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How to compare number with all number in List not loop List?

I have a problem in compare number with all item in List.
Because in bellow code. You can see when i = 1, the if-else statement will fail. Because it is foreach all values.
First values are 0. I don't want this case.
Have any method to compare i variable to with all values?
Thanks.
List<int> values = new List<int>();
private void CheckCondition()
{
bool flag = false;
for (int i = 0; i < 5; i++)
{
foreach (int num in values)
{
if (i == num || i<num)
continue;
else
{
values.Add(i);
flag = true;
break;
}
}
if (flag)
break;
else
continue;
}
}

This code will look for each value of i in your collection of values
List<int> values = new List<int> {1,2,3,4,5};
for(int i = 0; i< num; i++)
{
if(values.Contains(i))
{
//Do the code you want
}
}
If you want to check some other condition regarding the list and the i you can:
List<int> values = new List<int> {1,2,3,4,5};
for(int i = 0; i< num; i++)
{
if(values.Any(item => item == i || item > i))
{
//Do the code you want
}
}

Do you mean you want to add i to values if there is no i in that list?
Use FindIndex to check if list contains i it will return the index. If it is not, it will return -1.
List<int> values = new List<int>();
private void CheckCondition()
{
for (int i = 0; i < 5; i++)
{
int idx = values.FindIndex(item => item == i || item > i);
if(idx == -1)
{
values.Add(i);
break;
}
}
}

Describe what you want. I do can answer youre question regarding to check if an item is in the List
List<int> values = new List<int> {1, 2, 3, 4};
bool containsItem = values.Contains(2);

c# list permutations with limited length

I have a list of Offers, from which I want to create "chains" (e.g. permutations) with limited chain lengths.
I've gotten as far as creating the permutations using the Kw.Combinatorics project.
However, the default behavior creates permutations in the length of the list count. I'm not sure how to limit the chain lengths to 'n'.
Here's my current code:
private static List<List<Offers>> GetPerms(List<Offers> list, int chainLength)
{
List<List<Offers>> response = new List<List<Offers>>();
foreach (var row in new Permutation(list.Count).GetRows())
{
List<Offers> innerList = new List<Offers>();
foreach (var mix in Permutation.Permute(row, list))
{
innerList.Add(mix);
}
response.Add(innerList);
innerList = new List<Offers>();
}
return response;
}
Implemented by:
List<List<AdServer.Offers>> lst = GetPerms(offers, 2);
I'm not locked in KWCombinatorics if someone has a better solution to offer.

Here's another implementation which I think should be faster than the accepted answer (and it's definitely less code).
public static IEnumerable<IEnumerable<T>> GetVariationsWithoutDuplicates<T>(IList<T> items, int length)
{
if (length == 0 || !items.Any()) return new List<List<T>> { new List<T>() };
return from item in items.Distinct()
from permutation in GetVariationsWithoutDuplicates(items.Where(i => !EqualityComparer<T>.Default.Equals(i, item)).ToList(), length - 1)
select Prepend(item, permutation);
}
public static IEnumerable<IEnumerable<T>> GetVariations<T>(IList<T> items, int length)
{
if (length == 0 || !items.Any()) return new List<List<T>> { new List<T>() };
return from item in items
from permutation in GetVariations(Remove(item, items).ToList(), length - 1)
select Prepend(item, permutation);
}
public static IEnumerable<T> Prepend<T>(T first, IEnumerable<T> rest)
{
yield return first;
foreach (var item in rest) yield return item;
}
public static IEnumerable<T> Remove<T>(T item, IEnumerable<T> from)
{
var isRemoved = false;
foreach (var i in from)
{
if (!EqualityComparer<T>.Default.Equals(item, i) || isRemoved) yield return i;
else isRemoved = true;
}
}
On my 3.1 GHz Core 2 Duo, I tested with this:
public static void Test(Func<IList<int>, int, IEnumerable<IEnumerable<int>>> getVariations)
{
var max = 11;
var timer = System.Diagnostics.Stopwatch.StartNew();
for (int i = 1; i < max; ++i)
for (int j = 1; j < i; ++j)
getVariations(MakeList(i), j).Count();
timer.Stop();
Console.WriteLine("{0,40}{1} ms", getVariations.Method.Name, timer.ElapsedMilliseconds);
}
// Make a list that repeats to guarantee we have duplicates
public static IList<int> MakeList(int size)
{
return Enumerable.Range(0, size/2).Concat(Enumerable.Range(0, size - size/2)).ToList();
}
Unoptimized
GetVariations 11894 ms
GetVariationsWithoutDuplicates 9 ms
OtherAnswerGetVariations 22485 ms
OtherAnswerGetVariationsWithDuplicates 243415 ms
With compiler optimizations
GetVariations 9667 ms
GetVariationsWithoutDuplicates 8 ms
OtherAnswerGetVariations 19739 ms
OtherAnswerGetVariationsWithDuplicates 228802 ms

