I have a string from OCR which contains some errors.
For example "2SQ41S" in place of "250415", i have a dictionary for the possible replacements:
O/Q can be replaced by 0,
S can be replaced by 5...
I can calculate the checksum to be sure that the good word is found.
Here is the function recursive which doesn't work, it will be stopped when startPosition>=6, it's before the correct word was found:
public void CombinaisonTest()
{
string date = "2SO41S";
Dictionary<char, String[]> replaceDictionary= new Dictionary<char, String[]>()
{
{'O', new []{"Q", "0"}},
{'S', new []{"8", "5", "B"}}
};
String result = "";
var r = combinations2(date, 0, replaceDictionary);
Console.WriteLine("Date: " + date);
Console.WriteLine("R: " + r);
}
public string combinations2(string date, int startPosition, Dictionary<char, String[]> dictionary)
{
Console.WriteLine("Call function " + date + ", " + startPosition);
if (string.Join("", date).Equals("250415")) //need to calculate checksum
{
Console.WriteLine("Found: " + date);
return date;
}
if (startPosition >= date.Length)
{
Console.WriteLine("Not Found: ");
return "";
}
for (int i = startPosition; i < date.Length; i++)
{
if (dictionary.ContainsKey(date.ToCharArray()[i]))
{
foreach (var value in dictionary[date.ToCharArray()[i]])
{
return combinations2(date.Remove(i, 1).Insert(i, value), startPosition + 1, dictionary);
}
}
else
{
return combinations2(date, i + 1, dictionary);
}
}
return combinations2(date, startPosition + 1, dictionary);
}
Do you have any ideas for the corrections, please?
Thank you.
There are a couple of issues with the code. The first is that when iterating through the values in the dictionary, it returns after checking the first one, so it will only ever try and substitute Q for 0 and 8 for S. The second is that you are attempting two methods of processing the characters in the string: iterative AND recursive. You don't need to iterate over the index i with a for loop and also use recursion.
Another issue (which isn't a problem in your use case but stops the algorithm being more generic) is that the algorithm attempts to do a substitution in every case where an ambiguous character is encountered, as well as iterating over each value in the dictionary you should also consider the case where the character is left unmodified.
The function can be changed to remove the outer for loop and iterate over the values in the dictionary, test each one (recursing over the remainder of the string) and only return if a match is found. A simple way to do this is to store the result in a string and only return it if it is not the empty string (since your function returns the empty string when no match is found). If all the values in the dictionary have been tried and no match has been found, then it tries recursing without modifying the string.
public string combinations2(string date, int startPosition, Dictionary<char, String[]> dictionary)
{
Console.WriteLine("Call function " + date + ", " + startPosition);
if (string.Join("", date).Equals("250415")) //need to calculate checksum
{
Console.WriteLine("Found: " + date);
return date;
}
if (startPosition >= date.Length)
{
Console.WriteLine("Not Found: ");
return "";
}
if (dictionary.ContainsKey(date.ToCharArray()[startPosition]))
{
foreach (var value in dictionary[date.ToCharArray()[startPosition]])
{
string result = combinations2(date.Remove(startPosition, 1).Insert(startPosition, value), startPosition + 1, dictionary);
if(result != "")
return result;
}
}
return combinations2(date, startPosition + 1, dictionary);
}
I'm creating a WebSocket protocol of my own, and thought to have a text key/value header part, ending at two consecutive newlines, followed by a binary tail.
Turns out, splitting a ByteString in half (at the two newlines) is really tedious. There is no built-in .split method, for one. And no .indexOf for finding a binary fingerprint.
What would you use for this? Is there an easier way for me to build such a protocol?
