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Printing even numbers with commas except the last in C# not working

I have been trying to fix this issue for some days now but can't find the error. It seems as simple as an if statement for the code to print commas for all but the last number. It worked for me putting random numbers but when i put the specific numbers (24,7,35,2,27,7,89)it prints the comma at the end.
Print even numbers with commas
This is my code, but i tried multiple other ways.
using System.Collections.Generic;
using System.Text;
using System.Transactions;
namespace ArrayExercises
{
class TaskFour
{
public static void FindEvenNumbers()
{
int[] input = new int[7];
int count = 1;
string comma = ",";
Console.WriteLine("[== Please Enter 7 numbers ==]");
Console.WriteLine();
for (int i = 0; i < input.Length; i++)
{
Console.WriteLine($"Enter number {count}:");
input[i] = int.Parse(Console.ReadLine());
count++;
}
Console.WriteLine("The even numbers in this array are: ");
for (int i = 0; i < input.Length; i++)
{
if (input[i] % 2 == 0)
{
if (i < input.Length)
{
Console.Write(input[i] + comma);
}
else
{
Console.Write(input[i]);
}
}
}
}
}
}
Thanks in advance :)

You can use below code without introducing any extra space using for loop.
for (int i = 0; i < input.Length; i++)
{
if (input[i] % 2 == 0)
{
if (i != 0) Console.Write(comma);
Console.Write(input[i]);
}
}
Or use the inbuilt string.Join function.
Console.WriteLine(string.Join(',', input.Where(i => i % 2 == 0)));

This answer assumes that the above is homework/tutorial, there are more efficient means that what I am posting here
The problem is this segment:
if (i < input.Length)
{
Console.Write(input[i] + comma);
}
The problem is that you are always outputting a comma, regardless of what is to follow. I think that the easiest approach would be to add the comma before hand, meaning, while you are going through the second loop, if you have already printed a number before, you pre-pend a comma and print the number you have, if you did not print any numbers before (that is, you are about to print the first number) then you do not prepend the comma.

i will always be less than the length the input due to the loop's condition.
Instead of implementing this yourself, you could let string.join do the heavy lifting for you:
string result = string.Join(comma, input.Where(i => i % 2 == 0));
Console.WriteLine(result);

Your else condition will never execute. If you are rookie in C# then I would suggest get all even numbers first and store it in the list. Iterate over again to print with comma and just skip the last comma.
//Your existing source code
List<int> evenNumber = new List<int>();
for (int i = 0; i < input.Length; i++)
{
if (input[i] % 2 == 0)
{
evenNumber.Add(input[i]);
}
}
for(int i = 0; i < evenNumber.Length; i++)
{
//Print just number if it is last, otherwise print number with comma
if(i == evenNumber.Length - 1)
Console.Write(evenNumber[i]);
else
Console.Write(evenNumber[i]+comma);
}
If you know string.Join() then use linq to get list of even numbers and print list of even numbers with ',' as a delimiter
var evenNumbers = input.Where(x => x % 2 ==0); //Filter all even numbers
Console.WriteLine(string.Join(",", evenNumbers); //Print using string.Join

Yet another version:
string result = "";
for (int i = 0; i < input.Length; i++)
{
if (input[i] % 2 == 0)
{
result += $",{input[i]}";
}
}
Console.WriteLine( result.Length > 0 ? result.Substring(1).ToString() : "");

Finding the longest word (and write it out) in string without split, distinct and foreach [duplicate]

This question already has answers here:
Finding longest word in string
(4 answers)
Closed 3 years ago.
I got an assignment to make a method to find the longest word in a string without split, distinct and foreach.
I was able to split the words and count the length but I am stuck on how can I actually compare and write them out.
static void Main(string[] args)
{
String s1 = "Alex has 2 hands.";
longestWord(s1);
Console.
}
static void longestWord(String s1)
{
char emp = ' ';
int count = 0;
char[] ee = s1.ToCharArray();
for (int i = 0; i < ee.Length; i++)
{
if (ee[i] == emp || ee[i] == '.')
{
count++;
Console.Write(" " + (count-1));
Console.WriteLine();
count = 0;
}
else
{
Console.Write(ee[i]);
count++;
}
}
}
The output right now looks like this:
Alex 4
has 3
2 1
hands 5
I am pretty sure I would be able to get only the longest number to show by comparing count before reset with temp int but how to write out the word with it.
Or if there is a easier way which probably is.