You're not looking for a permutation, but for a variation. Here is a possible algorithm. I prefer iterator methods for functions that can potentially return very many elements. This way, the caller can decide if he really needs all elements:
IEnumerable<IList<T>> GetVariations<T>(IList<T> offers, int length)
{
var startIndices = new int[length];
var variationElements = new HashSet<T>(); //for duplicate detection
while (startIndices[0] < offers.Count)
{
var variation = new List<T>(length);
var valid = true;
for (int i = 0; i < length; ++i)
{
var element = offers[startIndices[i]];
if (variationElements.Contains(element))
{
valid = false;
break;
}
variation.Add(element);
variationElements.Add(element);
}
if (valid)
yield return variation;
//Count up the indices
startIndices[length - 1]++;
for (int i = length - 1; i > 0; --i)
{
if (startIndices[i] >= offers.Count)
{
startIndices[i] = 0;
startIndices[i - 1]++;
}
else
break;
}
variationElements.Clear();
}
}
The idea for this algorithm is to use a number in offers.Count base. For three offers, all digits are in the range 0-2. We then basically increment this number step by step and return the offers that reside at the specified indices. If you want to allow duplicates, you can remove the check and the HashSet<T>.
Update
Here is an optimized variant that does the duplicate check on the index level. In my tests it is a lot faster than the previous variant:
IEnumerable<IList<T>> GetVariations<T>(IList<T> offers, int length)
{
var startIndices = new int[length];
for (int i = 0; i < length; ++i)
startIndices[i] = i;
var indices = new HashSet<int>(); // for duplicate check
while (startIndices[0] < offers.Count)
{
var variation = new List<T>(length);
for (int i = 0; i < length; ++i)
{
variation.Add(offers[startIndices[i]]);
}
yield return variation;
//Count up the indices
AddOne(startIndices, length - 1, offers.Count - 1);
//duplicate check
var check = true;
while (check)
{
indices.Clear();
for (int i = 0; i <= length; ++i)
{
if (i == length)
{
check = false;
break;
}
if (indices.Contains(startIndices[i]))
{
var unchangedUpTo = AddOne(startIndices, i, offers.Count - 1);
indices.Clear();
for (int j = 0; j <= unchangedUpTo; ++j )
{
indices.Add(startIndices[j]);
}
int nextIndex = 0;
for(int j = unchangedUpTo + 1; j < length; ++j)
{
while (indices.Contains(nextIndex))
nextIndex++;
startIndices[j] = nextIndex++;
}
break;
}
indices.Add(startIndices[i]);
}
}
}
}
int AddOne(int[] indices, int position, int maxElement)
{
//returns the index of the last element that has not been changed
indices[position]++;
for (int i = position; i > 0; --i)
{
if (indices[i] > maxElement)
{
indices[i] = 0;
indices[i - 1]++;
}
else
return i;
}
return 0;
}

If I got you correct here is what you need
this will create permutations based on the specified chain limit
public static List<List<T>> GetPerms<T>(List<T> list, int chainLimit)
{
if (list.Count() == 1)
return new List<List<T>> { list };
return list
.Select((outer, outerIndex) =>
GetPerms(list.Where((inner, innerIndex) => innerIndex != outerIndex).ToList(), chainLimit)
.Select(perms => (new List<T> { outer }).Union(perms).Take(chainLimit)))
.SelectMany<IEnumerable<IEnumerable<T>>, List<T>>(sub => sub.Select<IEnumerable<T>, List<T>>(s => s.ToList()))
.Distinct(new PermComparer<T>()).ToList();
}
class PermComparer<T> : IEqualityComparer<List<T>>
{
public bool Equals(List<T> x, List<T> y)
{
return x.SequenceEqual(y);
}
public int GetHashCode(List<T> obj)
{
return (int)obj.Average(o => o.GetHashCode());
}
}
and you'll call it like this
List<List<AdServer.Offers>> lst = GetPerms<AdServer.Offers>(offers, 2);
I made this function is pretty generic so you may use it for other purpose too
eg
List<string> list = new List<string>(new[] { "apple", "banana", "orange", "cherry" });
List<List<string>> perms = GetPerms<string>(list, 2);
result