References:
akka ByteString
Using akka-http 10.1.0-RC1, akka 2.5.8
One approach would be to first create sliding pairs from an indexedSeq of the ByteString, then split the ByteString using the identified indexes of the delimiter-pair, as in the following example:
import akka.util.ByteString
val bs = ByteString("aa\nbb\n\nxyz")
// bs: akka.util.ByteString = ByteString(97, 97, 10, 98, 98, 10, 10, 120, 121, 122)
val delimiter = 10
// Create sliding pairs from indexedSeq of the ByteString
val slidingList = bs.zipWithIndex.sliding(2).toList
// slidingList: List[scala.collection.immutable.IndexedSeq[(Byte, Int)]] = List(
// Vector((97,0), (97,1)), Vector((97,1), (10,2)), Vector((10,2), (98,3)),
// Vector((98,3), (98,4)), Vector((98,4), (10,5)), Vector((10,5), (10,6)),
// Vector((10,6), (120,7)), Vector((120,7), (121,8)), Vector((121,8), (122,9))
// )
// Get indexes of the delimiter-pair
val dIndex = slidingList.filter{
case Vector(x, y) => x._1 == delimiter && y._1 == delimiter
}.flatMap{
case Vector(x, y) => Seq(x._2, y._2)
}
// Split the ByteString list
val (bs1, bs2) = ( bs.splitAt(dIndex(0))._1, bs.splitAt(dIndex(1))._2.tail )
// bs1: akka.util.ByteString = ByteString(97, 97, 10, 98, 98)
// bs2: akka.util.ByteString = ByteString(120, 121, 122)
I came up with this. Haven't tested it in practise, yet.
#tailrec
def peelMsg(bs: ByteString, accHeaderLines: Seq[String]): Tuple2[Seq[String],ByteString] = {
val (a: ByteString, tail: ByteString) = bs.span(_ != '\n')
val b: ByteString = tail.drop(1)
if (a.isEmpty) { // end marker - empty line
Tuple2(accHeaderLines,b)
} else {
val acc: Seq[String] = accHeaderLines :+ a.utf8String // append
peelMsg(b,acc)
}
}
val (headerLines: Seq[String], value: ByteString) = peelMsg(bs,Seq.empty)
My code, for now:
// Find the index of the (first) double-newline
//
val n: Int = {
val bsLen: Int = bs.length
val tmp: Int = bs.zipWithIndex.find{
case ('\n',i) if i<bsLen-1 && bs(i+1)=='\n' => true
case _ => false
}.map(_._2).getOrElse{
throw new RuntimeException("No delimiter found")
}
tmp
}
val (bs1: ByteString, bs2: ByteString) = bs.splitAt(n) // headers, \n\n<binary>
Influenced by #leo-c's answer, but using a normal .find instead of the sliding window. Realised that since ByteString allows random access, I can combine a streaming search with that condition.
this is my line of code.
budgetLabel.text = String((budgetLabel.text)!.toInt()! - (budgetItemTextBox.text)!.toInt()!)
the code works, but when I try to input a floating value into the textbox the program crashes. I am assuming the strings need to be converted to a float/double data type. I keep getting errors when i try to do that.
In Swift 2 there are new failable initializers that allow you to do this in more safe way, the Double("") returns an optional in cases like passing in "abc" string the failable initializer will return nil, so then you can use optional-binding to handle it like in the following way:
let s1 = "4.55"
let s2 = "3.15"
if let n1 = Double(s1), let n2 = Double(s2) {
let newString = String( n1 - n2)
print(newString)
}
else {
print("Some string is not a double value")
}
If you're using a version of Swift < 2, then old way was:
var n1 = ("9.99" as NSString).doubleValue // invalid returns 0, not an optional. (not recommended)
// invalid returns an optional value (recommended)
var pi = NSNumberFormatter().numberFromString("3.14")?.doubleValue
Fixed: Added Proper Handling for Optionals
let budgetLabel:UILabel = UILabel()
let budgetItemTextBox:UITextField = UITextField()
budgetLabel.text = ({
var value = ""
if let budgetString = budgetLabel.text, let budgetItemString = budgetItemTextBox.text
{
if let budgetValue = Float(budgetString), let budgetItemValue = Float(budgetItemString)
{
value = String(budgetValue - budgetItemValue)
}
}
return value
})()
You need to be using if let. In swift 2.0 it would look something like this:
if let
budgetString:String = budgetLabel.text,
budgetItemString:String = budgetItemTextBox.text,
budget:Double = Double(budgetString),
budgetItem:Double = Double(budgetItemString) {
budgetLabel.text = String(budget - budgetItem)
} else {
// If a number was not found, what should it do here?
}
i would like to explain my problem by the following example.
assume the word: abc
a has variants: ä, à
b has no variants.
c has variants: ç
so the possible words are:
abc
äbc
àbc
abç
äbç
àbç
now i am looking for the algorithm that prints all word variantions for abritray words with arbitray lettervariants.
