Chapter 6: Arrays

Swift Programming from Scratch

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Chapter 6: Arrays

Introduction

Often when you’re dealing with data you don’t just have a fixed amount of elements. Take for example a program where you compute the average of multiple grades in a class:

var grade1 = 4
var grade2 = 3

var average = Double(grade1 + grade2) / 2.0
print("Average grade: \(average)")

What if we wanted the program to also work when we have 3 grades?
We’d have to change our program to work with 3 grades.

var grade1 = 4
var grade2 = 3
var grade3 = 5

var average = Double(grade1 + grade2 + grade3) / 3.0
print("Average grade: \(average)")

After doing this it will no longer work with 2 grades. What if we wanted our program to work with any number of grades between 1 grade and 10 grades.

It’s not practical to write a separate program for each case. We would like to have something like a list of grades. This list would contain any number of grades. This is where arrays come in.

What is an array?

An array is an ordered collection that stores multiple values of the same type. That means that an array of Int can only store Int values. And you can only insert Int values in it.

Declaring Arrays

To declare an array you can use the square brackets syntax([Type]).

var arrayOfInts: [Int]

You can initialize an array with an array literal. An array literal is a list of values, separated by commas, surrounded by a pair of square brackets:

[value, value, ...]

var arrayOfInts: [Int] = [1, 2, 3]
var arrayOfStrings: [String] = ["We", "❤", "Swift"]

` Keep in mind that you can create empty arrays if you don’t write any values.

var emptyArray: [Int] = []

Getting values

To get all the values from an array you can use the for-in syntax. This is called iterating through an array.

var listOfNumbers = [1, 2, 3, 10, 100] // an array of numbers
var listOfNames = ["Andrei", "Silviu", "Claudiu"] // an array of strings

for number in listOfNumbers {
    print(number)
}
// 1
// 2
// 3
// 10
// 100

for name in listOfNames {
    print("Hello " + name + "!")
}
// Hello Andrei!
// Hello Silviu!
// Hello Claudiu!

To get the number of elements in an array you can use the count property.

var listOfNumbers = [1, 2, 3, 10, 100] // an array of numbers

print(listOfNumbers.count) // 5

You can access specific elements from an array using the subscript syntax. To do this pass the index of the value you want to retrieve within square brackets immediately after the name of the array. Also you can get a subsequence from the array if you pass a range instead of an index.
Element in an array are indexed from 0 to the number of elements minus one. So an array with 3 elements will have elements at index 0, 1 and 2.

var listOfNumbers = [1, 2, 3, 10, 100] // an array of numbers

listOfNumbers[0] // 1
listOfNumbers[1] // 2
listOfNumbers[2] // 3
listOfNumbers[3] // 10
listOfNumbers[4] // 100
//listOfNumbers[5]// this gives an error uncomment this line to see it

listOfNumbers[1...2] // [2, 3] this is a subsequence of the original array

Adding values

You can add elements to the end of an array using the append method.

// create a empty array of integers
var numbers: [Int] = []

for i in 1...5 {
    numbers.append(i)
    print(numbers)
    // [1]
    // [1, 2]
    // [1, 2, 3]
    // [1, 2, 3, 4]
    // [1, 2, 3, 4, 5]
}

print(numbers)
// [1, 2, 3, 4, 5]

To insert an item into the array at a specified index, call the array’s insert(at:) method.

var numbers: [Int] = [1, 2, 3]

numbers.insert(0, at: 0) // numbers will be [0, 1, 2, 3]
numbers.insert(9, at: 1) // numbers will be [0, 9, 1, 2, 3]

You can also append another array using the += operator.

var numbers: [Int] = [1, 2, 3]

numbers += [4, 5, 6] // numbers will be [1, 2, 3, 4, 5, 6]

// or just one value
numbers += [7] // numbers will be [1, 2, 3, 4, 5, 6, 7]

Removing Values

To remove an item from a specific index call the remove(at:) method.

var numbers: [Int] = [1, 2, 3]

numbers.remove(at: 0) // numbers will be [2, 3]

Changing values

To change a value use the assignment operator (=) after the subscript syntax.

var numbers: [Int] = [1, 2, 3]

numbers[0] = 7 // numbers will be [7, 2, 3]
numbers[1] = 5 // numbers will be [7, 5, 3]
numbers[2] = 4 // numbers will be [7, 5, 4]

Or you could replace a subsequence of values using range subscripting.

var numbers: [Int] = [1, 2, 3, 4, 5, 6]

numbers[2...4] = [0, 0] // numbers will now be [1, 2, 0, 0, 6].