How to get the most common value in an Int array? (C#)

How to get the most common value in an Int array using C#
eg: Array has the following values: 1, 1, 1, 2
Ans should be 1

var query = (from item in array
group item by item into g
orderby g.Count() descending
select new { Item = g.Key, Count = g.Count() }).First();
For just the value and not the count, you can do
var query = (from item in array
group item by item into g
orderby g.Count() descending
select g.Key).First();
Lambda version on the second:
var query = array.GroupBy(item => item).OrderByDescending(g => g.Count()).Select(g => g.Key).First();

Some old fashioned efficient looping:
var cnt = new Dictionary<int, int>();
foreach (int value in theArray) {
if (cnt.ContainsKey(value)) {
cnt[value]++;
} else {
cnt.Add(value, 1);
}
}
int mostCommonValue = 0;
int highestCount = 0;
foreach (KeyValuePair<int, int> pair in cnt) {
if (pair.Value > highestCount) {
mostCommonValue = pair.Key;
highestCount = pair.Value;
}
}
Now mostCommonValue contains the most common value, and highestCount contains how many times it occured.

I know this post is old, but someone asked me the inverse of this question today.
LINQ Grouping
sourceArray.GroupBy(value => value).OrderByDescending(group => group.Count()).First().First();
Temp Collection, similar to Guffa's:
var counts = new Dictionary<int, int>();
foreach (var i in sourceArray)
{
if (!counts.ContainsKey(i)) { counts.Add(i, 0); }
counts[i]++;
}
return counts.OrderByDescending(kv => kv.Value).First().Key;

public static int get_occure(int[] a)
{
int[] arr = a;
int c = 1, maxcount = 1, maxvalue = 0;
int result = 0;
for (int i = 0; i < arr.Length; i++)
{
maxvalue = arr[i];
for (int j = 0; j <arr.Length; j++)
{
if (maxvalue == arr[j] && j != i)
{
c++;
if (c > maxcount)
{
maxcount = c;
result = arr[i];
}
}
else
{
c=1;
}
}
}
return result;
}

Maybe O(n log n), but fast:
sort the array a[n]
// assuming n > 0
int iBest = -1; // index of first number in most popular subset
int nBest = -1; // popularity of most popular number
// for each subset of numbers
for(int i = 0; i < n; ){
int ii = i; // ii = index of first number in subset
int nn = 0; // nn = count of numbers in subset
// for each number in subset, count it
for (; i < n && a[i]==a[ii]; i++, nn++ ){}
// if the subset has more numbers than the best so far
// remember it as the new best
if (nBest < nn){nBest = nn; iBest = ii;}
}
// print the most popular value and how popular it is
print a[iBest], nBest

Yet another solution with linq:
static int[] GetMostCommonIntegers(int[] nums)
{
return nums
.ToLookup(n => n)
.ToLookup(l => l.Count(), l => l.Key)
.OrderBy(l => l.Key)
.Last()
.ToArray();
}
This solution can handle case when several numbers have the same number of occurences:
[1,4,5,7,1] => [1]
[1,1,2,2,3,4,5] => [1,2]
[6,6,6,2,2,1] => [6]

How to get all subsets of an array?

Given an array: [dog, cat, mouse]
what is the most elegant way to create:
[,,]
[,,mouse]
[,cat,]
[,cat,mouse]
[dog,,]
[dog,,mouse]
[dog,cat,]
[dog,cat,mouse]
I need this to work for any sized array.
This is essentially a binary counter, where array indices represent bits. This presumably lets me use some bitwise operation to count, but I can't see a nice way of translating this to array indices though.