I would recommend you to solve this recursively. Here's some Java code for you to get started:
static Map<Character, char[]> variants = new HashMap<Character, char[]>() {{
put('a', new char[] {'ä', 'à'});
put('b', new char[] { });
put('c', new char[] { 'ç' });
}};
public static Set<String> variation(String str) {
Set<String> result = new HashSet<String>();
if (str.isEmpty()) {
result.add("");
return result;
}
char c = str.charAt(0);
for (String tailVariant : variation(str.substring(1))) {
result.add(c + tailVariant);
for (char variant : variants.get(c))
result.add(variant + tailVariant);
}
return result;
}
Test:
public static void main(String[] args) {
for (String str : variation("abc"))
System.out.println(str);
}
Output:
abc
àbç
äbc
àbc
äbç
abç
A quickly hacked solution in Python:
def word_variants(variants):
print_variants("", 1, variants);
def print_variants(word, i, variants):
if i > len(variants):
print word
else:
for variant in variants[i]:
print_variants(word + variant, i + 1, variants)
variants = dict()
variants[1] = ['a0', 'a1', 'a2']
variants[2] = ['b0']
variants[3] = ['c0', 'c1']
word_variants(variants)
Common part:
string[] letterEquiv = { "aäà", "b", "cç", "d", "eèé" };
// Here we make a dictionary where the key is the "base" letter and the value is an array of alternatives
var lookup = letterEquiv
.Select(p => p.ToCharArray())
.SelectMany(p => p, (p, q) => new { key = q, values = p }).ToDictionary(p => p.key, p => p.values);
A recursive variation written in C#.
List<string> resultsRecursive = new List<string>();
// I'm using an anonymous method that "closes" around resultsRecursive and lookup. You could make it a standard method that accepts as a parameter the two.
// Recursive anonymous methods must be declared in this way in C#. Nothing to see.
Action<string, int, char[]> recursive = null;
recursive = (str, ix, str2) =>
{
// In the first loop str2 is null, so we create the place where the string will be built.
if (str2 == null)
{
str2 = new char[str.Length];
}
// The possible variations for the current character
var equivs = lookup[str[ix]];
// For each variation
foreach (var eq in equivs)
{
// We save the current variation for the current character
str2[ix] = eq;
// If we haven't reached the end of the string
if (ix < str.Length - 1)
{
// We recurse, increasing the index
recursive(str, ix + 1, str2);
}
else
{
// We save the string
resultsRecursive.Add(new string(str2));
}
}
};
// We launch our function
recursive("abcdeabcde", 0, null);
// The results are in resultsRecursive
A non-recursive version
List<string> resultsNonRecursive = new List<string>();
// I'm using an anonymous method that "closes" around resultsNonRecursive and lookup. You could make it a standard method that accepts as a parameter the two.
Action<string> nonRecursive = (str) =>
{
// We will have two arrays, of the same length of the string. One will contain
// the possible variations for that letter, the other will contain the "current"
// "chosen" variation of that letter
char[][] equivs = new char[str.Length][];
int[] ixes = new int[str.Length];
for (int i = 0; i < ixes.Length; i++)
{
// We start with index -1 so that the first increase will bring it to 0
equivs[i] = lookup[str[i]];
ixes[i] = -1;
}
// The current "workin" index of the original string
int ix = 0;
// The place where the string will be built.
char[] str2 = new char[str.Length];
// The loop will break when we will have to increment the letter with index -1
while (ix >= 0)
{
// We select the next possible variation for the current character
ixes[ix]++;
// If we have exausted the possible variations of the current character
if (ixes[ix] == equivs[ix].Length)
{
// Reset the current character to -1
ixes[ix] = -1;
// And loop back to the previous character
ix--;
continue;
}
// We save the current variation for the current character
str2[ix] = equivs[ix][ixes[ix]];
// If we are setting the last character of the string, then the string
// is complete
if (ix == str.Length - 1)
{
// And we save it
resultsNonRecursive.Add(new string(str2));
}
else
{
// Otherwise we have to do everything for the next character
ix++;
}
}
};
// We launch our function
nonRecursive("abcdeabcde");
// The results are in resultsNonRecursive
Both heavily commented.
What is the best way to join a list of strings into a combined delimited string. I'm mainly concerned about when to stop adding the delimiter. I'll use C# for my examples but I would like this to be language agnostic.
EDIT: I have not used StringBuilder to make the code slightly simpler.
