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Home/ Questions/Q 413171
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T18:09:28+00:00 2026-05-12T18:09:28+00:00

class CounterDict<TKey> { public Dictionary<TKey, int> _dict = new Dictionary<TKey, int>(); public void Add(TKey

  • 0
class CounterDict<TKey>
{
    public Dictionary<TKey, int> _dict = new Dictionary<TKey, int>();

    public void Add(TKey key)
    {
        if(_dict.ContainsKey(key))
            _dict[key]++;
        else
        {
            _dict.Add(key, 1);
        }
    }
}

class Program
{
    static void Main(string[] args)
    {
        string line =  "The woods decay the woods decay and fall.";

        CounterDict<string> freq = new CounterDict<string>();
        foreach (string item in line.Split())
        {
            freq.Add(item.Trim().ToLower());
        }

        foreach (string key in freq._dict.Keys)
        {
            Console.WriteLine("{0}:{1}",key,freq._dict[key]);
        }           
    }
}

I want to calculate number of occurences of all the words in a string.
I think above code will be slow at this task because of (look into the Add function) :

    if(_dict.ContainsKey(key))
    _dict[key]++;
    else
    {
        _dict.Add(key, 1);
    }

Also, is keeping _dict__ public good practice? (I don’t think it is.)

How should I modify this or change it totally to do the job?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-12T18:09:29+00:00Added an answer on May 12, 2026 at 6:09 pm

    How about this:

    Dictionary<string, int> words = new Dictionary<string, int>();
    string input = "The woods decay the woods decay and fall.";
    foreach (Match word in Regex.Matches(input, @"\w+", RegexOptions.ECMAScript))
    {
        if (!words.ContainsKey(word.Value))
        {
            words.Add(word.Value, 1);
        }
        else
        {
            words[word.Value]++;
        }
    }
    

    Principal point was replacing .Split by a regular expression, so you don’t need to keep a big string array in memory and you can work with one item at time.

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