Keep in mind that you don’t need to replace a sequence with another sequence with the same number of elements. In the example above numbers had 6 elements and after the replacement of the subsequence 2...4 ([3, 4, 5]) it had 5.

Type Inference

Thanks to Swift’s type inference, you don’t have to declare the type of an array if you initialize it with something other than an empty array literal([]).

// arrayOfNumbers will be of type [Int]
var arrayOfNumbers = [1, 2, 3] 

// arrayOfStrings will be of type [String]
var arrayOfStrings = ["We", "❤", "Swift"] 

// arrayOfBools will be of type [Bool]
var arrayOfBools = [true, false, true, true, false] 

// this is the proper way of declaring a empty array of Int - [Int]
var emptyArrayOfInts: [Int] = [] 

// this will infer into [Int] because the right hand side of the 
// assignment has a known type [Int]
var anotherEmptyArray = emptyArrayOfInts

Copy Behavior

Swift’s Array types are implemented as structures. This means that arrays are copied when they are assigned to a new constant or variable, or when they are passed to a function or method.

var numbers = [1, 2, 3]
var otherNumbers = numbers // this will create a copy of numbers

// this will append 4 to otherNumbers but not to numbers
otherNumbers.append(4) 

// numbers = [1, 2, 3]
// otherNumbers = [1, 2, 3, 4]

Mutability

If you create an array and assign it to a variable, the collection that is created will be mutable. This means that you can change (or mutate) the collection after it is created. Changes can be done by adding, removing, or changing items in the collection. Conversely, if you assign an array to a constant, that array is immutable, and its size and contents cannot be changed. In other words if you want to be able to change an array declare it using the varkeyword, and if you don’t want to be able to change it use the let keyword.

var numbers = [1, 2, 3, 4, 5, 6]

numbers.append(7) // [1, 2, 3, 4, 5, 6, 7]
numbers.remove(at: 0) // [2, 3, 4, 5, 6, 7]

let strings = ["We", "❤", "Swift"]

// the next lines will not compile!
strings.append("!") // this will give an error because strings is immutable
strings.remove(at: 0) // this will give a similar error

6.1 Max

Print the maximum value from listOfNumbers.

Example 1

Input:

var listOfNumbers = [1, 2, 3, 10, 100]

Output:

[collapse]

Example 2

Input:

var listOfNumbers = [10, 12, 33, 11, 1]

Output:

[collapse]

Hint

Assume that the first element of the array is also the largest.

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Solution

var listOfNumbers = [1, 2, 3, 10, 100]

var maxVal = listOfNumbers[0]

for number in listOfNumbers {
    if maxVal < number {
        maxVal = number
    }
}

print(maxVal)

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6.2 Odd

Print all the odd numbers from listOfNumbers.

Example 1

Input:

var listOfNumbers = [1, 2, 3, 10, 100]

Output:

1
3

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Example 2

Input:

var listOfNumbers = [10, 12, 33, 11, 1]

Output:

33
11
1

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Solution

var listOfNumbers = [1, 2, 3, 10, 100]

for number in listOfNumbers {
    if number % 2 != 0 {
        print(number)
    }
}

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6.3 Sum

Print the sum of all the numbers from listOfNumbers.

Example 1

Input:

var listOfNumbers = [1, 2, 3, 10, 100]

Output:

[collapse]

Example 2

Input:

var listOfNumbers = [10, 12, 33, 11, 1]

Output:

[collapse]

Hint

Store the sum in a variable. Keep increasing it.

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Solution

var listOfNumbers = [1, 2, 3, 10, 100]

var sum = 0

for number in listOfNumbers {
    sum += number
}

print(sum)

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6.4 Odd Index

Print all the numbers from listOfNumbers that are located at odd indexes.

