Given a dictionary and a list of letters find all valid words that can be built with the letters

Question

The brute force way can solve the problem in O(n!), basically calculating all the permutations and checking the results in a dictionary. I am looking for ways to improve the com

Answer 1

Swift 4

func findValidWords(in dictionary: [String], with letters: [Character]) -> [String] {
    var validWords = [String]()

    for word in dictionary {
        var temp = word
        for char in letters {
            temp = temp.filter({ $0 != char })

            if temp.isEmpty {
                validWords.append(word)
                break
            }
        }
    }

    return validWords
}

print(findValidWords(in: ["ape", "apples", "orange", "elapse", "lap", "soap", "bar", "sole"], with: ["a","p","l","e","s","o"]))

Output => ["ape", "apples", "elapse", "lap", "soap", "sole"]