C # hyphenation of words in a string with a breakdown to a certain length

Question

There is a line: Quisque velit nisi, pretium ut lacinia in, elementum id enim. Nulla porttitor accumsan tincidunt. Quisque velit nisi, pretium ut lacinia in, elementum id enim. Nulla porttitor accumsan tincidunt.

It is necessary to break it into three parts, but save whole words and move it to the next line only after the end of the word. For example, we break into three lines:

 Quisque velit nisi, pretium ut lacinia in, elementum id enim. Nulla porttitor accumsan tincidunt.

Now I use this simple code, but it cuts whole words:

 int chunkSize = string.Length / 3; int stringLength = string.Length; for (int i = 0; i < stringLength; i += chunkSize) { if (i + chunkSize > stringLength) chunkSize = stringLength - i; Console.WriteLine(string.Substring(i, chunkSize)); }

It turns out at the output:

 Quisque velit nisi, pretium ut l acinia in, elementum id enim. Nu lla porttitor accumsan tincidunt .

Tell me how can this be easier to implement?

UPDATE

Important: the original string may differ from the above. May contain any characters.

Found a simple solution here https://stackoverflow.com/a/17571171/2127124

 public static class ExtensionMethods { public static string[] Wrap(this string text, int max) { var charCount = 0; var lines = text.Split(new[] { ' ' }, StringSplitOptions.RemoveEmptyEntries); return lines.GroupBy(w => (charCount += (((charCount % max) + w.Length + 1 >= max) ? max - (charCount % max) : 0) + w.Length + 1) / max) .Select(g => string.Join(" ", g.ToArray())) .ToArray(); } }

If the line contains N words (or N - 1 separator ' ' ) and it needs to be divided into K parts, then the number of possible partitions is (N - K + 2) * (N - K + 1) / 2 provided that N >= K .
Or do you need the most optimal (with the lengths of the pieces closest to str.Length / K )?
I thought about splitting the source line into spaces through split, and then forming the necessary set of the resulting list.
I would break into words, then find the number of words in the line and bring it all out
either it is banal to go in the usual cycle, to look, if they come to the end but the word has not ended - we go to the end of the word and carry over, and then again also

Andrey NOP Andrey NOP 25k 4 gold marks 25 silver marks 57 bronze marks · Answer 1 · 2017-08-19T12:52:22

The idea is to find the indexes of the characters that need to be broken ideally (into equal parts). Find all possible splits of the string and select the option with the smallest error. For an error I decided to accept the sum of the absolute values of the deviations of the selected indices from the ideal ones.

 static int Abs(int x) => x >= 0 ? x : -x; static void Main(string[] args) { // Входной текст var str = "Quisque velit nisi, pretium ut lacinia in, elementum id enim. Nulla porttitor accumsan tincidunt."; // Количество выходных строк var numOfParts = 3; // Перечень разделителей var delimeters = " "; // Индексы, по которым надо разбить в идеале var idealParts = Enumerable.Range(1, numOfParts - 1) .Select(x => ((str.Length - numOfParts + 1) * x + numOfParts / 2) / numOfParts) .ToArray(); // индексы всех имеющихся мест, по которым можно разбить var indices = str.Select((c, i) => (c: c, i: i)) .Where(t => delimeters.Contains(tc)) .Select(t => ti) .ToArray(); // indices.Length ^ numOfParts var numOfSp = Enumerable.Range(0, numOfParts - 1) .Aggregate(1, (m, x) => m * indices.Length); // Генерируем все возможные комбинации var splits = new int[numOfSp][]; for (int d = 0; d < numOfSp; ++d) { var z = d; splits[d] = new int[numOfParts - 1]; for (int i = numOfParts - 2; i >= 0; --i) { splits[d][i] = z % indices.Length; z /= indices.Length; } } // Для каждой комбинации подсчитываем ошибку var spWithEr = splits.Select(s => (s: s, er: s.Select((x, i) => Abs(indices[x] - idealParts[i])).Sum())) .ToArray(); // Находим минимальную ошибку var minEr = spWithEr.Min(t1 => t1.er); // Выбираем комбинацию с наименьшей ошибкой var minSp = spWithEr.First(t => t.er == minEr).s; // Отбираем индексы var sp = new[] { -1 }.Concat(minSp.Select(x => indices[x]) .Concat(new[] { str.Length })) .ToArray(); // Разбиваем на строки по индексам из sp for (int i = 0; i < sp.Length - 1; ++i) Console.WriteLine(str.Substring(sp[i] + 1, sp[i + 1] - sp[i] - 1)); Console.ReadLine(); }

Conclusion:

 Quisque velit nisi, pretium ut lacinia in, elementum id enim. Nulla porttitor accumsan tincidunt.

Is it overhead - it's up to you

Thanks for the answer and the code! You are right, for my task it is redundant. - pwb

Andrey NOP 25k 4 gold marks 25 silver marks 57 bronze marks · Answer 2 · 2017-08-19T17:13:26

We break a line into words using the Split method and set the maximum number of characters in a line. Then we build each line of words, separating them with spaces and checking whether the current line has not gotten the maximum length.
The code is quite compact, and let it speak for itself)

 var inputString = @"Quisque velit nisi, pretium ut lacinia in, elementum id enim. Nulla porttitor accumsan tincidunt"; var words = inputString.Split(new Char[] { ' ' }); var maxLengthString = 53; int wordIndex = 0; var spaceLetter = " "; var currentLine = new StringBuilder(); while (true) { if (currentLine.Length + words[wordIndex].Length + 1 > maxLengthString)// Определяем не привысила ли текущая строка максимальную длину { Console.WriteLine(currentLine); currentLine.Remove(0, currentLine.Length); } currentLine.Append(words[wordIndex]); currentLine.Append(spaceLetter); wordIndex++; if(wordIndex == words.Length) { Console.WriteLine(currentLine); break; } }

Nevertheless, the question was asked and the answers were heard :)
Tell me, what about commas and other characters in your example?
My output is without commas and some characters, for example, cutting the symbol / :(
Added punctuation to regular expression var words = new Regex (@ "\ w (? <! \ D) [\ w '-,;.?!] *"). Matches (inputString);
For example Cheese, soft Serbian Brynza 35% 250g Kuč mlekara, Serbia For a good text should be the same.

C # hyphenation of words in a string with a breakdown to a certain length

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