You are already on a good way. Instead of directly printing the words, store the length and position of the longest word and print it at the end. Like so:
static void longestWord(String s1)
{
char emp = ' ';
int longestLength = 0;
int longestStart = 0;
int currentStart = 0;
for (int i = 0; i < s1.Length; i++)
{
if (s1[i] == emp || s1[i] == '.')
{
// calculate the current word length
int currentLength = i - currentStart;
// test if this is longer than the currently longest
if(currentLength > longestLength)
{
longestLength = currentLength;
longestStart = currentStart;
}
// a new word starts at the next character
currentStart = i + 1;
}
}
// print the longest word
Console.WriteLine($"Longest word has length {longestLength}: \"{s1.Substring(longestStart, longestLength)}\"");
}
There is no need for the .ToCharArray(). You can access the string directly.

I will question whether you are actually supposed to treat "2" as a word and count it at all. Using regular expressions will allow you to approach the problem using a LINQ one-liner:
static void Main(string[] args)
{
String s1 = "Alex has 2 hands.";
var word = longestWord(s1);
Console.WriteLine(word);
//Console.ReadLine();
}
static string longestWord(string s1) {
return Regex.Matches(s1,"[A-Za-z]+") // find all sequences containing alphabetical characters, you can add numberas as well: [A-Za-z0-9]
.Cast<Match>() // cast results to Enumberable<Match> so we can apply LINQ to it
.OrderByDescending(m => m.Length) // luckily for us, Match comes with Length, so we just sort our results by it
.Select(m => m.Value) // instead of picking up everything, we only want the actual word
.First(); // since we ordered by descending - first item will be the longest word
}

You can store for every word the chars in new list of chars (list for dynamic length)
and if the new word is longer of the prev long word convert it to string.
If you have two word in same length it will take the first.
If you want the last change the "maxLength < count" to "maxLength <= count"
static string longestWord(String s1)
{
char emp = ' ';
int count = 0;
int maxLength = 0;
string maxWord = string.Empty;
List<char> newWord = new List<char>();
char[] ee = s1.ToCharArray();
for (int i = 0; i < ee.Length; i++)
{
if (ee[i] == emp || ee[i] == '.')
{
if (maxLength < count)
{
maxLength = count;
maxWord = new string(newWord.ToArray());
}
count = 0;
newWord = new List<char>();
}
else
{
newWord.Add(ee[i]);
count++;
}
}
return maxWord;
}

How to get every other character in a string in c#

so I'm working on this problem: https://www.hackerrank.com/challenges/30-review-loop/problem (it's in C#)
and so far I'm just trying to break it down piece by piece, and so far I was able to get it to show every other character, but I'm not sure how to concatenate each letter into a new string.
My code for the problem is as follows I've commented out the two for loops, because I felt like there was a more elegant solution to this than what I had, but I didn't want to lose where I was in case another path proved to be more challenging.
using System;
using System.Collections.Generic;
using System.IO;
class Solution {
static void Main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution */
int inputQTY = Int32.Parse(Console.ReadLine());
string input = Console.ReadLine(); // The example gives us the first word to be Hacker then the next word Rank on the next line, so the outputs would be Hce akr, and Rn ak respectively.
int strLen = input.Length;
char[] inputCharArray = input.ToCharArray();
string output = "";
/*
for (int j = 0; j < inputQTY; j++){
for (int i = 0; i < strLen; i++) {
if (j % 2 == 0 && i % 2 == 0) {
Console.WriteLine(inputCharArray[i]);
output = new string (new char[] {inputCharArray[i]});
Console.WriteLine(output);
Console.WriteLine("This is i: {0}", i);
Console.WriteLine("This is j: {0}", j);
Console.WriteLine("--------------");
Console.WriteLine("");
}
else {
Console.WriteLine("This is the next j part hopefully: {0}", j);
}
}
}*/
}
}
Like I understand I need to first step through the word, grab every other letter, then step through the word again and grab the remaining letters, then concatenate those letters into words, and concatenate those words into a sentence, so j would be the loop giving me the two words where I is the loop getting the two words put together..... but I'm having a difficult time wrapping my head around where I'm going wrong here. On top of this, I feel like there's another approach entirely that I'm missing, using commands I may not even know about.
Anyhoo any help is appreciated, hopefully I won't be so green after this. Thanks!
Ok so I ended up solving it with the following code, thank you for all your help everyone!
I ended up solving it with the following code (in C#):
using System;
using System.Collections.Generic;
using System.IO;
using System.Text;
class Solution {
static void Main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution */
int count = Int32.Parse(Console.ReadLine());
for (int k = 0; k < count; k++) {
char[] word = Console.ReadLine().ToCharArray();
StringBuilder sb1 = new StringBuilder();
StringBuilder sb2 = new StringBuilder();
for (int i = 0; i < word.Length; i+=2) {
sb1.Append(word[i]);
}
for (int j = 1; j < word.Length; j+=2) {
sb2.Append(word[j]);
}
Console.WriteLine(sb1 + " " + sb2);
}
}
}