Elegant? Why not Linq it.
public static IEnumerable<IEnumerable<T>> SubSetsOf<T>(IEnumerable<T> source)
{
if (!source.Any())
return Enumerable.Repeat(Enumerable.Empty<T>(), 1);
var element = source.Take(1);
var haveNots = SubSetsOf(source.Skip(1));
var haves = haveNots.Select(set => element.Concat(set));
return haves.Concat(haveNots);
}

string[] source = new string[] { "dog", "cat", "mouse" };
for (int i = 0; i < Math.Pow(2, source.Length); i++)
{
string[] combination = new string[source.Length];
for (int j = 0; j < source.Length; j++)
{
if ((i & (1 << (source.Length - j - 1))) != 0)
{
combination[j] = source[j];
}
}
Console.WriteLine("[{0}, {1}, {2}]", combination[0], combination[1], combination[2]);
}

You can use the BitArray class to easily access the bits in a number:
string[] animals = { "Dog", "Cat", "Mouse" };
List<string[]> result = new List<string[]>();
int cnt = 1 << animals.Length;
for (int i = 0; i < cnt; i++) {
string[] item = new string[animals.Length];
BitArray b = new BitArray(i);
for (int j = 0; j < item.Length; j++) {
item[j] = b[j] ? animals[j] : null;
}
result.Add(item);
}

static IEnumerable<IEnumerable<T>> GetSubsets<T>(IList<T> set)
{
var state = new BitArray(set.Count);
do
yield return Enumerable.Range(0, state.Count)
.Select(i => state[i] ? set[i] : default(T));
while (Increment(state));
}
static bool Increment(BitArray flags)
{
int x = flags.Count - 1;
while (x >= 0 && flags[x]) flags[x--] = false ;
if (x >= 0) flags[x] = true;
return x >= 0;
}
Usage:
foreach(var strings in GetSubsets(new[] { "dog", "cat", "mouse" }))
Console.WriteLine(string.Join(", ", strings.ToArray()));

Guffa's answer had the basic functionality that I was searching, however the line with
BitArray b = new BitArray(i);
did not work for me, it gave an ArgumentOutOfRangeException. Here's my slightly adjusted and working code:
string[] array = { "A", "B", "C","D" };
int count = 1 << array.Length; // 2^n
for (int i = 0; i < count; i++)
{
string[] items = new string[array.Length];
BitArray b = new BitArray(BitConverter.GetBytes(i));
for (int bit = 0; bit < array.Length; bit++) {
items[bit] = b[bit] ? array[bit] : "";
}
Console.WriteLine(String.Join("",items));
}

Here's a solution similar to David B's method, but perhaps more suitable if it's really a requirement that you get back sets with the original number of elements (even if empty):.
static public List<List<T>> GetSubsets<T>(IEnumerable<T> originalList)
{
if (originalList.Count() == 0)
return new List<List<T>>() { new List<T>() };
var setsFound = new List<List<T>>();
foreach (var list in GetSubsets(originalList.Skip(1)))
{
setsFound.Add(originalList.Take(1).Concat(list).ToList());
setsFound.Add(new List<T>() { default(T) }.Concat(list).ToList());
}
return setsFound;
}
If you pass in a list of three strings, you'll get back eight lists with three elements each (but some elements will be null).

Here's an easy-to-follow solution along the lines of your conception:
private static void Test()
{
string[] test = new string[3] { "dog", "cat", "mouse" };
foreach (var x in Subsets(test))
Console.WriteLine("[{0}]", string.Join(",", x));
}
public static IEnumerable<T[]> Subsets<T>(T[] source)
{
int max = 1 << source.Length;
for (int i = 0; i < max; i++)
{
T[] combination = new T[source.Length];
for (int j = 0; j < source.Length; j++)
{
int tailIndex = source.Length - j - 1;
combination[tailIndex] =
((i & (1 << j)) != 0) ? source[tailIndex] : default(T);
}
yield return combination;
}
}