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Example 1

Input:

var listOfNumbers = [1, 2, 3, 10, 100]

Output:

2
10

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Example 2

Input:

var listOfNumbers = [1, 10, 12, 33, 11, 1]

Output:

10
33
1

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Hint

Use a while loop and the index subscripts.

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Solution 1

var listOfNumbers = [1, 2, 3, 10, 100]

var i = 1

while i < listOfNumbers.count {
    print(listOfNumbers[i])
    i += 2
}
// 2
// 10

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Solution 2

var listOfNumbers = [1, 2, 3, 10, 100]

for var i = 1; i < listOfNumbers.count; i += 2 {
    print(listOfNumbers[i])
}

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6.5 Going back

Print the numbers from listOfNumbers in reverse order on separate lines.

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Example 1

Input:

var listOfNumbers = [1, 2, 3, 10, 100]

Output:

[collapse]

Example 2

Input:

var listOfNumbers = [12, 33, 11, 1]

Output:

[collapse]

Hint

Use a loop that counts from the last index down to the first one.

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Solution 1

var listOfNumbers = [1, 2, 3, 10, 100, 2]

var i = listOfNumbers.count - 1

while i >= 0 {
    print(listOfNumbers[i])
    i -= 1
}

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Solution 2

var listOfNumbers = [1, 2, 3, 10, 100, 2]

for i in 1...listOfNumbers.count {
    print(listOfNumbers[listOfNumbers.count - i])
}

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Solution 3

var listOfNumbers = [1, 2, 3, 10, 100, 2]

for var i = listOfNumbers.count - 1; i >= 0; --i {
    print(listOfNumbers[i])
}

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6.6 Reverse

Reverse the order of the elements in listOfNumbers without creating any additional arrays.

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Example 1

Input:

var listOfNumbers = [1, 2, 3]

Expected value:

listOfNumbers = [3, 2, 1]

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Example 2

Input:

var listOfNumbers = [12, 33, 11, 1]

Expected value:

listOfNumbers = [1, 11, 33, 12]

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Hint 1

Use 2 indices.

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Hint 2

At each step advance with the indices one step closer to the middle of thearray.

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Solution

var listOfNumbers = [1, 2, 3, 10, 100]

var firstIndex = 0
var lastIndex = listOfNumbers.count - 1

while firstIndex < lastIndex {
    // swap
    var tmp = listOfNumbers[firstIndex]
    listOfNumbers[firstIndex] = listOfNumbers[lastIndex]
    listOfNumbers[lastIndex] = tmp

    // go to next pair
    firstIndex += 1
    lastIndex -= 1
}

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6.7 Sorting

Sort the values in listOfNumbers in descending order.

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Solution 1: Insertion Sort

var listOfNumbers = [1, 2, 3, 10, 100]

var nElements = listOfNumbers.count

for fixedIndex in 0..<nElements {
    for i in fixedIndex+1..<nElements {
        if listOfNumbers[fixedIndex] < listOfNumbers[i] {
            var tmp = listOfNumbers[fixedIndex]
            listOfNumbers[fixedIndex] = listOfNumbers[i]
            listOfNumbers[i] = tmp
        }
    }
}

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Solution 2: Bubble Sort

var listOfNumbers = [1, 2, 3, 10, 100]

var nElements = listOfNumbers.count

var didSwap = true

while didSwap {
    didSwap = false

    for i in 0..<nElements - 1 {
        if listOfNumbers[i] < listOfNumbers[i+1] {
            var tmp = listOfNumbers[i]
            listOfNumbers[i] = listOfNumbers[i+1]
            listOfNumbers[i+1] = tmp
            didSwap = true
        }
    }
}

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Solution 3: The Swift way

In real life people don’t implement sorting anymore, they just call it.

array.sort(by: <) // will sort the array in ascending order
array.sort(by: >) // will sort the array in descending order

// you can also use the sorted method to create a sorted copy
let sortedArray = array.sorted(by: <)

var listOfNumbers = [3, 2, 100, 10, 1]

listOfNumbers.sort(by: <)

print(listOfNumbers)
// [1, 2, 3, 10, 100]

listOfNumbers.sort(by: >)

print(listOfNumbers)
// [100, 10, 3, 2, 1]

So to solve the problem we would write:

var listOfNumbers = [3, 2, 100, 10, 1]

listOfNumbers.sort(by: <)

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6.8 Search

Find out if x appears in listOfNumbers. Print yes if true and no otherwise.

