'Find Duplicate Elements In Array Using Swift
How to find Duplicate Elements in Array? I have array of phone numbers so in the phone numbers i should start searching from the right side to the left side and find similar 6 integers. then i should print them out.
Solution 1:[1]
To find duplicates, you could build cross reference by phone number, then filter that down to duplicates only. For example, consider:
let contacts = [
Contact(name: "Rob", phone: "555-1111"),
Contact(name: "Richard", phone: "555-2222"),
Contact(name: "Rachel", phone: "555-1111"),
Contact(name: "Loren", phone: "555-2222"),
Contact(name: "Mary", phone: "555-3333"),
Contact(name: "Susie", phone: "555-2222")
]
In Swift 4, you can build the cross reference dictionary with:
let crossReference = Dictionary(grouping: contacts, by: { $0.phone })
Or, in Swift 5.2 (thanks to SE-0249), you can do:
let crossReference = Dictionary(grouping: contacts, by: \.phone)
Or
let crossReference: [String: [Contact]] = contacts.reduce(into: [:]) {
$0[$1.phone, default: []].append($1)
}
Then, to find the duplicates:
let duplicates = crossReference
.filter { $1.count > 1 } // filter down to only those with multiple contacts
.sorted { $0.1.count > $1.1.count } // if you want, sort in descending order by number of duplicates
Clearly use whatever model types make sense for you, but the above uses the following Contact type:
struct Contact {
let name: String
let phone: String
}
There are many, many ways to implement this, so I wouldn't focus on the implementation details of the above, but rather focus on the concept: Build cross reference original array by some key (e.g. phone number) and then filter results down to just those keys with duplicate values.
It sounds like you want to flatten this structure that reflects the duplicates, into a single array of contacts (I'm not sure why you'd want to do that, as you lose the structure identifying which are duplicates of each other), but if you want to do that, you can flatMap it:
let flattenedDuplicates = crossReference
.filter { $1.count > 1 } // filter down to only those with multiple contacts
.flatMap { $0.1 } // flatten it down to just array of contacts that are duplicates of something else
For Swift 2 or 3 renditions, see previous renditions of this answer.
Solution 2:[2]
Feeling ~clever~. Given an array of Ints
let x = [1, 1, 2, 3, 4, 5, 5]
let duplicates = Array(Set(x.filter({ (i: Int) in x.filter({ $0 == i }).count > 1})))
// [1, 5]
Please note, this is horrendously inefficient for everyone involved, including the compiler, and you.
I'm just showing off.
Edit: lol someone downvoted this, which leads me to reiterate, just in case: please DO NOT USE THIS in production or anywhere else.
Solution 3:[3]
Swift 4+
2 line, fast solution:
var numbers = [1,2,3,4,5,6,6,6,7,8,8]
let dups = Dictionary(grouping: numbers, by: {$0}).filter { $1.count > 1 }.keys
//Results: [6, 8]
Solution 4:[4]
Entirely derived from Rob's very neat answer. I've made it in to an Array extension and given names to the intermediate steps for clarity:
extension Array where Element: Hashable {
func duplicates() -> Array {
let groups = Dictionary(grouping: self, by: {$0})
let duplicateGroups = groups.filter {$1.count > 1}
let duplicates = Array(duplicateGroups.keys)
return duplicates
}
}
[1, 2, 2, 3, 1].duplicates() -> [1, 2]
Solution 5:[5]
You could implement it using "Merge sort", but you need to make one modification, during the merge step you should ignore the duplicates.
The easiest way to find duplicate elements is if the phone number is just a 6-digit number and has type Int, you could sort the array of phone numbers and than filter that to find duplicates.
var phoneNumbers = [123456, 234567, 345678, 123456, 456789, 135790, 456789, 142638]
func findDuplicates(sortedArray array: [Int]) -> [Int]
{
var duplicates: [Int] = []
var prevItem: Int = 0
var addedItem: Int = 0
for item in array
{
if(prevItem == item && addedItem != item)
{
duplicates.append(item)
addedItem = item
}
prevItem = item
}
return duplicates
}
func sortPhoneNumbers(phoneNumbers: [Int]) -> [Int]
{
return phoneNumbers.sorted({ return $0<$1 })
}
sortPhoneNumbers(phoneNumbers)
findDuplicates(sortPhoneNumbers(phoneNumbers))
In addition, you could implement the findDuplicates method in different ways:
Using Set (Swift 1.2+):
func findDuplicates(array: [Int]) -> [Int]
{
var duplicates = Set<Int>()
var prevItem = 0
for item in array
{
if(prevItem == item)
{
duplicates.insert(item)
}
prevItem = item
}
return Array(duplicates)
}
And so on.
