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Alien dictionary - code review May 4, 2018
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace _269AlienDictionary_Review
{
class Program
{
/// <summary>
/// code review May 4, 2018
/// Leetcode: Alien Dictionary
/// </summary>
/// <param name="args"></param>
static void Main(string[] args)
{
string[] words = { "wrt", "wrf", "er", "ett", "rftt" };
// verify result manually here: "wertf"
// 5 minutes warmup the algorithm of topological sort with a dictionary of words:
// "wrt" -> "wrf", t -> f, so f's dependency list {t}
// "wrf" -> "er", w -> e, so e's dependency list {w}
// "er" -> "ett", r -> t, so t's dependency list {r}
// "ett" -> "rftt", e -> r, so r's dependency list {e}
// right now w's dependency list is empty, indegree is 0. => remove w first
// update e's dependency list and e's indegree from 1 to 0.
// continue to work on rest...
var result = AlienOrder(words);
Debug.Assert(result.CompareTo("wertf") == 0);
}
/// <summary>
/// code review on May 4, 2018
/// </summary>
/// <param name="words"></param>
/// <returns></returns>
public static string AlienOrder(string[] words)
{
if (words == null || words.Length <= 1)
{
return "";
}
// Graph prensentation - 3 variables at least
// nodes, node's dependency list and inDegree array
// nodes - function getNodes()
var nodes = getNodes(words);
var dependencyList = new Dictionary<char, HashSet<char>>();
var inDegree = new int[26];
graphSetup(words, dependencyList, inDegree);
return topologicalSort(words, dependencyList, inDegree, nodes);
}
/// <summary>
/// code review on May 4, 2018
/// every char is a node in the graph
/// getNodes - get all chars in the words
/// </summary>
/// <param name="words"></param>
/// <returns></returns>
private static HashSet<char> getNodes(string[] words)
{
var hashset = new HashSet<char>();
foreach (string s in words)
{
hashset.UnionWith(s.ToList());
}
return hashset;
}
/*
* Explain the function graphSetup design using the most simple example:
* set up graph for {"wrt","wrf"}
*
* Here are the task to complete:
* 1. Find the first different char - actually, it is an edge in the graph,
* here is t-> f
* 2. Need to handle case - no edge
* 3. At most one edge for two consecutive words
* 4. Only handle the array of string by two neighboring nodes - transmitive
*
* what to construct in the graph:
* 1. Add 't' to dependencyList, and also add 't' as char, its neighbors {'f'}, 'f' depends on 't'.
* 2. Work on inDegree, increment inDegree['f'-'a']++, 'f' has one more indegree from 't'.
* 3. need to get the first different char from two strings.
*
* precondition:
* words's length >=2
* base case <=1 is handled in callee function
*/
private static void graphSetup(
string[] words,
Dictionary<char, HashSet<char>> dependencyList,
int[] inDegree
)
{
var length = words.Length;
// go over each word
for (int i = 1; i < length; i++)
{
var previous = words[i - 1];
var current = words[i];
int firstUnmatched = 0;
while (previous[firstUnmatched] == current[firstUnmatched])
{
firstUnmatched++;
}
// no edge
if (firstUnmatched >= Math.Min(previous.Length, current.Length))
continue;
// work on style - double checking - using correct variable
char edgeFrom = previous[firstUnmatched];
char edgeTo = current[firstUnmatched];
if (!dependencyList.ContainsKey(edgeFrom))
{
dependencyList.Add(edgeFrom, new HashSet<char>());
}
if (!dependencyList[edgeFrom].Contains(edgeTo))
{
dependencyList[edgeFrom].Add(edgeTo);
inDegree[edgeTo - 'a']++;
}
// skip if edgeTo is in the hashset.
}
}
/*
* Two tasks:
* 1. get nodes with indegree 0 - enqueue those nodes
* 2. play with queue
* a node is dequeue, append to the output stringBuilder
* adjust its dependency list - inDegree value - decrement one
* if the neighbor node is with 0 indegree value, push it to the queue
*
* Checking list about queue:
* 1. First queue is not empty - add nodes into queue first - with indegree 0 nodes
* 2. when node is dequeued, the indegree is updated accordingly, then, more nodes will be
* added to queue when its indegree is 0
* 3. Every node in the queue is with indegree 0 - this is a fact.
*/
public static string topologicalSort(
string[] words,
Dictionary<char, HashSet<char>> dependencyList,
int[] inDegree,
HashSet<char> nodes
)
{
var queue = new Queue<char>();
// add those nodes with inDegree value 0 to the queue
for (int i = 0; i < 26; i++)
{
var visit = (char)(i + 'a');
if (!nodes.Contains(visit))
{
continue;
}
if (inDegree[i] == 0)
{
queue.Enqueue(visit);
}
}
var stringBuilder = new StringBuilder();
while (queue.Count > 0)
{
var visit = queue.Dequeue();
stringBuilder.Append(visit);
if (!dependencyList.ContainsKey(visit)) // bug001 - forget these 2 lines, runtime execution error
{
continue;
}
var neighbors = dependencyList[visit];
foreach (char node in neighbors)
{
int index = node - 'a';
inDegree[index]--;
// update queue
if (inDegree[index] == 0)
{
queue.Enqueue(node);
}
}
}
// edge case:
return stringBuilder.Length < nodes.Count ? "" : stringBuilder.ToString();
}
}
}
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