Ivan

Posted on Nov 26, 2018

Daily Coding Problem #3

#dev #practice #algorithms #challenge

For quite some time I found my time procastinating after getting home from work. Untill a few days ago I stumbled upon the Daily Coding Problem (DCP) and decided to give it a shot.
The code is in C#.

How it works?

The folks at DCP send you a problem that was asked at a top company everyday for you to solve. The premium membership also offers the opportunity to verify your solution.

Problem #2

This problem was asked by Google.

Given the root to a binary tree, implement serialize(root), which serializes the tree into a string, and deserialize(s), which deserializes the string back into the tree.

For example, given the following Node class

class Node: def init(self, val, left=None, right=None): self.val = val self.left = left self.right = right The following test should pass:

node = Node('root', Node('left', Node('left.left')), Node('right')) assert deserialize(serialize(node)).left.left.val == 'left.left'

My solution

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace Task03
{
    public class Program
    {
        private const string Empty_Marker = "1";

        public static void Main(string[] args)
        {
            var node = new Node("root", new Node("left", new Node("left.left")), new Node("right"));

            var serialized = Serialize(node);

            Console.WriteLine(serialized);
            node = Deserialize(serialized);

            Console.WriteLine(node.Left.Left.Value);
        }

        public static string Serialize(Node node)
        {
            if (node == null)
            {
                return Empty_Marker + '-';
            }

            var builder = new StringBuilder();

            builder.Append($"{node.Value}-");

            builder.Append(Serialize(node.Left));
            builder.Append(Serialize(node.Right));

            return builder.ToString();
        }

        public static Node Deserialize(string serializedNode)
        {
            var nodes = serializedNode
                .Split('-', StringSplitOptions.RemoveEmptyEntries)
                .ToArray();

            var queue = new Queue<string>(nodes);

            var node = DeserializeNode(queue);

            return node;
        }

        private static Node DeserializeNode(Queue<string> nodes)
        {
            if (nodes.Peek() != null)
            {
                var nextNode = nodes.Dequeue();

                if (nextNode == Empty_Marker)
                {
                    return null;
                }

                var node = new Node(nextNode);

                node.Left = DeserializeNode(nodes);

                node.Right = DeserializeNode(nodes);

                return node;
            }

            return null;
        }
    }

Explanation

For the serialization, we traverse the tree depth first and add the current node's value to the serialized string. When we reach a node which is null (basically it's parent doesn't have a child), we put an Empty_Marker to signal that there is no node here. This works assuming the symbols for Empty_Marker and '-' are not going to be used in any node's value.

For the deserialize, we split the given string and put the results in a queue. Then build the tree depth first, getting through the queue's first element and stopping when we reach an Empty_Marker

I will try to do the daily problem everyday, but sometimes life gets in the way :)
Solutions are posted in my github account

cwetanow / DailyCodingProblem

DailyCodingProblem

This repository contains solutions of the Daily Coding Problem tasks

View on GitHub

Feel free to leave a like and let me know in the comments if I should continue to post my solutions.

If you also have any better solution in mind, by all means share it, so we can learn from each other.

Top comments (10)

Tiago Romero Garcia • Dec 5 '18 • Edited

In Javascript, JSON serialization/unserialization is already supported, so this felt like cheating:

class Node {
    constructor(val, left = null, right = null) {
        this.val = val;
        this.left = left;
        this.right = right;
    }

    toString() {
        return JSON.stringify(this);
    }
};

const serialize = node => node.toString();
const deserialize = string => JSON.parse(string);

const node = new Node('root', new Node('left', new Node('left.left')), new Node('right'));

console.log(deserialize(serialize(node)).left.left.val); // prints "left.left"

Levi Ermonaites De Freitas • Feb 18 '21

Exactly man, I looked at it, and thought, what?? They're asking one thing that already exists?

E. Choroba • Nov 26 '18 • Edited

The tricky part is to make it work for arbitrary values, i.e. drop the assumption the symbols for Empty_Marker and '-' are not going to be used in any node's value.

