Created
December 12, 2017 18:29
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Anagrams interview test with some broken features
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public static void printAnagramList(String[] words) | |
{ | |
// Store anagram groups | |
var groupings = new Dictionary<string, string>(); | |
for (int i = 1; i < words.Length; i++) | |
{ | |
var ws = GetSortedWord(words[i]); | |
for (int j = (i + 1); j < words.Length; j++) | |
{ | |
char[] word2 = words[j].ToCharArray(); | |
Array.Sort(word2); | |
// Helpful to compare things later | |
var wsss = (new string(word2)); | |
// If the sorted string are equal, they have the same length and letter | |
// occurrences. However, we don't want to deal with them if they are the | |
// exact same word, which can happen in subsequent loops | |
if (ws.Equals(wss) && !words[i].Equals(words[j])) | |
{ | |
// if our groupings already contain the key, append our current word | |
if (groupings.ContainsKey(ws)) | |
{ | |
// if the word is already in the group, don't add it again | |
if (!groupings[ws].Split(',').Contains(words[j])) | |
{ | |
groupings[ws] += "," + words[j]; | |
} | |
else | |
{ | |
// we don't have a new key/group started yet, let's create one | |
groupings.Add(ws, words[i] + "," + words[j]); | |
} | |
} | |
} | |
} | |
// output all groupings on their own line | |
foreach (var key in groupings.Keys) | |
{ | |
Console.WriteLine(groupings[key]); | |
}} | |
} | |
public static string GetSortedWord(string word) | |
{ | |
char[] word_char = word.ToCharArray(); | |
Array.Sort(word_char); | |
return (new string(word_char)); | |
} |
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