This is a small change to Mehrdad's solution above:
static IEnumerable<T[]> GetSubsets<T>(T[] set) {
bool[] state = new bool[set.Length+1];
for (int x; !state[set.Length]; state[x] = true ) {
yield return Enumerable.Range(0, state.Length)
.Where(i => state[i])
.Select(i => set[i])
.ToArray();
for (x = 0; state[x]; state[x++] = false);
}
}
or with pointers
static IEnumerable<T[]> GetSubsets<T>(T[] set) {
bool[] state = new bool[set.Length+1];
for (bool *x; !state[set.Length]; *x = true ) {
yield return Enumerable.Range(0, state.Length)
.Where(i => state[i])
.Select(i => set[i])
.ToArray();
for (x = state; *x; *x++ = false);
}
}

I'm not very familiar with C# but I'm sure there's something like:
// input: Array A
foreach S in AllSubsetsOf1ToN(A.Length):
print (S.toArray().map(lambda x |> A[x]));
Ok, I've been told the answer above won't work. If you value elegance over efficiency, I would try recursion, in my crappy pseudocode:
Array_Of_Sets subsets(Array a)
{
if (a.length == 0)
return [new Set();] // emptyset
return subsets(a[1:]) + subsets(a[1:]) . map(lambda x |> x.add a[0])
}

Here is a variant of mqp's answer, that uses as state a BigInteger instead of an int, to avoid overflow for collections containing more than 30 elements:
using System.Numerics;
public static IEnumerable<IEnumerable<T>> GetSubsets<T>(IList<T> source)
{
BigInteger combinations = BigInteger.One << source.Count;
for (BigInteger i = 0; i < combinations; i++)
{
yield return Enumerable.Range(0, source.Count)
.Select(j => (i & (BigInteger.One << j)) != 0 ? source[j] : default);
}
}

Easy to understand version (with descriptions)
I assumed that source = {1,2,3,4}
public static IEnumerable<IEnumerable<T>> GetSubSets<T>(IEnumerable<T> source)
{
var result = new List<IEnumerable<T>>() { new List<T>() }; // empty cluster added
for (int i = 0; i < source.Count(); i++)
{
var elem = source.Skip(i).Take(1);
// for elem = 2
// and currently result = [ [],[1] ]
var matchUps = result.Select(x => x.Concat(elem));
//then matchUps => [ [2],[1,2] ]
result = result.Concat(matchUps).ToList();
// matchUps and result concat operation
// finally result = [ [],[1],[2],[1,2] ]
}
return result;
}

The way this is written, it is more of a Product (Cartesian product) rather than a list of all subsets.
You have three sets: (Empty,"dog"), (Empty,"cat"),(Empty,"mouse").
There are several posts on general solutions for products. As noted though, since you really just have 2 choices for each axis a single bit can represent the presence or not of the item.
So the total set of sets is all numbers from 0 to 2^N-1. If N < 31 an int will work.

How do I remove duplicates from a C# array?

I have been working with a string[] array in C# that gets returned from a function call. I could possibly cast to a Generic collection, but I was wondering if there was a better way to do it, possibly by using a temp array.
What is the best way to remove duplicates from a C# array?

You could possibly use a LINQ query to do this:
int[] s = { 1, 2, 3, 3, 4};
int[] q = s.Distinct().ToArray();

Here is the HashSet<string> approach:
public static string[] RemoveDuplicates(string[] s)
{
HashSet<string> set = new HashSet<string>(s);
string[] result = new string[set.Count];
set.CopyTo(result);
return result;
}
Unfortunately this solution also requires .NET framework 3.5 or later as HashSet was not added until that version. You could also use array.Distinct(), which is a feature of LINQ.

The following tested and working code will remove duplicates from an array. You must include the System.Collections namespace.
string[] sArray = {"a", "b", "b", "c", "c", "d", "e", "f", "f"};
var sList = new ArrayList();
for (int i = 0; i < sArray.Length; i++) {
if (sList.Contains(sArray[i]) == false) {
sList.Add(sArray[i]);
}
}
var sNew = sList.ToArray();
for (int i = 0; i < sNew.Length; i++) {
Console.Write(sNew[i]);
}
You could wrap this up into a function if you wanted to.

If you needed to sort it, then you could implement a sort that also removes duplicates.
Kills two birds with one stone, then.