Here's such a solution in Perl, tests included:

#!/usr/bin/perl
use warnings;
use strict;

package Node {
    use Moo;

    has value => (is => 'ro', required => 1);
    has left  => (is => 'ro', isa => sub { shift->isa('Node') or die });
    has right => (is => 'ro', isa => sub { shift->isa('Node') or die });

    my $_escape_value = sub {
        my ($self) = @_;
        my $value = $self->value;
        $value =~ s/$_->[0]/\\$_->[1]/g
            for [qr/\\/ => '\\'],
                [qr/"/  => '\''],
                [qr/\(/ => '['],
                [qr/\)/ => ']'];

        return $value
    };

    my $unescape = sub {
        my ($value) = @_;

        $value =~ s/(?<!\\)\\$_->[0]/$_->[1]/g
            for [qr/\]/ => ')'],
                [qr/\[/ => '('],
                [qr/'/  => '"'];
        $value =~ s/\\\\/\\/g;

        return $value
    };

    my $parse = sub {
        my ($rest) = @_;
        my $pos = '(' eq substr $rest, 0, 1;
        my $rank = $pos;
        until (0 == $rank) {
            ++$rank if '(' eq substr $rest, $pos, 1;
            --$rank if ')' eq substr $rest, $pos, 1;
            ++$pos;
        }

        my $left  = substr $rest, 0, $pos;
        my $right = substr $rest, $pos + ($pos != length $rest);

        length $_ and $_  = __PACKAGE__->deserialize($_)
            for $left, $right;

        return $left, $right
    };

    sub serialize {
        my ($self) = @_;

        my $value = $self->$_escape_value;
        my ($left, $right) = map $_ ? $_->serialize : "",
                                 $self->left,
                                 $self->right;

        return '(' . join(',', qq("$value"), $left, $right) . ')'
    }

    sub deserialize {
        my ($class, $string) = @_;

        return undef unless length $string;

        my ($value, $rest) = $string =~ /^\("([^"]*)",(.*)\)$/sg;

        $value = $unescape->($value);
        my ($left, $right) = $parse->($rest);

        return $class->new(value => $value,
                           (left  => $left)  x !! ref $left,
                           (right => $right) x !! ref $right)
    }

};

use Test::More tests => 7;

my $node1 = 'Node'->new(
    value => 'root',
    left  => 'Node'->new(
        value => 'left',
        left => 'Node'->new(value => my $left_left = 'left.left')),
    right => 'Node'->new(value => 'right'),
);

is 'Node'->deserialize($node1->serialize)->left->left->value, $left_left;

my $node2 = 'Node'->new(
    value => my $empty = "",
    right => 'Node'->new(value => my $special_chars = '("[,\'\\]"))',
                         right => 'Node'->new(
                             value => my $newline = "1\n2")));

is 'Node'->deserialize($node2->serialize)->value,
    $empty;
is 'Node'->deserialize($node2->serialize)->right->value,
    $special_chars;
is 'Node'->deserialize($node2->serialize)->right->right->value,
    $newline;

my $node3 = 'Node'->new(value => $node1->serialize,
                        left  => 'Node'->new(value => $node2->serialize));

is 'Node'->deserialize($node3->serialize)->value,
    $node1->serialize;
is 'Node'->deserialize($node3->serialize)->left->value,
    $node2->serialize;

my $super = $node3->serialize;

$super = 'Node'->new(
    value => 'Node'->deserialize($super)->serialize
)->serialize
    for 1 .. 6;

my $root = $super;
$root = 'Node'->deserialize($root)->value for 1 .. 8;
is $root, 'root';

Benjamin Braatz • Nov 28 '18 • Edited

Python solution defining __repr__, which is meant for exactly this purpose:

class Node:
    def __init__(self, val, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

    def __repr__(self):
        return ( 'Node(' + repr(self.val) + ', '
                         + repr(self.left) + ', '
                         + repr(self.right) + ')' )

    def __eq__(self, other):
        if isinstance(other, Node):
            return ( self.val == other.val and
                     self.left == other.left and
                     self.right == other.right )
        return False

    def __hash__(self):
        return hash((self.val, self.left, self.right))

serialise = repr
deserialise = eval

node = Node('root', Node('left', Node('left.left')), Node('right'))
assert deserialise(serialise(node)).left.left.val == 'left.left'
assert deserialise(serialise(node)) == node
assert hash(deserialise(serialise(node))) == hash(node)

Edit: Included test from problem statement in the code, so that it should now be ready for copy and paste into file and python3 -i <file>. While we are at it, implemented __eq__ and __hash__, so that we can assert even stronger tests.

Solomon Li ✊YOLO🙃 • Mar 10 '19

Quite impressive, thank you!

