Algorithms for Anagrams output

funWithAnagrams.swift

// brute force

func funWithAnagrams1(text: [String]) -> [String] {
    var result = [String]()
    
    var lookup =  [Int: Set<Character>]()
    
    for string in text {
        if let set = lookup[string.count],
           set.isSuperset(of: Array(string)) {
            continue
        } else {
            lookup[string.count] = Set(string)
            result.append(string)
        }
    }    
    return result.sorted()
}

// more detail and explore and use Set to reduce time of complexity.

func funWithAnagrams(text: [String]) -> [String] {
    var result = [String]()
    
    var lookup =  [Int: [Set<Character>]]()
    
    for string in text {
        let array: Array<Character> = Array(string)
        let currentStringSet: Set<Character> = Set(array)
        if let sets = lookup[currentStringSet.count], sets.contains(where: { $0.count == currentStringSet.count})  {
            for interSet in sets {
                for target in string where !interSet.contains(target) {
                    break
                }
            }
        } else {
            if lookup[currentStringSet.count] != nil {
                lookup[currentStringSet.count]?.append(currentStringSet)
            } else {
                lookup[currentStringSet.count] = [currentStringSet]
            }
            result.append(string)
        }
    }
    return result.sorted()
}