Solution 6:[6]
To filter an array based on properties, you can use this method:
extension Array {
func filterDuplicates(@noescape includeElement: (lhs:Element, rhs:Element) -> Bool) -> [Element]{
var results = [Element]()
forEach { (element) in
let existingElements = results.filter {
return includeElement(lhs: element, rhs: $0)
}
if existingElements.count == 0 {
results.append(element)
}
}
return results
}
}
Which you can call as followed, based on the contacts example of Rob:
let filteredContacts = myContacts.filterDuplicates { $0.name == $1.name && $0.phone == $1.phone }
Solution 7:[7]
I've found a way by using reduce, here is the code(Swift 4):
let testNumbers = [1,1,2,3,4,5,2]
let nondupicate = testNumbers.reduce(into: [Int]()) {
if !$0.contains($1) {
$0.append($1)
} else {
print("Found duplicate: \($1)")
}
}
As a side effect, it returns an array with no duplicated elements.
You can easily modify it for counting duplicated elements numbers, checking string arrays etc.
Solution 8:[8]
Same as in @tikhop's answer, but as Array extension (Swift 3):
extension Array where Element: Comparable & Hashable {
public var duplicates: [Element] {
let sortedElements = sorted { $0 < $1 }
var duplicatedElements = Set<Element>()
var previousElement: Element?
for element in sortedElements {
if previousElement == element {
duplicatedElements.insert(element)
}
previousElement = element
}
return Array(duplicatedElements)
}
}
Solution 9:[9]
Antoine's solution in Swift 3+ syntax
extension Array {
func filterDuplicates(includeElement: @escaping (_ lhs: Element, _ rhs: Element) -> Bool) -> [Element] {
var results = [Element]()
forEach { (element) in
let existingElements = results.filter {
return includeElement(element, $0)
}
if existingElements.count == 0 {
results.append(element)
}
}
return results
}
}
Solution 10:[10]
let inputArray = [9820213496, 9546533545, 9820213496, 995543567]
var outputArray = [Int]()
for element in inputArray{
if outputArray.contains(element){
print("\(element) is Duplicate")
}else{
outputArray.append(element)
}
}
print(outputArray) // print Array without duplication
Solution 11:[11]
A very simple answer which preserves all duplicates
let originalNums = [5, 3, 2, 3 , 7 , 5,3]
var nums = Array(originalNums)
let numSet = Set(nums)
for num in numSet {
if let index = nums.index(of: num) {
nums.remove(at: index)
}
}
output
[3, 5, 3]
Solution 12:[12]
I also had a similar problem and have overcome in the following way. (Xcode 8.3.2)
let a = [123456, 234567, 345678, 123456, 456789, 135790, 456789, 142638]
var b = a // copy-on-write so that "a" won't be modified
while let c = b.popLast() {
b.forEach() {
if $0 == c {
Swift.print("Duplication: \(c)")
}
}
}
// Duplication: 456789
// Duplication: 123456
The point is that the number of comparison. It would be smaller than others.
Assume that the number of items in the array is N. In each loop, the number will be decrementing by one. So, the total number will be (N-1) + (N-2) + (N-3) + ... + 2 + 1 = N * (N-1) / 2 When N = 10, that will be 9 + 8 + ... = 45
In contrast, that of some algorithms might be N * N. When N = 10 that will be 100.
In spite of that, taking into account of the cost of deep-copy or shallow-copy, I agree that @Patrick Perini's brilliant way would be better than this in some situations even the number of that would be N * N.
