Find the repeated sequence in the line that go in a row - swift

Question

Given a string of arbitrary length. I need to find 1 subsequences of identical characters that go in a row.
My function (there are two of them, but these are two parts of the same function) turned out to be complex and cumbersome and did not fit because of this. The function I need should be simple and not too long.
Example:
Input : str = "abcabc"
Output : abc
Input : str = "aa"
Output : a
Input : str = "abcbabcb"
Output : abcb
Input : str = "abcbca"
Output : bcbc
Input : str = "cbabc"
Output :
Input : str = "acbabc"
Output :
My unsuccessful function:
func findRepetition(_ p: String) -> [String:Int] {
var repDict: [String:Int] = [:]
var p = p
while p.count != 0 {
for i in 0...p.count-1 {
repDict[String(Array(p)[0..<i]), default: 0] += 1
}
p = String(p.dropFirst())
}
return repDict
}
var correctWords = [String]()
var wrongWords = [String]()
func getRepeats(_ p: String) -> Bool {
let p = p
var a = findRepetition(p)
for i in a {
var substring = String(Array(repeating: i.key, count: 2).joined())
if p.contains(substring) {
wrongWords.append(p)
return false
}
}
correctWords.append(p)
return true
}
I will be very grateful for your help!

Answer 1

Here's a solution using regular expression. I used a capture group that tries to match as many characters as possible such that the whole group repeats at least once.
import Foundation
func findRepetition(_ s: String) -> String? {
if s.isEmpty { return nil }
let pattern = "([a-z]+)\\1+"
let regex = try? NSRegularExpression(pattern: pattern, options: [])
if let match = regex?.firstMatch(in: s, options: [], range:
NSRange(location: 0, length: s.utf16.count)) {
let unitRange = match.range(at: 1)
return (s as NSString).substring(with: unitRange)
}
return nil
}
print(findRepetition("abcabc")) //prints abc
print(findRepetition("aa")) //prints a
print(findRepetition("abcbabcb")) //prints abcb
print(findRepetition("abcbca")) //prints bc
print(findRepetition("cbabc")) //prints nil
print(findRepetition("acbabc")) //prints nil

Answer 2

func findRepetitions(_ p : String) -> [String: Int]{
let half = p.count / 2 + 1
var result : [String : Int] = [:]
for i in 1..<half {
for j in 0...(p.count-i) {
let sub = (p as! NSString).substring(with: NSRange.init(location: j, length: i))
if let val = result[sub] {
result[sub] = val + 1
}else {
result[sub] = 1
}
}
}
return result
}
This is for finding repetitions of possible substrings in your string. Hope it can help

Answer 3

Here is a solution that is based on the Suffix Array Algorithm, that finds the longest substring that is repeated (contiguously):
func longestRepeatedSubstring(_ str: String) -> String {
let sortedSuffixIndices = str.indices.sorted { str[$0...] < str[$1...] }
let lcsArray = [0]
+
sortedSuffixIndices.indices.dropFirst().map { index in
let suffix1 = str[sortedSuffixIndices[index]...]
let suffix2 = str[sortedSuffixIndices[index - 1]...]
let commonPrefix = suffix1.commonPrefix(with: suffix2)
let count = commonPrefix.count
let repeated = suffix1.dropFirst(count).commonPrefix(with: commonPrefix)
return count == repeated.count ? count : 0
}
let maxRepeated = zip(sortedSuffixIndices.indices,lcsArray).max(by: { $0.1 < $1.1 })
if let tuple = maxRepeated, tuple.1 != 0 {
let suffix1 = str[sortedSuffixIndices[tuple.0 - 1]...]
let suffix2 = str[sortedSuffixIndices[tuple.0]...]
let longestRepeatedSubstring = suffix1.commonPrefix(with: suffix2)
return longestRepeatedSubstring
} else {
return ""
}
}
Here is an easy to understand tutorial about such an algorithm.
It works for these examples:
longestRepeatedSubstring("abcabc") //"abc"
longestRepeatedSubstring("aa") //"a"
longestRepeatedSubstring("abcbabcb") //"abcd"
longestRepeatedSubstring("abcbca") //"bcbc"
longestRepeatedSubstring("cbabc") //""
longestRepeatedSubstring("acbabc") //""
As well as these:
longestRepeatedSubstring("a😍ca😍c") //"a😍c"
longestRepeatedSubstring("Ab cdAb cd") //"Ab cd"
longestRepeatedSubstring("aabcbc") //"bc"
Benchmarks
Here is a benchmark that clearly shows that the Suffix Array algorithm is much faster than using a regular expression.
The result is:
Regular expression: 7.2 ms
Suffix Array : 0.1 ms