Sajjad Dehghani • Mar 9 '20

I think this is a minimal and fast solution :

import re


class Node:
    def __init__(self, val, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


def serialize(root: Node) -> str:
    if root is None:
        return ""
    return f"({root.val},{serialize(root.left)},{serialize(root.right)})"


def deserialize(serialized: str) -> Node:
    if serialized is None:
        return None
    match = re.search(r"\(([\w\.]+),(\(.*\))?,(\(.*\))?\)", serialized)
    if match:
        val, left, right = match.groups()
        return Node(val=val, left=deserialize(left), right=deserialize(right))


node = Node("root", Node("left", Node("left.left")), Node("right"))
assert deserialize(serialize(node)).left.left.val == "left.left"

Jay • Nov 28 '18

Go solution. Its a bit verbose as it does not have optional/default params.

package main

import (
    "fmt"
    "strings"
)

type Node struct {
    val   string
    left  *Node
    right *Node
}

func (node *Node) serialize() string {
    if node == nil {
        return "#"
    }

    leftSide := node.left.serialize()
    rightSide := node.right.serialize()

    ret := fmt.Sprintf("%s->%s->%s", node.val, leftSide, rightSide)
    return ret
}

func deserialize(str string) Node {
    node, _ := deserailizeNodes(strings.Split(str, "->"))
    return node
}

func deserailizeNodes(nodes []string) (Node, []string) {

    if len(nodes) == 0 {
        return Node{"", nil, nil}, nil
    }

    nextNode := nodes[0]

    if nextNode == "#" {
        return Node{"", nil, nil}, nodes[1:]
    }

    left, rem := deserailizeNodes(nodes[1:])
    right, _ := deserailizeNodes(rem)

    if left.val == "" && right.val == "" {
        return Node{nextNode, nil, nil}, nil
    }
    if left.val == "" {
        return Node{nextNode, nil, &right}, nil
    }
    if right.val == "" {
        return Node{nextNode, &left, nil}, nil
    }

    node := Node{val: nextNode, left: &left, right: &right}
    return node, nil

}

func main() {
    node := Node{"root", &Node{"left", &Node{"left.left", nil, nil}, nil}, &Node{"right", nil, nil}}
    fmt.Println(node.serialize())
    // root->left->left.left->#->#->#->right->#->#
    n := deserialize(node.serialize())
    fmt.Println(n.left.left.val == "left.left") // true
}

Stefano Prosperi • Nov 27 '18 • Edited

Here's my C# solution

using System;

namespace DailyCodingProblem3
{
    class Node
    {
        public string Val { get; set; }
        public Node Left { get; set; }
        public Node Right { get; set; }

        public Node(string val, Node left = null, Node right = null)
        {
            Val = val;
            Left = left;
            Right = right;
        }
    }

    class Program
    {
        public static string Serialize(Node node)
        {
            if (node == null)
            {
                return "-";
            }

            string ret = $"{node.Val}-";
            ret += Serialize(node.Left);
            ret += Serialize(node.Right);

            return ret;
        }

        public static Node Deserialize(ref string str)
        {
            if (str.Length == 0 || str.StartsWith("-"))
            {
                return null;
            }

            var val = str.Split('-')[0];

            str = str.Substring(str.IndexOf('-') + 1);
            var left = Deserialize(ref str);

            str = str.Substring(str.IndexOf('-') + 1);
            var right = Deserialize(ref str);

            return new Node(val, left, right);
        }

        static void Main()
        {
            var node = new Node("root", new Node("left", null, new Node("left.right", new Node("left.right.left"))), new Node("right"));
            var ret = Serialize(node);

            Console.WriteLine(ret);
            Console.WriteLine(Serialize(Deserialize(ref ret)));
        }
    }
}

Geoffrey Olson Jr. • Aug 20 '19 • Edited

Python solution using recursion.

class Node:
    def __init__(self, val, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Stringifier():
    def __init__(self):
        self.current = 0

    def serialize(self, root, strTree = []):
        if root is None:
            return
        strTree.append('{')
        strTree.append('\"' + root.val + '\"')
        self.serialize(root.left, strTree)
        self.serialize(root.right, strTree)
        strTree.append('}')
        return "".join(strTree)

    def _deserialize(self, strTree):
        left = None
        right = None
        if strTree[self.current] == '{' :
            self.current += 1
            val = ""
            if strTree[self.current] == '\"':
                self.current += 1
            while strTree[self.current] != '\"':
                val += strTree[self.current]
                self.current += 1
            self.current += 1
            if strTree[self.current] == '{':
                left = self._deserialize(strTree)
            if strTree[self.current] == '{':
                right = self._deserialize(strTree)
            if strTree[self.current] == '}':
                self.current += 1
                return Node(val, left, right)

    def deserialize(self, strTree):
        self.current = 0
        return self._deserialize(strTree)

if __name__ == "__main__":
    stringifier = Stringifier()
    node = Node('root', Node('left', Node('left.left')), Node('right'))
    assert stringifier.deserialize(stringifier.serialize(node)).left.left.val == 'left.left'

Solomon Li ✊YOLO🙃 • Mar 10 '19

My first time to see someone using Lua, thanks for the code, cool!

DEV Community

Daily Coding Problem #3

How it works?

Problem #2

My solution

Explanation

cwetanow / DailyCodingProblem

DailyCodingProblem

Top comments (10)

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