LINQ version, updated to fix index error:
output = $"{new string(s.Where((x,i) => i % 2 == 0).ToArray())} {new string(s.Where((x,i) => i % 2 != 0).ToArray())}";
To explain, you're grabbing every character whose index in the string is evenly divisible by 2 and printing it, then every character in the string whose index is not evenly divisible by 2 and printing it.
Update:
Since I was asked for further explanation. First, here's the full code that runs successfully in the HackerRank challenge:
using System;
using System.Collections.Generic;
using System.Linq;
class Solution
{
static void Main(String[] args)
{
List<string> tests = new List<string>();
var testCount = int.Parse(Console.ReadLine());
for (var i = 0; i < testCount; i++)
{
tests.Add(Console.ReadLine());
}
foreach (var s in tests)
{
Console.WriteLine($"{new string(s.Where((x, i) => i % 2 == 0).ToArray())} {new string(s.Where((x, i) => i % 2 != 0).ToArray())}");
}
}
}
Regarding what each section of the code does:
i % 2 == 0
This is a test to see if a number is evenly divisible by two, or an even number.
s.Where((x,i) => i % 2 == 0)
This says, for the array of characters that make up the string 's', return all characters (the result is an IEnumerable) where that character's index (location in the string) is an even number.
new string(s.Where((x,i) => i % 2 == 0).ToArray())
This says to take that IEnumerable of characters with even numbered indexes and return them to an Array of characters. Then, create a new string out of that array of characters.
For the odd numbers, it's the same, but you use != 0 in the mod.

I used this simple method of appending to two StringBuilder objects
var sb1 = new StringBuilder();
var sb2 = new StringBuilder();
int i = 0;
foreach (char c in input)
{
var sb = (i % 2 == 0 ? sb1 : sb2);
sb.Append(c);
i = i + 1;
}
output = sb1.ToString() + " " + sb2.ToString();

this is the long way..
int count = int.Parse(Console.ReadLine());
for(int k = 0; k < count; k++){
char[] inputChars = Console.ReadLine().ToCharArray();
char[] evenChars = new char[inputChars.Length % 2 == 0 ? inputChars.Length / 2 : (inputChars.Length + 1) / 2];
char[] oddChars = new char[inputChars.Length - evenChars.Length];
int evenIndex=0,oddIndex = 0;
for(int i = 0; i < inputChars.Length;i++)
if(i % 2 == 0)
evenChars[evenIndex++] = inputChars[i];
else
oddChars[oddIndex++] = inputChars[i];
Console.WriteLine(string.Format("{0} {1}",string.Concat(evenChars),string.Concat(oddChars)));
}
an alternative..
int count = int.Parse(Console.ReadLine());
for(int k = 0; k < count; k++){
string input = Console.ReadLine();
Enumerable.Range(0, input.Length)
.OrderBy(o => o % 2 != 0)
.Select(o => {
if(o == 1)
Console.Write(" ");
Console.Write(input[o]);
return input[o];
}).ToArray();
Console.Write("\n");
}

What's the fastest way to remove characters from an alpha-numeric string?