EDIT:
Alternative way with IteratorProtocol
let a = [123456, 234567, 345678, 123456, 456789, 135790, 456789, 142638]
var i = a.makeIterator()
while let c = i.next() {
var j = i
while let d = j.next() {
if c == d {
Swift.print("Duplication: \(c)")
}
}
}
// Duplication: 123456
// Duplication: 456789
That looks more complex, but uses the same idea as before. This does not have unnecessary memory allocations or copies.
My concern is efficiency, i.e. quicker UI response, longer battery life, smaller memory footprint, etc. Avoiding unnecessary memory allocations and/or memory copies which are automatically done by Swift in the behind scene would be crucial if we are providing competitive products. (-;
Solution 13:[13]
// find duplicate number in an array
var arrNum = [1, 2, 3 , 3, 2, 5, 6, 2]
let setOfNum = Set(Array(arrNum))
print(setOfNum)
Output: [6, 3, 5, 1, 2]
// find duplicate string in an array
var arrStr = ["1", "2", "3" , "3", "2", "5", "6", "2"]
let setOfStr = Set(Array(arrStr))
print(setOfNum)
Output: [6, 3, 5, 1, 2]
Solution 14:[14]
Here's an efficient, O(n) method to do it. Some of the other answers here use .filter on a duplicates array or even the return value array, which makes the operation work in O(n^2) (using .contains is the same). Using a Set to store the duplicates we can make it O(n).
The other method that's been shown here is using a dictionary to first store the array elements. The idea being that a dictionary can't have duplicate elements. However, this doesn't guarantee the original order of the array to be preserved, so we need a different method.
Here's an Array extension that adds a removeDuplicates method that is efficient and guarantees the same result order as the original array's order.
extension Array where Iterator.Element == Int {
func removeDuplicates() -> [Int] {
var retVal: [Int] = []
var duplicates: Set<Int> = []
for number in self {
if !duplicates.contains(number) {
duplicates.insert(number)
retVal.append(number)
}
}
return retVal
}
}
If you want to return the duplicate elements, just reverse some of the checks in the for loop (Still O(n)).
extension Array where Iterator.Element == Int {
func findDuplicates() -> [Int] {
var retVal: [Int] = []
var duplicates: Set<Int> = []
for number in self {
if duplicates.contains(number) {
retVal.append(number)
} else {
duplicates.insert(number)
}
}
return retVal
}
}
Solution 15:[15]
There's still quite a bit of useful reusable stuff missing from Swift to make this simple, but OrderedCollections, which have not been used by the other answers yet, make it easier to get the duplicates "in order".
XCTAssertEqual(
.init("????????????????????".duplicates),
"??????"
)
import OrderedCollections
public extension Sequence where Element: Hashable {
/// The non-unique elements of this collection, in the order of their first occurrences.
var duplicates: OrderedSet<Element> {
OrderedDictionary(bucketing: self).filter { $1 > 1 }.keys
}
}
import struct OrderedCollections.OrderedDictionary
public protocol DictionaryProtocol {
associatedtype Key
associatedtype Value
init<KeysAndValues: Sequence>(
_: KeysAndValues,
uniquingKeysWith: (Value, Value) throws -> Value
) rethrows where KeysAndValues.Element == (Key, Value)
}
extension Dictionary: DictionaryProtocol { }
extension OrderedDictionary: DictionaryProtocol { }
public extension DictionaryProtocol where Value == Int {
/// Create "buckets" from a sequence of keys,
/// such as might be used for a histogram.
init<Keys: Sequence>(bucketing unbucketedKeys: Keys)
where Keys.Element == Key {
self.init(zip(unbucketedKeys, 1), uniquingKeysWith: +)
}
}
/// `zip` a sequence with a single value, instead of another sequence.
@inlinable public func zip<Sequence: Swift.Sequence, Constant>(
_ sequence: Sequence, _ constant: Constant
) -> LazyMapSequence<
LazySequence<Sequence>.Elements,
(LazySequence<Sequence>.Element, Constant)
> {
sequence.lazy.map { ($0, constant) }
}
Solution 16:[16]
extension Array where Element: Hashable {
func similar() -> Self {
var used = [Element: Bool]()
return self.filter { used.updateValue(true, forKey: $0) != nil }
}
}
Sources
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