After hours of Googling, I'm still at a standstill. I would appreciate if someone would point out the error in my formula or coding choice. Please keep in mind I'm new to Swift. I'm not used to non C-style for loops.
if textField.text != "" {
input = Double(textField.text!)! // parse input
// return if number less than 2 entered
if input < 2 {
resultLabel.text = "Enter a number greater than or equal to 2."
return;
}
// get square root of input and parse to int
inputSquared = Int(sqrt(input));
// loop from 2 to input iterating by 1
for i in stride(from: 2, through: input, by: 1) {
if inputSquared % Int(i) == 0 {
resultLabel.text = "\(Int(input)) is not a prime number."
}
else {
resultLabel.text = "\(Int(input)) is a prime number!"
}
}
}
I didn't know the formula on how to find a prime number. After looking up multiple formulas I have sorta settled on this one. Every result is a prime number, however. So my if condition is wrong. I just don't know how to fix it.
Check my algorithm.It works.But I'm not sure this is an effective algorithm for prime number
var input:Int = 30
var isPrime:Bool = true
if(input == 2){
print("Input value 2 is prim number")
}
else if(input < 2){
print("Input value must greater than 2")
}
else{
for i in 2...input-1{
if((input%i) == 0){
isPrime = false
break;
}
}
if(isPrime){
print("Your Input Value \(input) is Prime!")
}
}
A couple of solutions that work have been posted, but none of them explain why yours doesn't. Some of the comments get close, however.
Your basic problem is that you take the square root of input, then iterate from 2 to the input checking if the integer part of the square root is divisible by i. You got that the wrong way round. You need to iterate from 2 to the square root and check that the input is divisible by i. If it is, you stop because input is not prime. If you get to the end without finding a divisor, you have a prime.
try this code in playground you will get this better idea and try to use playground when you try the swift as you are not familiar with swift playground is best.
let input = 13 // add your code that take value from textfield
var prime = 1
// throw error less than 2 entered
if input < 2 {
assertionFailure("number should be 2 or greater")
}
// loop from 2 to input iterating by 1
for i in stride(from: 2, to: input, by: 1) {
if input % i == 0{
prime = 0
}
}
if prime == 1 {
print("\(input) number is prime")
} else {
print("\(input) number is not prime")
}
Related
I have been working on a hacker rank problem where I have to print a number which is a factorial of 25. Here is the code I used.
func extraLongFactorials(n: Int) -> Void {
let factorialNumber = factorial(number: n)
var arrayForStorage: [Int] = []
var loop = factorialNumber
while (loop > 0) {
let digit = loop.truncatingRemainder(dividingBy: 10)
arrayForStorage.append(Int(digit))
loop /= 10
}
arrayForStorage = arrayForStorage.reversed()
var returnString = ""
for element in arrayForStorage {
returnString = "\(returnString)\(element)"
}
print(returnString)
}
func factorial(number: Int) -> Double {
if number == 0 || number == 1 {
return 1
} else if number == 2 {
return 2
} else {
return Double(number) * factorial(number: number - 1)
}
}
But when I try to print the factorial number it just prints 0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000015511210043330982408266888 when it should print
15511210043330985984000000.
I think for a Double number truncatingRemainder(dividingBy: 10) method is not giving me the exact number of the remainder. Because when I tried to print the truncatingRemainder of 15511210043330985984000000 it is giving me as 8. Here is the code.
let number: Double = 15511210043330985984000000
print(number.truncatingRemainder(dividingBy: 10))
So finally I didn't find any solution for the problem of how to split the large number and add it into an array. Looking forward for the solution.
Type Double stores a number as a mantissa and an exponent. The mantissa represents the significant figures of the number, and the exponent represents the magnitude of the number. A Double can only represent about 16 significant figures, and your number has 26 digits, so you can't accurately store 15511210043330985984000000 in a Double.
let number1: Double = 15511210043330985984000000
let number2: Double = 15511210043330985984012345
if number1 == number2 {
print("they are equal")
}
they are equal
You will need another approach to find large factorials like the one shown in this answer.
I am working on Euler problem 5 which is:
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
I am having trouble with my nested loops. I have a for loop within in a while loop. My logic is I
check a number (currentNumber) against 1-20 (i), if currentNumber is not a multiple of i (checked using modular arithmatic)
then it breaks out of that loop and trys the next largest number.
My issue is I cannot figure out how to jump out of only my inner loop and not my outer loop. Here is my code:
class Five {
init() {
var currentNumber = 1
while true {
for i in 1...20 {
if currentNumber % i != 0 {
currentNumber += 1
continue
}
}
break
}
print("the smallest positive number that is evenly divisible " +
"by all of the numbers from 1 to 20 is \(currentNumber)")
}
}
You already got a good and correct answer. Just as an add-on, for the
sake of completeness:
An alternative to labeled continue statements is to move the inner loop into a separate function from which you can “early return”:
func isDivisibleBy1To20(_ number: Int) -> Bool {
for j in 2...20 {
if number % j != 0 {
return false
}
}
return true
}
var currentNumber = 1
while !isDivisibleBy1To20(currentNumber) {
currentNumber += 1
}
print("solution:", currentNumber)
Using functional methods this can be simplified to
func isDivisibleBy1To20(_ number: Int) -> Bool {
return !(2...20).contains(where: { number % $0 != 0 })
}
let solution = (1...).first(where: isDivisibleBy1To20)!
print("solution:", solution)
(Remark: There are other, much faster methods to solve this problem.)