Say we have the following strings that we pass as parameters to the function below:
string sString = "S104";
string sString2 = "AS105";
string sString3 = "ASRVT106";
I want to be able to extract the numbers from the string to place them in an int variable. Is there a quicker and/or more efficient way of removing the letters from the strings than the following code?: (*These strings will be populated dynamically at runtime - they are not assigned values at construction.)
Code:
public GetID(string sCustomTag = null)
{
m_sCustomTag = sCustomTag;
try {
m_lID = Convert.ToInt32(m_sCustomTag); }
catch{
try{
int iSubIndex = 0;
char[] subString = sCustomTag.ToCharArray();
//ITERATE THROUGH THE CHAR ARRAY
for (int i = 0; i < subString.Count(); i++)
{
for (int j = 0; j < 10; j++)
{
if (subString[i] == j)
{
iSubIndex = i;
goto createID;
}
}
}
createID: m_lID = Convert.ToInt32(m_sCustomTag.Substring(iSubIndex));
}
//IF NONE OF THAT WORKS...
catch(Exception e)
{
m_lID = 00000;
throw e;
}
}
}
}
I've done things like this before, but I'm not sure if there's a more efficient way to do it. If it was just going to be a single letter at the beginning, I could just set the subStringIndex to 1 every time, but the users can essentially put in whatever they want. Generally, they will be formatted to a LETTER-then-NUMBER format, but if they don't, or they want to put in multiple letters like sString2 or sString3, then I need to be able to compensate for that. Furthermore, if the user puts in some whacked-out, non-traditional format like string sString 4 = S51A24;, is there a way to just remove any and all letters from the string?
I've looked about, and can't find anything on MSDN or Google. Any help or links to it are greatly appreciated!

You can use a regular expression. It's not necessarily faster, but it's more concise.
string sString = "S104";
string sString2 = "AS105";
string sString3 = "ASRVT106";
var re = new Regex(#"\d+");
Console.WriteLine(re.Match(sString).Value); // 104
Console.WriteLine(re.Match(sString2).Value); // 105
Console.WriteLine(re.Match(sString3).Value); // 106

You can use a Regex, but it's probably faster to just do:
public int ExtractInteger(string str)
{
var sb = new StringBuilder();
for (int i = 0; i < str.Length; i++)
if(Char.IsDigit(str[i])) sb.Append(str[i]);
return int.Parse(sb.ToString());
}
You can simplify further with some LINQ at the expense of a small performance penalty:
public int ExtractInteger(string str)
{
return int.Parse(new String(str.Where(c=>Char.IsDigit(c)).ToArray()));
}
Now, if you only want to parse the first sequence of consecutive digits, do this instead:
public int ExtractInteger(string str)
{
return int.Parse(new String(str.SkipWhile(c=>!Char.IsDigit(c)).TakeWhile(c=>Char.IsDigit(c)).ToArray()));
}

Fastest is to parse the string without removing anything:
var s = "S51A24";
int m_lID = 0;
for (int i = 0; i < s.Length; i++)
{
int d = s[i] - '0';
if ((uint)d < 10)
m_lID = m_lID * 10 + d;
}
Debug.Print(m_lID + ""); // 5124

string removeLetters(string s)
{
for (int i = 0; i < s.Length; i++)
{
char c = s[i];
if (IsEnglishLetter(c))
{
s = s.Remove(i, 1);
}
}
return s;
}
bool IsEnglishLetter(char c)
{
return (c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z');
}