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Example 1

Input:

var listOfNumbers = [1, 2, 3, 10, 100]
var x = 3

Output:

yes

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Example 2

Input:

var listOfNumbers = [1, 2, 3, 10, 100]
var x = 5

Output:

no

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Hint

First assume that the element does not appear in the array. Store that state in a boolean variable.

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Solution

var listOfNumbers = [1, 2, 3, 10, 100]

var x = 10

var xAppears = false

for number in listOfNumbers {
    if number == x {
        xAppears = true
    }
}

if xAppears {
    print("yes")
} else {
    print("no")
}

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6.9 Intersection

Print all the elements from otherNumbers that appear in listOfNumbers. Don’t print anything if listOfNumbers andotherNumbers have no common elements.

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Example 1

Input:

var listOfNumbers = [1, 2, 3, 10, 100]
var otherNumbers = [1, 2, 3, 4, 5, 6]

Output:

1
2
3

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Example 2

Input:

var listOfNumbers = [1, 2, 3, 10, 100]
var otherNumbers = [5, 2, 3, 10, 13]

Output:

2
3
10

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Example 3

Input:

var listOfNumbers = [1, 2, 3, 10, 100]
var otherNumbers = [5, 6]

Output:

[collapse]

Hint

Use an approach similar to the Search problem for each element fromotherNumbers.

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Solution

var listOfNumbers = [1, 2, 3, 10, 100]

var otherNumbers = [1, 2, 3, 4, 5, 6]

for otherNumber in otherNumbers {
    for number in listOfNumbers {
        if number == otherNumber {
            print(number)
            break
        }
    }
}

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6.10 Divisors

Print all the numbers from listOfNumbers that are divisible by at least one number from divisors.

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Example 1

Input:

var listOfNumbers = [1, 2, 3, 10, 100]
var divisors = [2, 5]

Output:

2
10
100

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Example 2

Input:

var listOfNumbers = [1, 2, 3, 10, 100]
var divisors = [5, 3]

Output:

3
10
100

[collapse]

Example 3

Input:

var listOfNumbers = [1, 2, 3, 10, 100]
var divisors = [7, 9, 13]

Output:

[collapse]

Hint

Try solving the problem for the case when listOfNumbers contains a single element.

[collapse]

Solution

var listOfNumbers = [1, 2, 3, 10, 100]
var divisors = [7, 5]

for number in listOfNumbers {
    for divisor in divisors {
        if number % divisor == 0 {
            print(number)
            break
        }
    }
}

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6.11 Greatest divisor of all

Find and print the greatest common divisor of all the numbers in numbers. A common divisor of a list of numbers is a number that divides all of them.

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Example 1

Input:

var numbers = [12, 36, 720, 18]

Output:

[collapse]

Example 2

Input:

var numbers = [3, 12, 36, 18, 7]

Output:

[collapse]

Hint

Use an approach similar to the case with only 2 numbers.

[collapse]

Solution

var numbers = [12, 36, 720, 18]

// find the minimum value in numbers
var maxDiv = numbers[0]

for number in numbers {
    if number < maxDiv {
        maxDiv = number
    }
}

var gcd = 1

// find the biggest number that divides all the numbers
for divisor in 1...maxDiv {
    // we assume that divisor divides all numbers
    var dividesAll = true
    for number in numbers {
        // if we find one that does not divide by divisor
        if number % divisor != 0 {
            // we remeber and stop searching
            dividesAll = false
            break
        }
    }

    // if divisor divides all numbers then it's the biggest one so far
    if dividesAll {
        gcd = divisor
    }
}

print(gcd)

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6.12 Fibonacci

Generate the first N numbers in the fibonacci sequence. Store them in an array named fibonacci and print them one on each line.

