individual numbers printed; how can they be summed?)
Just a little walk through about the process. First you will need a variable which can hold the value of your sum, whenever loop will get execute. You can define an optional variable of type Int or initialize it with a default value same as I have done in the first line. Every time the loop will execute, i which is either multiple of 3 or 5 will be added to the totalSum and after last iteration you ll get your result.
var totalSum = 0
for i in 0...10000 {
if i % 3 == 0 || i % 5 == 0
{
print(i)
totalSum = totalSum + i
}
}
print (totalSum)
In Swift you can do it without a repeat loop:
let numberOfDivisiblesBy3And5 = (0...10000).filter{ $0 % 3 == 0 || $0 % 5 == 0 }.count
Or to get the sum of the items:
let sumOfDivisiblesBy3And5 = (0...10000).filter{ $0 % 3 == 0 || $0 % 5 == 0 }.reduce(0, {$0 + $1})
range : to specify the range of numbers for operation to act on.
here we are using filter method to filter out numbers that are multiple of 3 and 5 and then sum the filtered values.
(reduce(0,+) does the job)
let sum = (3...n).filter({($0 % 3) * ($0 % 5) == 0}).reduce(0,+)
You just need to sum the resulting i like below
var sum = 0
for i in 0...10000 {
if i % 3 == 0 || i % 5 == 0 {
sum = sum + i
print(i)
}
}
Now sum contains the Sum of the values
Try this:
var sum = 0
for i in 0...10000 {
if i % 3 == 0 || i % 5 == 0 {
sum = sum + i
print(i)
}
}
print(sum)
In the Bottom line, this should to be working.
var sum = 0
for i in 0...10000 {
if i % 3 == 0 || i % 5 == 0 {
sum += i
print(i)
}
}
print(sum)