While you asked "what's the fastest way to remove characters..." what you really appear to be asking is "how do I create an integer by extracting only the digits from the string"?
Going with this assumption, your first call to Convert.ToInt32 will be slow for the case where you have other than digits because of the exception throwing.
Let's try another approach. Let's think about each of the cases.
The string always starts with a series of digits (e.g. 123ABC => 123)
The string always ends with a series of digits (e.g. ABC123 => 123)
A string has a series of contiguous digits in the middle (e.g. AB123C ==> 123)
The digits are possibly noncontiguous (e.g. A77C12 => 7712)
Case 4 is the "safest" assumption (after all, it is a superset of Case 1, 2 and 3. So, we need an algorithm for that. As a bonus I'll provide algorithms specialized to the other cases.
The Main Algorithm, All Cases
Using in-place unsafe iteration of the characters of the string, which uses fixed, we can extract digits and convert them to a single number without the data copy in ToCharArray(). We can also avoid the allocations of, say, a StringBuilder implementation and a possibly slow regex solution.
NOTE: This is valid C# code though it's using pointers. It does look like C++, but I assure you it's C#.
public static unsafe int GetNumberForwardFullScan(string s)
{
int value = 0;
fixed (char* pString = s)
{
var pChar = pString;
for (int i = 0; i != s.Length; i++, pChar++)
{
// this just means if the char is not between 0-9, we exit the loop (i.e. stop calculating the integer)
if (*pChar < '0' || *pChar > '9')
continue;
// running recalculation of the integer
value = value * 10 + *pChar - '0';
}
}
return value;
}
Running this against any of the inputs: "AS106RVT", "ASRVT106", "106ASRVT", or "1AS0RVT6" results in pulling out 1, 0, 6 and calculating on each digit as
0*10 + 1 == 1
1*10 + 0 == 10
10*10 + 6 == 106
Case 1 Only Algorithm (Digits at Start of String)
This algorithm is identical to the one above, but instead of continue we can break as soon as we reach a non-digit. This would be much faster if we can assume all the inputs start with digits and the strings are long.
Case 2 Only Algorithm (Digits at End of String)
This is almost the same as Case 1 Only except you have to
iterate from the end of the string to the beginning (aka backwards) stopping on the first non-digit
change the calculation to sum up powers of ten.
Both of those are a bit tricky, so here's what that looks like
public static unsafe int GetNumberBackward(string s)
{
int value = 0;
fixed (char* pString = s)
{
char* pChar = pString + s.Length - 1;
for (int i = 0; i != -1; i++, pChar--)
{
if (*pChar < '0' || *pChar > '9')
break;
value = (*pChar - '0') * (int)Math.Pow(10, i) + value;
}
}
return value;
}
So each of the iteration of the calculation looks like
6*100 + 0 == 6
0*101 + 6 == 6
1*102 + 6 == 106
While I used Math.Pow in these examples, you can find integer only versions that might be faster.
Cases 1-3 Only (i.e. All Digits Contiguous Somewhere in the String
This algorithm says to
Scan all non-digits
Then scan only digits
First non-digit after that, stop
It would look like
public static unsafe int GetContiguousDigits(string s)
{
int value = 0;
fixed (char* pString = s)
{
var pChar = pString;
// skip non-digits
int i = 0;
for (; i != s.Length; i++, pChar++)
if (*pChar >= '0' && *pChar <= '9')
break;
for (; i != s.Length; i++, pChar++)
{
if (*pChar < '0' || *pChar > '9')
break;
value = value * 10 + *pChar - '0';
}
}
return value;
}

Find all possible combinations of word with and without hyphens

For a string that may have zero or more hyphens in it, I need to extract all the different possibilities with and without hyphens.
For example, the string "A-B" would result in "A-B" and "AB" (two possibilities).
The string "A-B-C" would result in "A-B-C", "AB-C", "A-BC" and "ABC" (four possibilities).
The string "A-B-C-D" would result in "A-B-C-D", "AB-C-D", "A-BC-D", "A-B-CD", "AB-CD", "ABC-D", "A-BCD" and "ABCD" (eight possibilities).
...etc, etc.
I've experimented with some nested loops but haven't been able to get anywhere near the desired result. I suspect I need something recursive unless there is some simple solution I am overlooking.
NB. This is to build a SQL query (shame that SQL Server does't have MySQL's REGEXP pattern matching).
Here is one attempt I was working on. This might work if I do this recursively.
string keyword = "A-B-C-D";
List<int> hyphens = new List<int>();
int pos = keyword.IndexOf('-');
while (pos != -1)
{
hyphens.Add(pos);
pos = keyword.IndexOf('-', pos + 1);
}
for (int i = 0; i < hyphens.Count(); i++)
{
string result = keyword.Substring(0, hyphens[i]) + keyword.Substring(hyphens[i] + 1);
Response.Write("<p>" + result);
}
A B C D are words of varying length.

Take a look at your sample cases. Have you noticed a pattern?
With 1 hyphen there are 2 possibilities.
With 2 hyphens there are 4 possibilities.
With 3 hyphens there are 8 possibilities.
The number of possibilities is 2n.
This is literally exponential growth, so if there are too many hyphens in the string, it will quickly become infeasible to print them all. (With just 30 hyphens there are over a billion combinations!)
That said, for smaller numbers of hyphens it might be interesting to generate a list. To do this, you can think of each hyphen as a bit in a binary number. If the bit is 1, the hyphen is present, otherwise it is not. So this suggests a fairly straightforward solution:
Split the original string on the hyphens
Let n = the number of hyphens
Count from 2n - 1 down to 0. Treat this counter as a bitmask.
For each count begin building a string starting with the first part.
Concatenate each of the remaining parts to the string in order, preceded by a hyphen only if the corresponding bit in the bitmask is set.
Add the resulting string to the output and continue until the counter is exhausted.
Translated to code we have:
public static IEnumerable<string> EnumerateHyphenatedStrings(string s)
{
string[] parts = s.Split('-');
int n = parts.Length - 1;
if (n > 30) throw new Exception("too many hyphens");
for (int m = (1 << n) - 1; m >= 0; m--)
{
StringBuilder sb = new StringBuilder(parts[0]);
for (int i = 1; i <= n; i++)
{
if ((m & (1 << (i - 1))) > 0) sb.Append('-');
sb.Append(parts[i]);
}
yield return sb.ToString();
}
}
Fiddle: https://dotnetfiddle.net/ne3N8f