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Example 1

Input:

var N = 6

Expected value:

fibonacci = [1, 1, 2, 3, 5, 8]

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Example 2

Input:

var N = 9

Expected value:

fibonacci = [1, 1, 2, 3, 5, 8, 13, 21, 34]

[collapse]

Approach 1

Use append to add the next numbers

var N = 30
var fibonacci = [1, 1]

// your code here

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Solution 1

var N = 30

var fibonacci = [1, 1]

for i in 2...N - 1 {
    fibonacci.append(fibonacci[i-1] + fibonacci[i-2])
}

for number in fibonacci {
    print(number)
}

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Approach 2

Create an array with ones and compute all the numbers

let N = 30
var fib = [Int](repeating: 1, count: N)

// your code here

[collapse]

Solution 2

let N = 30
var fib = [Int](repeating: 1, count: N)

for i in 2..<N {
    fib[i] = fib[i-1] + fib[i-2]
}

for number in fib {
    print(number)
}

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6.13 Divisors

Given a number find and store all it’s divisors in an array called divisors, then print the divisors in ascending order on separate lines.

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Example 1

Input:

var number = 6

Expected value:

divisors = [1, 2, 3, 6]

Output:

[collapse]

Example 2

Input:

var number = 30

Expected value:

divisors = [1, 2, 3, 5, 6, 10, 15, 30]

Output:

[collapse]

Hint

Any value between 1 and number can be a divisor of number.

[collapse]

Solution

var number = 60
var divisors: [Int] = []

for divisor in 1...number {
    if number % divisor == 0 {
        divisors.append(divisor)
    }
}

for divisor in divisors {
    print(divisor)
}

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6.14 Digits

Find and store the digits of number from left to right in an array digits, and then print the digits on separate lines.

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Example 1

Input:

var number = 12345

Expected value:

digits = [1, 2, 3, 4, 5]

Output:

[collapse]

Example 2

Input:

var number = 232121

Expected value:

digits = [2, 3, 2, 1, 2, 1]

Output:

[collapse]

Hint

Store the digits from right to left if you find it easier. The digits from left to right are the reversed array.

[collapse]

Solution

var number = 12345
var digits: [Int] = []

while number > 0 {
    var digit = number % 10

    digits = [digit] + digits

    number /= 10 // 12345 -> 1234 -> 123 -> 12 -> 1
}

for digit in digits {
    print(digit)
}

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6.15 Unique

Create a list unique with all the unique numbers from listOfNumbers, and then print the numbers on separate lines.

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Example 1

Input:

var listOfNumbers = [1, 2, 3, 1, 2, 10, 100]

Expected value:

unique = [1, 2, 3, 10, 100]

Output:

[collapse]

Example 2

Input:

var listOfNumbers = [2, 3, 1, 1, 1, 2, 2, 2, 10]

Expected value:

unique = [2, 3, 1, 10]

Output:

[collapse]

Solution

var listOfNumbers = [1, 2, 3, 1, 2, 10, 100]

var unique: [Int] = []

for number in listOfNumbers {
    var numberIsNew = true

    for otherNumber in unique {
        if number == otherNumber {
            numberIsNew = false
            break
        }
    }

    if numberIsNew {
        unique.append(number)
    }
}

for number in unique {
    print(number)
}

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Swift Programming from Scratch

33 comments for “Chapter 6: Arrays”