Use a For Loop
for(int i=0; i < list.Length; i++)
{
result += list[i];
if(i != list.Length - 1)
result += delimiter;
}
Use a For Loop setting the first item previously
result = list[0];
for(int i = 1; i < list.Length; i++)
result += delimiter + list[i];
These won't work for an IEnumerable where you don't know the length of the list beforehand so
Using a foreach loop
bool first = true;
foreach(string item in list)
{
if(!first)
result += delimiter;
result += item;
first = false;
}
Variation on a foreach loop
From Jon's solution
StringBuilder builder = new StringBuilder();
string delimiter = "";
foreach (string item in list)
{
builder.Append(delimiter);
builder.Append(item);
delimiter = ",";
}
return builder.ToString();
Using an Iterator
Again from Jon
using (IEnumerator<string> iterator = list.GetEnumerator())
{
if (!iterator.MoveNext())
return "";
StringBuilder builder = new StringBuilder(iterator.Current);
while (iterator.MoveNext())
{
builder.Append(delimiter);
builder.Append(iterator.Current);
}
return builder.ToString();
}
What other algorithms are there?
It's impossible to give a truly language-agnostic answer here as different languages and platforms handle strings differently, and provide different levels of built-in support for joining lists of strings. You could take pretty much identical code in two different languages, and it would be great in one and awful in another.
In C#, you could use:
StringBuilder builder = new StringBuilder();
string delimiter = "";
foreach (string item in list)
{
builder.Append(delimiter);
builder.Append(item);
delimiter = ",";
}
return builder.ToString();
This will prepend a comma on all but the first item. Similar code would be good in Java too.
EDIT: Here's an alternative, a bit like Ian's later answer but working on a general IEnumerable<string>.
// Change to IEnumerator for the non-generic IEnumerable
using (IEnumerator<string> iterator = list.GetEnumerator())
{
if (!iterator.MoveNext())
{
return "";
}
StringBuilder builder = new StringBuilder(iterator.Current);
while (iterator.MoveNext())
{
builder.Append(delimiter);
builder.Append(iterator.Current);
}
return builder.ToString();
}
EDIT nearly 5 years after the original answer...
In .NET 4, string.Join was overloaded pretty significantly. There's an overload taking IEnumerable<T> which automatically calls ToString, and there's an overload for IEnumerable<string>. So you don't need the code above any more... for .NET, anyway.
In .NET, you can use the String.Join method:
string concatenated = String.Join(",", list.ToArray());
Using .NET Reflector, we can find out how it does it:
public static unsafe string Join(string separator, string[] value, int startIndex, int count)
{
if (separator == null)
{
separator = Empty;
}
if (value == null)
{
throw new ArgumentNullException("value");
}
if (startIndex < 0)
{
throw new ArgumentOutOfRangeException("startIndex", Environment.GetResourceString("ArgumentOutOfRange_StartIndex"));
}
if (count < 0)
{
throw new ArgumentOutOfRangeException("count", Environment.GetResourceString("ArgumentOutOfRange_NegativeCount"));
}
if (startIndex > (value.Length - count))
{
throw new ArgumentOutOfRangeException("startIndex", Environment.GetResourceString("ArgumentOutOfRange_IndexCountBuffer"));
}
if (count == 0)
{
return Empty;
}
int length = 0;
int num2 = (startIndex + count) - 1;
for (int i = startIndex; i <= num2; i++)
{
if (value[i] != null)
{
length += value[i].Length;
}
}
length += (count - 1) * separator.Length;
if ((length < 0) || ((length + 1) < 0))
{
throw new OutOfMemoryException();
}
if (length == 0)
{
return Empty;
}
string str = FastAllocateString(length);
fixed (char* chRef = &str.m_firstChar)
{
UnSafeCharBuffer buffer = new UnSafeCharBuffer(chRef, length);
buffer.AppendString(value[startIndex]);
for (int j = startIndex + 1; j <= num2; j++)
{
buffer.AppendString(separator);
buffer.AppendString(value[j]);
}
}
return str;
}
There's little reason to make it language-agnostic when some languages provide support for this in one line, e.g., Python's
",".join(sequence)
See the join documentation for more info.
For python be sure you have a list of strings, else ','.join(x) will fail.
For a safe method using 2.5+
delimiter = '","'
delimiter.join(str(a) if a else '' for a in list_object)
The "str(a) if a else ''" is good for None types otherwise str() ends up making then 'None' which isn't nice ;)
In PHP's implode():
$string = implode($delim, $array);
I'd always add the delimeter and then remove it at the end if necessary. This way, you're not executing an if statement for every iteration of the loop when you only care about doing the work once.
