Comparing quantities of unique elements contained in large data sets [duplicate] - ios

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Refactored Solution In Swift
(2 answers)
Closed 6 years ago.
I'm attempting to solve HackerRank's Hash Table Ransom Note challenge. There are 19 test cases and I'm passing all but two of time due to timeout on larger data sets (10,000-30,000 entries).
I'm given:
1) an array of words contained in a magazine and
2) an array of words for a ransom note. My objective is to determine if the words in the magazine can be used to construct a ransom note.
I need to have enough unique elements in the magazineWords to satisfy the quantity needed by noteWords.
I'm using this code to make that determination...and it takes FOREVER...
for word in noteWordsSet {
// check if there are enough unique words in magazineWords to put in the note
if magazineWords.filter({$0==word}).count < noteWords.filter({$0==word}).count {
return "No"
}
}
What is a faster way to accomplish this task?
Below is my complete code for the challenge:
import Foundation
var magazineWords = // Array of 1 to 30,000 strings
var noteWords = // Array of 1 to 30,000 strings
enum RegexString: String {
// Letters a to z, A to Z, 1 to 5 characters long
case wordCanBeUsed = "([a-zA-Z]{1,5})"
}
func matches(for regexString: String, in text: String) -> [String] {
// Hat tip MartinR for this
do {
let regex = try NSRegularExpression(pattern: regexString)
let nsString = text as NSString
let results = regex.matches(in: text, range: NSRange(location: 0, length: nsString.length))
return results.map { nsString.substring(with: $0.range)}
} catch let error {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
func canCreateRansomNote(from magazineWords: [String], for noteWords: [String]) -> String {
// figure out what's unique
let magazineWordsSet = Set(magazineWords)
let noteWordsSet = Set(noteWords)
let intersectingValuesSet = magazineWordsSet.intersection(noteWordsSet)
// constraints specified in challenge
guard magazineWords.count >= 1, noteWords.count >= 1 else { return "No" }
guard magazineWords.count <= 30000, noteWords.count <= 30000 else { return "No" }
// make sure there are enough individual words to work with
guard magazineWordsSet.count >= noteWordsSet.count else { return "No" }
guard intersectingValuesSet.count == noteWordsSet.count else { return "No" }
// check if all the words can be used. assume the regex method works perfectly
guard noteWords.count == matches(for: RegexString.wordCanBeUsed.rawValue, in: noteWords.joined(separator: " ")).count else { return "No" }
// FIXME: this is a processor hog. I'm timing out when I get to this point
// need to make sure there are enough magazine words to write the note
// compare quantity of word in magazine with quantity of word in note
for word in noteWordsSet {
// check if there are enough unique words in magazineWords to put in the note
if magazineWords.filter({$0==word}).count < noteWords.filter({$0==word}).count {
return "No"
}
}
return "Yes"
}
print(canCreateRansomNote(from: magazineWords, for: noteWords))

Answer 1

I don't know how to read from the test case on the contest website or what frameworks you are allowed. If Foundation is allowed, you can use NSCountedSet
import Foundation
let fileContent = try! String(contentsOf: URL(fileURLWithPath: "/path/to/file.txt"))
let scanner = Scanner(string: fileContent)
var m = 0
var n = 0
scanner.scanInt(&m)
scanner.scanInt(&n)
var magazineWords = NSCountedSet(capacity: m)
var ransomWords = NSCountedSet(capacity: n)
for i in 0..<(m+n) {
var word: NSString? = nil
scanner.scanUpToCharacters(from: .whitespacesAndNewlines, into: &word)
if i < m {
magazineWords.add(word!)
} else {
ransomWords.add(word!)
}
}
var canCreate = true
for w in ransomWords {
if ransomWords.count(for: w) > magazineWords.count(for: w) {
canCreate = false
break
}
}
print(canCreate ? "Yes" : "No")
It works by going through the input file one word at a time, counting how many times that word appears in the magazine and in the ransom note. Then if any word appear more frequently in the ransom note than in the magazine, it fails the test immediately. Run the 30,000 words test case in less than 1 second on my iMac 2012.