HansHorstGlatt
December 19, 2014 at 6:03 pm
Hi. It is wellknown, that “Andrei” and “Silviu” are part of the “We ❤︎ Swift” team! But who is “Claudiu”?
Reply
- Andrei Puni
  December 19, 2014 at 7:51 pm
  Claudiu is our intern :). We hired him to learn programming in Swift using the exercises from Swift programming from Scratch. We know programming for so long that it’s hard to imagine how would it be not to. He helped us understand a lot about what a non-programmer mind thinks when seeing some code or learning a new concept.
  Reply
HansHorstGlatt
December 21, 2014 at 11:14 am
There is obviously an “printing” error in the solution to problem 5.6. Instead of ++lastIndex one should read –last index.
Reply
- HansHorstGlatt
  December 21, 2014 at 11:15 am
  sorry: –lastIndex
  Reply
  - HansHorstGlatt
    December 21, 2014 at 11:19 am
    funny mistake! of course minus minus lastIndex (lastIndex = lastIndex – 1)
    Reply
IK
March 7, 2015 at 1:58 am
6.6 Reverse can be accomplished with much less code:
for x in listOfNumbers {
listOfNumbers.append(listOfNumbers[0])
listOfNumbers.removeAtIndex(0)
}
Reply
- IK
  March 7, 2015 at 2:01 am
  Edit sorry I misread the question, please delete these two comments.
  Reply
  - SimNico
    March 16, 2015 at 9:15 pm
    You inspired me to try something similar… and there you go
    for i in 0..<listOfNumbers.count {
    listOfNumbers.insert(listOfNumbers[listOfNumbers.count – 1], atIndex: i)
    listOfNumbers.removeLast()
    }
    Reply
    - SimNico
      March 16, 2015 at 9:20 pm
      You can even replace « listOfNumbers[listOfNumbers.count – 1] » by « listOfNumbers.last! » for more clarity.
      Reply
- Bhavik patel
  March 7, 2017 at 1:25 am
  actually it can be done with a function listOfNumbers.reverse()
  Reply
IK
March 7, 2015 at 2:18 am
6.6 Alternate
var theCount = listOfNumbers.count
var index = 0
for x in listOfNumbers {
listOfNumbers.insert(listOfNumbers[theCount – 1], atIndex: index)
listOfNumbers.removeLast()
index += 1
}
// result is [100,10,3,2,1]
Reply
anonimousy
March 8, 2015 at 8:30 am
how do you access the last element of a string array?
Reply
- Andrei Puni
  March 8, 2015 at 5:07 pm
  var strings: [String] = [“Hello”, “there”]
  let lastIndex = strings.count – 1
  strings[lastIndex] // there
  // or
  strings.last // Optional(“there”)
  Reply
Andy Dent
April 29, 2015 at 7:05 pm
6.8 Search
Using reduce with a flag instead of a loop
var listOfNumbers = [1, 2, 3, 10, 100]
var x = 10
let xAppears = listOfNumbers.reduce(false) {flag, num in flag || num==x}
println(xAppears ? “yes” : “no”)
Reply
Pearson
July 9, 2015 at 6:24 pm
Hello,
How do I iterate through an array and convert each value to a string?
Thank you
Reply
- Silviu Pop
  July 14, 2015 at 10:41 am
  Hi, you can use string interpolation to create a string out of any element of an array.
  for value in array {
  let stringValue = “\(value)”
  }
  Reply
waleed
July 21, 2015 at 5:02 am
Why do we need the break in Q6.9 and what does it do exactly?
Reply
- Andrei Puni
  July 23, 2015 at 8:15 pm
  break stops a loop.
  for i in 1…3 {
  print(i)
  break
  }
  print(“go!”)
  will print 1 go! not 1 2 3 go!
  Reply
plan9fan
August 27, 2015 at 5:27 pm
I need to check an array of random integers (between 1 and 9) and see if any combination of them will add up to 10. Is there an equation for that calculation?
for example, if the array was [8, 7, 6] it would return false
if the array was [5, 6, 4, 2, 9] it would return true.
The amount of Ints the array would vary.
Reply
- Andrei Puni
  August 27, 2015 at 5:29 pm
  You are looking for the Knapsack Algorithm – https://en.wikipedia.org/wiki/Knapsack_problem
  Reply
sonat
January 15, 2016 at 2:05 pm
in 6.15 -> why cant i run this code properly?
for number in 0..<unique.count {
for i in 1..<unique.count {