StringBuilder sb = new StringBuilder();
foreach(string item in list){
sb.Append(item);
sb.Append(delimeter);
}
if (list.Count > 0) {
sb.Remove(sb.Length - delimter.Length, delimeter.Length)
}
I would express this recursively.
Check if the number of string arguments is 1. If it is, return it.
Otherwise recurse, but combine the first two arguments with the delimiter between them.
Example in Common Lisp:
(defun join (delimiter &rest strings)
(if (null (rest strings))
(first strings)
(apply #'join
delimiter
(concatenate 'string
(first strings)
delimiter
(second strings))
(cddr strings))))
The more idiomatic way is to use reduce, but this expands to almost exactly the same instructions as the above:
(defun join (delimiter &rest strings)
(reduce (lambda (a b)
(concatenate 'string a delimiter b))
strings))
List<string> aaa = new List<string>{ "aaa", "bbb", "ccc" };
string mm = ";";
return aaa.Aggregate((a, b) => a + mm + b);
and you get
aaa;bbb;ccc
lambda is pretty handy
In C# you can just use String.Join(separator,string_list)
The problem is that computer languages rarely have string booleans, that is, methods that are of type string that do anything useful. SQL Server at least has is[not]null and nullif, which when combined solve the delimiter problem, by the way: isnotnull(nullif(columnvalue, ""),"," + columnvalue))
The problem is that in languages there are booleans, and there are strings, and never the twain shall meet except in ugly coding forms, e.g.
concatstring = string1 + "," + string2;
if (fubar)
concatstring += string3
concatstring += string4 etc
I've tried mightily to avoid all this ugliness, playing comma games and concatenating with joins, but I'm still left with some of it, including SQL Server errors when I've missed one of the commas and a variable is empty.
Jonathan
Since you tagged this language agnostic,
This is how you would do it in python
# delimiter can be multichar like "| trlalala |"
delimiter = ";"
# sequence can be any list, or iterator/generator that returns list of strings
result = delimiter.join(sequence)
#result will NOT have ending delimiter
Edit: I see I got beat to the answer by several people. Sorry for dupication
I thint the best way to do something like that is (I'll use pseudo-code, so we'll make it truly language agnostic):
function concat(<array> list, <boolean> strict):
for i in list:
if the length of i is zero and strict is false:
continue;
if i is not the first element:
result = result + separator;
result = result + i;
return result;
the second argument to concat(), strict, is a flag to know if eventual empty strings have to be considered in concatenation or not.
I'm used to not consider appending a final separator; on the other hand, if strict is false the resulting string could be free of stuff like "A,B,,,F", provided the separator is a comma, but would instead present as "A,B,F".
that's how python solves the problem:
','.join(list_of_strings)
I've never could understand the need for 'algorithms' in trivial cases though
This is a Working solution in C#, in Java, you can use similar for each on iterator.
string result = string.Empty;
// use stringbuilder at some stage.
foreach (string item in list)
result += "," + item ;
result = result.Substring(1);
// output: "item,item,item"
If using .NET, you might want to use extension method so that you can do
list.ToString(",")
For details, check out Separator Delimited ToString for Array, List, Dictionary, Generic IEnumerable
// contains extension methods, it must be a static class.
public static class ExtensionMethod
{
// apply this extension to any generic IEnumerable object.
public static string ToString<T>(this IEnumerable<T> source,
string separator)
{
if (source == null)
throw new ArgumentException("source can not be null.");
if (string.IsNullOrEmpty(separator))
throw new ArgumentException("separator can not be null or empty.");
// A LINQ query to call ToString on each elements
// and constructs a string array.
string[] array =
(from s in source
select s.ToString()
).ToArray();
// utilise builtin string.Join to concate elements with
// customizable separator.
return string.Join(separator, array);
}
}
EDIT:For performance reasons, replace the concatenation code with string builder solution that mentioned within this thread.
Seen the Python answer like 3 times, but no Ruby?!?!?
the first part of the code declares a new array. Then you can just call the .join() method and pass the delimiter and it will return a string with the delimiter in the middle. I believe the join method calls the .to_s method on each item before it concatenates.
["ID", "Description", "Active"].join(",")
>> "ID, Description, Active"
this can be very useful when combining meta-programming with with database interaction.
does anyone know if c# has something similar to this syntax sugar?