You should be able to track each hyphen position, and basically say its either there or not there. Loop through all the combinations, and you got all your strings. I found the easiest way to track it was using a binary, since its easy to add those with Convert.ToInt32
I came up with this:
string keyword = "A-B-C-D";
string[] keywordSplit = keyword.Split('-');
int combinations = Convert.ToInt32(Math.Pow(2.0, keywordSplit.Length - 1.0));
List<string> results = new List<string>();
for (int j = 0; j < combinations; j++)
{
string result = "";
string hyphenAdded = Convert.ToString(j, 2).PadLeft(keywordSplit.Length - 1, '0');
// Generate string
for (int i = 0; i < keywordSplit.Length; i++)
{
result += keywordSplit[i] +
((i < keywordSplit.Length - 1) && (hyphenAdded[i].Equals('1')) ? "-" : "");
}
results.Add(result);
}

This works for me:
Func<IEnumerable<string>, IEnumerable<string>> expand = null;
expand = xs =>
{
if (xs != null && xs.Any())
{
var head = xs.First();
if (xs.Skip(1).Any())
{
return expand(xs.Skip(1)).SelectMany(tail => new []
{
head + tail,
head + "-" + tail
});
}
else
{
return new [] { head };
}
}
else
{
return Enumerable.Empty<string>();
}
};
var keyword = "A-B-C-D";
var parts = keyword.Split('-');
var results = expand(parts);
I get:
ABCD
A-BCD
AB-CD
A-B-CD
ABC-D
A-BC-D
AB-C-D
A-B-C-D

I've tested this code and it is working as specified in the question. I stored the strings in a List<string>.
string str = "AB-C-D-EF-G-HI";
string[] splitted = str.Split('-');
List<string> finalList = new List<string>();
string temp = "";
for (int i = 0; i < splitted.Length; i++)
{
temp += splitted[i];
}
finalList.Add(temp);
temp = "";
for (int diff = 0; diff < splitted.Length-1; diff++)
{
for (int start = 1, limit = start + diff; limit < splitted.Length; start++, limit++)
{
int i = 0;
while (i < start)
{
temp += splitted[i++];
}
while (i <= limit)
{
temp += "-";
temp += splitted[i++];
}
while (i < splitted.Length)
{
temp += splitted[i++];
}
finalList.Add(temp);
temp = "";
}
}

I'm not sure your question is entirely well defined (i.e. could you have something like A-BCD-EF-G-H?). For "fully" hyphenated strings (A-B-C-D-...-Z), something like this should do:
string toParse = "A-B-C-D";
char[] toParseChars = toPase.toCharArray();
string result = "";
string binary;
for(int i = 0; i < (int)Math.pow(2, toParse.Length/2); i++) { // Number of subsets of an n-elt set is 2^n
binary = Convert.ToString(i, 2);
while (binary.Length < toParse.Length/2) {
binary = "0" + binary;
}
char[] binChars = binary.ToCharArray();
for (int k = 0; k < binChars.Length; k++) {
result += toParseChars[k*2].ToString();
if (binChars[k] == '1') {
result += "-";
}
}
result += toParseChars[toParseChars.Length-1];
Console.WriteLine(result);
}
The idea here is that we want to create a binary word for each possible hyphen. So, if we have A-B-C-D (three hyphens), we create binary words 000, 001, 010, 011, 100, 101, 110, and 111. Note that if we have n hyphens, we need 2^n binary words.
Then each word maps to the output you desire by inserting the hyphen where we have a '1' in our word (000 -> ABCD, 001 -> ABC-D, 010 -> AB-CD, etc). I didn't test the code above, but this is at least one way to solve the problem for fully hyphenated words.
Disclaimer: I didn't actually test the code