if unique[number] == unique[i] { // xcode alerts me with the bad instructions in this line?
unique.removeAtIndex(i)
}
}
}
Reply
- Andrei Puni
  February 3, 2016 at 5:12 pm
  because you mutated the array while iterating over it. The first for loop will go from 0 to the nr of elements from unique-1. That will crash your code after at least one number is removed because the last.
  Looks like you think like a sculptor not a painter – you remove instead of adding. Here is a version of you solution that works:
  var listOfNumbers = [1, 2, 3, 1, 2, 10, 100]
  var unique: [Int] = listOfNumbers
  var foundTwoOfTheSame = true
  while foundTwoOfTheSame { foundTwoOfTheSame = false for number in 0..
  It stops the for loops every time they remove a number - it does that until all the numbers are unique.
  Reply
godhar
June 8, 2016 at 12:51 pm
Hi guys!
Great site…..what does this mean?
if number % 2 != 0 {
Cheers!
Reply
- Andrei Puni
  June 9, 2016 at 8:44 am
  You can find the answer in the first chapter under basic operators weheartswift.com/variables-constants-basic-operations/
  Reply
- taylor
  December 1, 2016 at 8:04 pm
  if number is NOT divisible by 2
  Reply
Duncan Rowland
November 20, 2016 at 6:56 pm
Recommend sets for interception:
var listOfNumbers: Set = [1, 2, 3, 10, 100]
var otherNumbers: Set = [1, 2, 3, 4, 5, 6]
// your code here
for i in listOfNumbers
.intersection(otherNumbers)
.sorted() {
print(i)
}
Reply
- Duncan Rowland
  November 20, 2016 at 6:57 pm
  Intersection (sorry, watching football!)
  Reply
taylor
December 1, 2016 at 8:02 pm
I’m getting a runtime error on the even test cases on problem 6.6 with this code:
var listOfNumbers = [2, 5, 7, 23]
// your code here
var j = listOfNumbers.count – 1
var i = 0
var tmp = 0
while i != j{
tmp = listOfNumbers[i];
listOfNumbers[i] = listOfNumbers[j]
listOfNumbers[j] = tmp
i += 1
j -= 1;
}
***End of Code***
The error happens on the line with “tmp = listOfNumbers[i]”
Any ideas on how to fix this? I implemented this algorithm in C using a for loop with relative ease. Wish swift 3 didn’t get rid of the C-style for-loop RIP
Reply
- taylor
  December 1, 2016 at 10:41 pm
  By even test cases I mean that the number of elements in listOfNumbers is even
  (i.e: listOfNumbers.count % 2 == 0 is True)
  Reply
  - taylor
    December 1, 2016 at 10:52 pm
    I got it. Here’s what i did instead:
    var j = listOfNumbers.count – 1
    var sizeOfArray = listOfNumbers.count
    for i in 0..<sizeOfArray/2 {
    var tmp = listOfNumbers[i]
    listOfNumbers[i] = listOfNumbers[j]
    listOfNumbers[j] = tmp
    j -= 1
    }
    **End**
    IMHO, I think this is a less elegant, less obvious solution to what could have been accomplished with a C-style loop, but maybe that's because they're teaching me C in my CS class right now lol.
    Do you know why they got rid of C loops in swift 3? If you know, that would be a good blog post, I believe.
    Reply
taylor
December 7, 2016 at 11:53 pm
Here’s my solution for 6.11, but I like the way you use booleans. Much cleaner code, imo.
var numbers = [12, 36, 720, 18]
// your code here
var gcd = 1
var min = numbers[0]
for i in numbers {
if i 0 {
if numbers[numbers.count – t] % i == 0 {
count += 1
}
if count == numbers.count {
gcd = i
}
t-=1
}}
print(gcd)
Reply
- taylor
  December 7, 2016 at 11:55 pm
  oops I think I didn’t copy/paste it all
  var gcd = 1
  var min = numbers[0]
  for i in numbers {
  if i 0 {
  if numbers[numbers.count – t] % i == 0 {
  count += 1
  }
  if count == numbers.count {
  gcd = i
  }
  t-=1
  }}
  print(gcd)
  Reply
  - taylor
    December 7, 2016 at 11:56 pm
    nvm it might be the comment box. last try:
    var gcd = 1
    var min = numbers[0]
    for i in numbers {
    if i 0 {
    if numbers[numbers.count – t] % i == 0 {
    count += 1 }
    if count == numbers.count {
    gcd = i }
    t-=1
    }}
    print(gcd)
    Reply