In Java 8 we can use:
List<String> list = Arrays.asList(new String[] { "a", "b", "c" });
System.out.println(String.join(",", list)); //Output: a,b,c
To have a prefix and suffix we can do
StringJoiner joiner = new StringJoiner(",", "{", "}");
list.forEach(x -> joiner.add(x));
System.out.println(joiner.toString()); //Output: {a,b,c}
Prior to Java 8 you can do like Jon's answer
StringBuilder sb = new StringBuilder(prefix);
boolean and = false;
for (E e : iterable) {
if (and) {
sb.append(delimiter);
}
sb.append(e);
and = true;
}
sb.append(suffix);
In .NET, I would use the String.join method if possible, which allows you to specify a separator and a string array. A list can be converted to an array with ToArray, but I don't know what the performance hit of that would be.
The three algorithms that you mention are what I would use (I like the second because it does not have an if statement in it, but if the length is not known I would use the third because it does not duplicate the code). The second will only work if the list is not empty, so that might take another if statement.
A fourth variant might be to put a seperator in front of every element that is concatenated and then remove the first separator from the result.
If you do concatenate strings in a loop, note that for non trivial cases the use of a stringbuilder will vastly outperform repeated string concatenations.
You could write your own method AppendTostring(string, delimiter) that appends the delimiter if and only if the string is not empty. Then you just call that method in any loop without having to worry when to append and when not to append.
Edit: better yet of course to use some kind of StringBuffer in the method if available.
string result = "";
foreach(string item in list)
{
result += delimiter + item;
}
result = result.Substring(1);
Edit: Of course, you wouldn't use this or any one of your algorithms to concatenate strings. With C#/.NET, you'd probably use a StringBuilder:
StringBuilder sb = new StringBuilder();
foreach(string item in list)
{
sb.Append(delimiter);
sb.Append(item);
}
string result = sb.ToString(1, sb.Length-1);
And a variation of this solution:
StringBuilder sb = new StringBuilder(list[0]);
for (int i=1; i<list.Count; i++)
{
sb.Append(delimiter);
sb.Append(list[i]);
}
string result = sb.ToString();
Both solutions do not include any error checks.
From http://dogsblog.softwarehouse.co.zw/post/2009/02/11/IEnumerable-to-Comma-Separated-List-(and-more).aspx
A pet hate of mine when developing is making a list of comma separated ids, it is SO simple but always has ugly code.... Common solutions are to loop through and put a comma after each item then remove the last character, or to have an if statement to check if you at the begining or end of the list. Below is a solution you can use on any IEnumberable ie a List, Array etc. It is also the most efficient way I can think of doing it as it relies on assignment which is better than editing a string or using an if.
public static class StringExtensions
{
public static string Splice<T>(IEnumerable<T> args, string delimiter)
{
StringBuilder sb = new StringBuilder();
string d = "";
foreach (T t in args)
{
sb.Append(d);
sb.Append(t.ToString());
d = delimiter;
}
return sb.ToString();
}
}
Now it can be used with any IEnumerable eg.
StringExtensions.Splice(billingTransactions.Select(t => t.id), ",")
to give us 31,32,35
For java a very complete answer has been given in this question or this question.
That is use StringUtils.join in Apache Commons
String result = StringUtils.join(list, ", ");
In Clojure, you could just use clojure.contrib.str-utils/str-join:
(str-join ", " list)
But for the actual algorithm:
(reduce (fn [res cur] (str res ", " cur)) list)
Groovy also has a String Object.join(String) method.
Java (from Jon's solution):
StringBuilder sb = new StringBuilder();
String delimiter = "";
for (String item : items) {
sb.append(delimiter).append(item);
delimeter = ", ";
}
return sb.toString();
Here is my humble try;
public static string JoinWithDelimiter(List<string> words, string delimiter){
string joinedString = "";
if (words.Count() > 0)
{
joinedString = words[0] + delimiter;
for (var i = 0; i < words.Count(); i++){
if (i > 0 && i < words.Count()){
if (joinedString.Length > 0)
{
joinedString += delimiter + words[i] + delimiter;
} else {
joinedString += words[i] + delimiter;
}
}
}
}
return joinedString;
}
Usage;
List<string> words = new List<string>(){"my", "name", "is", "Hari"};
Console.WriteLine(JoinWithDelimiter(words, " "));