I have following code that is supposed to do two things:
1) Fire blasters when temp is under 100 and blasters are not overheated.
2) Wait out until the temp is cooled down to zero.
I am not too sure why the codes below time out. I am a real beginner with Swift (I have an intermediate background in R.)
var shields = 5
var blastersOverheating = false
var blasterFireCount = 0
var temp = 0
while shields > 0 {
temp = temp + 1
blasterFireCount = blasterFireCount + 1
if temp == 100{
blastersOverheating = true
}
if blastersOverheating{
print("Sleeping to cool down to 0")
sleep(10)
temp = 0
blasterFireCount = 0
blastersOverheating = false
continue
}
print(temp)
}
If you run this code on a free online platform (like iswift.org/playground), your program will time out because it would run forever otherwise.
To counter this problem limit your loop so it only does, for example, 1000 cycles.
View the example online: http://swift.sandbox.bluemix.net/#/repl/59123b184ee0cd258050b2cd
var coolingDown = false
var blasterFireCount = 0
var temp = 0
for _ in 0..<1000 {
if coolingDown {
temp -= 1
print("cooling down \(temp)")
if temp == 0 { coolingDown = false }
} else {
temp += 1
blasterFireCount += 1
print("fire \(blasterFireCount)")
if temp == 100 { coolingDown = true }
}
}
Related
I'm writing a function thats supposed to loop though a 2D array diagonally(top left to bottom right). However, the code does not add to the outer while loop(i), it keeps it at 0. arr is 9X9
var i = 0
var j = 0
while(i < arr.count-1){
while (j < arr.count-1) {
print("i = \(i) --- j = \(j)")
if(i == j){
sumDiagonalLeft += arr[j][i]
print(sumDiagonalLeft)
if(arr[j][i] == 1){
informationUsed += 1
arr[i][j] = 2
}
}
j += 1
}
i += 1
}
Thank you for your time :)
try this,
var array:[[Int]] = []
array.append([1,2,3,4,5,6,7,8,9])
....
array.append([1,2,3,4,5,6,7,8,9])
for (index, element) in array.enumerated(){
for (innerIndex,innerElement) in element.enumerated(){
print(innerElement) // you can do your logics here
}
}
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
}
}
Here is my code.
import UIKit
var str = "Hello, playground"
//There are two sorted arrays nums1 and nums2 of size m and n respectively.
//Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
//Example 1:
//nums1 = [1, 3]
//nums2 = [2]
//
//The median is 2.0
//Example 2:
//nums1 = [1, 2]
//nums2 = [3, 4]
//
//The median is (2 + 3)/2 = 2.5
var num1 = [1,2,2,5]
var num2 = [2,3,9,9]
class Solution {
func findMedianSortedArrays(_ nums1: [Int], _ nums2: [Int]) -> Double {
var A = nums1
var B = nums2
var m = nums1.count
var n = nums2.count
var max_of_left : Int = 0
var min_of_right = 0
if n < m {
var temp : [Int]
var tempt : Int
temp = nums1
tempt = m
A = nums2
B = temp
m = n
n = tempt
}
if n == 0{
fatalError("Arrays must be fulfilled")
}
var imin = 0
var imax = m
let half_len = Int((m+n+1)/2)
while imin <= imax {
let i = Int((imin + imax) / 2)
let j = half_len - i
if i > 0 && A[i-1] > B[j]{
imax = i - 1
}
else if i < m && A[i] < B[j-1]{
imin = i + 1
}
else
{
if i == 0{
max_of_left = B[j-1]
}
else if j == 0{
max_of_left = A[i-1]
}
else
{
max_of_left = max(A[i-1], B[j-1])
}
if m+n % 2 == 1{
return Double(max_of_left)
}
if i==m{
min_of_right = B[j]
}
else if j == n{
min_of_right = A[i]
}
else{
min_of_right = min(A[i], B[j])
//editor indicates error here
}
return Double((Double(max_of_left+min_of_right) / 2.0))
}
}
}
}
var a = Solution()
print(a.findMedianSortedArrays(num1, num2))
error: day4_Median_of_Two_Sorted_Arrays.playground:86:13: error: missing return in a function expected to return 'Double'
Since I put my return out of if statement, I think it will be okay because it will stop while looping when it meets return.
But editor says it's not.
I want to know why. Please explain me why.
Every code path through your findMedianSortedArrays() must return a Double.
So you need a return of a Double placed outside of your while loop. Even if you had every code path within the while loop have a return double, if imin > imax you wouldn't even enter the while loop, and so would need a return of a double outside it.
I fixed it by putting another return out of while loop.
//: Playground - noun: a place where people can play
import UIKit
var str = "Hello, playground"
//There are two sorted arrays nums1 and nums2 of size m and n respectively.
//Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
//Example 1:
//nums1 = [1, 3]
//nums2 = [2]
//
//The median is 2.0
//Example 2:
//nums1 = [1, 2]
//nums2 = [3, 4]
//
//The median is (2 + 3)/2 = 2.5
var num1 = [1,2,2,5]
var num2 = [2,3,9,9]
class Solution {
func findMedianSortedArrays(_ nums1: [Int], _ nums2: [Int]) -> Double {
var A = nums1
var B = nums2
var m = nums1.count
var n = nums2.count
var max_of_left : Int = 0
var min_of_right = 0
if n < m {
var temp : [Int]
var tempt : Int
temp = nums1
tempt = m
A = nums2
B = temp
m = n
n = tempt
}
if n == 0{
fatalError("Arrays must be fulfilled")
}
var imin = 0
var imax = m
let half_len = Int((m+n+1)/2)
while imin <= imax {
let i = Int((imin + imax) / 2)
let j = half_len - i
if i > 0 && A[i-1] > B[j]{
imax = i - 1
}
else if i < m && A[i] < B[j-1]{
imin = i + 1
}
else
{
if i == 0{
max_of_left = B[j-1]
}
else if j == 0{
max_of_left = A[i-1]
}
else
{
max_of_left = max(A[i-1], B[j-1])
}
if m+n % 2 == 1{
return Double(max_of_left)
}
if i==m{
min_of_right = B[j]
}
else if j == n{
min_of_right = A[i]
}
else{
min_of_right = min(A[i], B[j])
}
return Double((Double(max_of_left+min_of_right) / 2.0))
}
}
return Double((Double(max_of_left+min_of_right) / 2.0))
}
}
var a = Solution()
print(a.findMedianSortedArrays(num1, num2))
For some reason, it always gave me the wrong result. It's always isItPrime = true no matter what number was assigned to the "number" variable.
This is my code:
let number = 6
var i = 1
var isItPrime: Bool?
while i < number {
if number % i == 0 {
isItPrime = false
} else {
isItPrime = true
}
i += 1
}
print(isItPrime)
Can somebody explain to me what's wrong with my code and why the isItPrime bool outputs always true ?
Problem 1
The last iteration of your while loop
while i < number {
if number % i == 0 {
isItPrime = false
} else {
isItPrime = true
}
i += 1
}
does overwrite the result.
So you always end up with the following result
if number % (number-1) == 0 {
isItPrime = false
} else {
isItPrime = true
}
Problem 2
Finally every number can be divided by 1, so you should start i from 2.
So
let number = 6
var i = 2
var isItPrime = true
while i < number {
if number % i == 0 {
isItPrime = false
break
}
}
print(isItPrime)
Refactoring
You can write a similar logic using Functional Programming
let number = 5
let upperLimit = Int(Double(number).squareRoot())
let isPrime = !(2...upperLimit).contains { number % $0 == 0 }
Because isItPrime is overwritten in subsequent iterations, the last number which is checked, which is number - 1 will always set isItPrime to true, because number and number - 1 are coprime.
Instead of saving the value to a boolean, just end the loop when you found out that the number is not a prime:
let number = 6
var isItPrime: Bool = true
for i in 2 ..< number {
if number % i == 0 {
isItPrime = false
break // end the loop, as we know that the number is not a prime.
}
}
print(isItPrime)
When dealing with problems like this, don't be afraid to take out a piece of paper and manually see what is going on in your loop.
Your loop will go from i = 1 to number = 5 (because of the < operator.
With that in mind, we perform each iteration manually.
for i = 1, number = 6
6 mod 1 = 0, isItPrime = false
for i = 2, number = 6
6 mod 2 = 0, isItPrime = false
for i = 3, number = 6
6 mod 3 = 0, isItPrime = false
for i = 4, number = 6
6 mod 4 = 2, isItPrime = true
Last iteration of the loop, for i = 5, number = 6
6 mod 5 = 1, isItPrime = true
There we can see that the problem is that the last iteration will always have a module of 1, therefore resulting in in your else clause getting executed.
It is always returning true because your while loop isn't working the way you want. Currently, it loops until i is 1 less than number. During that final run through the loop, number % i == 0 is false, so your code sets isItPrime to true.
To fix this problem, try this code:
let number = 6
var i = 2
var isItPrime: Bool?
while (i < number || isItPrime == false) {
if number % i == 0 {
isItPrime = false
} else {
isItPrime = true
}
i += 1
}
print(isItPrime)
You may have noticed I set i to 2, because any number modulo (%) 1 is 0
I think it's worth pointing out, however, that:
You should probably make this a method
If you initially set isItPrime to true, you can dispense with the else part of your if-else statement
Hope this helps!
// MARK: - Function
func primeNo(_ num:Int,_ divisor :Int = 2){
if divisor == num{
print("Num is prime")
}else{
if num%divisor != 0{
primeNo(num, divisor + 1)
}else{
print("num is not prime")
}
}
}
// MARK: - Use
primeNo(6)
out Put
num is not prime
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
}