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'm attempting to submit the HackerRank Day 6 Challenge for 30 Days of Code.
I'm able to complete the task without issue in an Xcode Playground, however HackerRank's site says there is no output from my method. I encountered an issue yesterday due to browser flakiness, but cleaning caches, switching from Safari to Chrome, etc. don't seem to resolve the issue I'm encountering here. I think my problem lies in inputString.
Task
Given a string, S, of length N that is indexed from 0 to N-1, print its even-indexed and odd-indexed characters as 2 space-separated strings on a single line (see the Sample below for more detail).
Input Format
The first line contains an integer, (the number of test cases).
Each line of the subsequent lines contain a String, .
Constraints
1 <= T <= 10
2 <= length of S < 10,000
Output Format
For each String (where 0 <= j <= T-1), print S's even-indexed characters, followed by a space, followed by S's odd-indexed characters.
This is the code I'm submitting:
import Foundation
let inputString = readLine()!
func tweakString(string: String) {
// split string into an array of lines based on char set
var lineArray = string.components(separatedBy: .newlines)
// extract the number of test cases
let testCases = Int(lineArray[0])
// remove the first line containing the number of unit tests
lineArray.remove(at: 0)
/*
Satisfy constraints specified in the task
*/
guard lineArray.count >= 1 && lineArray.count <= 10 && testCases == lineArray.count else { return }
for line in lineArray {
switch line.characters.count {
// to match constraint specified in the task
case 2...10000:
let characterArray = Array(line.characters)
let evenCharacters = characterArray.enumerated().filter({$0.0 % 2 == 0}).map({$0.1})
let oddCharacters = characterArray.enumerated().filter({$0.0 % 2 == 1}).map({$0.1})
print(String(evenCharacters) + " " + String(oddCharacters))
default:
break
}
}
}
tweakString(string: inputString)
I think my issue lies the inputString. I'm taking it "as-is" and formatting it within my method. I've found solutions for Day 6, but I can't seem to find any current ones in Swift.
Thank you for reading. I welcome thoughts on how to get this thing to pass.
readLine() reads a single line from standard input, which
means that your inputString contains only the first line from
the input data. You have to call readLine() in a loop to get
the remaining input data.
So your program could look like this:
func tweakString(string: String) -> String {
// For a single input string, compute the output string according to the challenge rules ...
return result
}
let N = Int(readLine()!)! // Number of test cases
// For each test case:
for _ in 1...N {
let input = readLine()!
let output = tweakString(string: input)
print(output)
}
(The forced unwraps are acceptable here because the format of
the input data is documented in the challenge description.)
Hi Adrian you should call readLine()! every row . Here an example answer for that challenge;
import Foundation
func letsReview(str:String){
var evenCharacters = ""
var oddCharacters = ""
var index = 0
for char in str.characters{
if index % 2 == 0 {
evenCharacters += String(char)
}
else{
oddCharacters += String(char)
}
index += 1
}
print (evenCharacters + " " + oddCharacters)
}
let rowCount = Int(readLine()!)!
for _ in 0..<rowCount {
letsReview(str:String(readLine()!)!)
}
any of you knows how can I check if the division remainder is integer or zero?
if ( integer ( 3/2))
You should use the modulo operator like this
// a,b are ints
if ( a % b == 0) {
// remainder 0
} else
{
// b does not divide a evenly
}
It sounds like what you are looking for is the modulo operator %, which will give you the remainder of an operation.
3 % 2 // yields 1
3 % 1 // yields 0
3 % 4 // yields 1
However, if you want to actually perform the division first, you may need something a bit more complex, such as the following:
//Perform the division, then take the remainder modulo 1, which will
//yield any decimal values, which then you can compare to 0 to determine if it is
//an integer
if((a / b) % 1 > 0))
{
//All non-integer values go here
}
else
{
//All integer values go here
}
Walkthrough
(3 / 2) // yields 1.5
1.5 % 1 // yields 0.5
0.5 > 0 // true
swift 3:
if a.truncatingRemainder(dividingBy: b) == 0 {
//All integer values go here
}else{
//All non-integer values go here
}
You can use the below code to know which type of instance it is.
var val = 3/2
var integerType = Mirror(reflecting: val)
if integerType.subjectType == Int.self {
print("Yes, the value is an integer")
}else{
print("No, the value is not an integer")
}
let me know if the above was useful.
Swift 5
if numberOne.isMultiple(of: numberTwo) { ... }
Swift 4 or less
if numberOne % numberTwo == 0 { ... }
Swift 2.0
print(Int(Float(9) % Float(4))) // result 1