This question is a follow up question to: this post

I'd love some advice on my implementation and the API of the Huffman class.

I'm also not sure how to test if my implementation is actually resulting in less bytes than a string. It seems that encoded.count (as a Data) is larger than word.utf8.count (as a String). Maybe I'm just not testing on large enough strings?

Also any thoughts on HuffData storing the code ex. ["00","01"] and the frequencyTable instead of the tree.

Here’s an example of how the API is used:

let encoded = try? Huffman.encode(word)
let decoded = try? Huffman.decode(encoded!)
XCTAssertEqual(decoded, word)

Here's the code:

import Foundation

struct HuffData: Codable {
    var code: [String]
    var frequencyTable: [String: String]

class Huffman {
    static func decode(_ data: Data) throws -> String {
        do {
            let huff = try JSONDecoder().decode(HuffData.self, from: data)
            let reverseTable = Dictionary(uniqueKeysWithValues: zip(huff.frequencyTable.values, huff.frequencyTable.keys))
            return huff.code.compactMap({ reverseTable[$0]}).joined()
        catch let error {
            throw error

    static func encode(_ input: String) throws -> Data {
        let frequencyTable = Huffman.buildFrequencyTable(for: input)
        let code = input.compactMap({frequencyTable[String($0)]})
        let huff = HuffData(code: code, frequencyTable: frequencyTable)
        do {
            let data = try JSONEncoder().encode(huff)
            return data
        catch let error {
            throw error

    static private func buildFrequencyTable(for input: String) -> [String: String] {
        // count letter frequency
        let sortedFrequency = input.reduce(into: [String: Int](), { freq, char in
            freq[String(char), default: 0] += 1
        // create queue of initial Nodes
        let queue ={ Node(name: $0.key, value: $0.value)}
        // generate key by traversing tree
        return Huffman.generateKey(for: Huffman.createTree(with: queue), prefix: "")

    static private func generateKey(for node: Node, prefix: String) -> [String: String] {
        var key = [String: String]()
        if let left = node.left, let right = node.right {
            key.merge(generateKey(for: left, prefix: prefix + "0"), uniquingKeysWith: {current,_ in current})
            key.merge(generateKey(for: right, prefix: prefix + "1"), uniquingKeysWith: {current,_ in current})
        }else {
            key[] = prefix
        return key

    static private func createTree(with queue: [Node]) -> Node {
        // initialize queue that sorts by decreasing count
        var queue = PriorityQueue(queue: queue)
        // until we have 1 root node, join subtrees of least frequency
        while queue.count > 1 {
            let node1 = queue.dequeue()
            let node2 = queue.dequeue()
            let rootNode = Huffman.createRoot(with: node1, and: node2)
            queue.enqueue(node: rootNode)
        return queue.queue[0]

    static private func createRoot(with first: Node, and second: Node) -> Node {
        return Node(name: "\(\(", value: first.value + second.value, left: first, right: second)


struct PriorityQueue {
    var queue: [Node]
    var count: Int {
        return queue.count
    mutating func enqueue(node: Node) {
        queue.insert(node, at: queue.index(where: {$0.value <= node.value}) ?? 0)
    mutating func dequeue() -> Node {
        return queue.removeLast()
    init(queue: [Node]){
        // assumes queue will always be sorted by decreasing count
        self.queue = queue.sorted(by: {$0.value > $1.value})

class Node: CustomStringConvertible {
    var description: String {
        return "\(name): \(value)"
    let name: String
    let value: Int
    let left: Node?
    let right: Node?

    init(name: String, value: Int, left: Node? = nil, right: Node? = nil) { = name
        self.value = value
        self.left = left
        self.right = right
  • 1
    I don't know Swift well enough to be sure, but it seems like your output is 8 times as large as it should be, by neglecting to pack the bits into bytes. That way, since a Huffman code is at least 1 bit, the result is guaranteed to be at least as large as the input. – harold Dec 6 at 17:48
  • oh no! I'm pretty sure I'm getting the correct zeros and ones from the tree so I'm not sure how the output is 8 times as large as it should be! I assumed joining the table with the code would add some overhead but would be worth it for strings that are large enough to consider encoding. – Turnipdabeets Dec 6 at 18:06
  • 1
    @harold is quite right: Each zero and one of the encoded sequence is stored as a "0" or "1" character and thus consumes 8 bits. – Martin R Dec 6 at 18:24

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