This might depend on how much you want to engineer the solution - if the array is never going to be that big and you don't care about sorting the list you might want to try something similar to the following:
public string[] RemoveDuplicates(string[] myList) {
System.Collections.ArrayList newList = new System.Collections.ArrayList();
foreach (string str in myList)
if (!newList.Contains(str))
newList.Add(str);
return (string[])newList.ToArray(typeof(string));
}

List<String> myStringList = new List<string>();
foreach (string s in myStringArray)
{
if (!myStringList.Contains(s))
{
myStringList.Add(s);
}
}
This is O(n^2), which won't matter for a short list which is going to be stuffed into a combo, but could be rapidly be a problem on a big collection.

-- This is Interview Question asked every time. Now i done its coding.
static void Main(string[] args)
{
int[] array = new int[] { 4, 8, 4, 1, 1, 4, 8 };
int numDups = 0, prevIndex = 0;
for (int i = 0; i < array.Length; i++)
{
bool foundDup = false;
for (int j = 0; j < i; j++)
{
if (array[i] == array[j])
{
foundDup = true;
numDups++; // Increment means Count for Duplicate found in array.
break;
}
}
if (foundDup == false)
{
array[prevIndex] = array[i];
prevIndex++;
}
}
// Just Duplicate records replce by zero.
for (int k = 1; k <= numDups; k++)
{
array[array.Length - k] = '\0';
}
Console.WriteLine("Console program for Remove duplicates from array.");
Console.Read();
}

Here is a O(n*n) approach that uses O(1) space.
void removeDuplicates(char* strIn)
{
int numDups = 0, prevIndex = 0;
if(NULL != strIn && *strIn != '\0')
{
int len = strlen(strIn);
for(int i = 0; i < len; i++)
{
bool foundDup = false;
for(int j = 0; j < i; j++)
{
if(strIn[j] == strIn[i])
{
foundDup = true;
numDups++;
break;
}
}
if(foundDup == false)
{
strIn[prevIndex] = strIn[i];
prevIndex++;
}
}
strIn[len-numDups] = '\0';
}
}
The hash/linq approaches above are what you would generally use in real life. However in interviews they usually want to put some constraints e.g. constant space which rules out hash or no internal api - which rules out using LINQ.

protected void Page_Load(object sender, EventArgs e)
{
string a = "a;b;c;d;e;v";
string[] b = a.Split(';');
string[] c = b.Distinct().ToArray();
if (b.Length != c.Length)
{
for (int i = 0; i < b.Length; i++)
{
try
{
if (b[i].ToString() != c[i].ToString())
{
Response.Write("Found duplicate " + b[i].ToString());
return;
}
}
catch (Exception ex)
{
Response.Write("Found duplicate " + b[i].ToString());
return;
}
}
}
else
{
Response.Write("No duplicate ");
}
}

Add all the strings to a dictionary and get the Keys property afterwards. This will produce each unique string, but not necessarily in the same order your original input had them in.
If you require the end result to have the same order as the original input, when you consider the first occurance of each string, use the following algorithm instead:
Have a list (final output) and a dictionary (to check for duplicates)
For each string in the input, check if it exists in the dictionary already
If not, add it both to the dictionary and to the list
At the end, the list contains the first occurance of each unique string.
Make sure you consider things like culture and such when constructing your dictionary, to make sure you handle duplicates with accented letters correctly.

The following piece of code attempts to remove duplicates from an ArrayList though this is not an optimal solution. I was asked this question during an interview to remove duplicates through recursion, and without using a second/temp arraylist:
private void RemoveDuplicate()
{
ArrayList dataArray = new ArrayList(5);
dataArray.Add("1");
dataArray.Add("1");
dataArray.Add("6");
dataArray.Add("6");
dataArray.Add("6");
dataArray.Add("3");
dataArray.Add("6");
dataArray.Add("4");
dataArray.Add("5");
dataArray.Add("4");
dataArray.Add("1");
dataArray.Sort();
GetDistinctArrayList(dataArray, 0);
}
private void GetDistinctArrayList(ArrayList arr, int idx)
{
int count = 0;
if (idx >= arr.Count) return;
string val = arr[idx].ToString();
foreach (String s in arr)
{
if (s.Equals(arr[idx]))
{
count++;
}
}
if (count > 1)
{
arr.Remove(val);
GetDistinctArrayList(arr, idx);
}
else
{
idx += 1;
GetDistinctArrayList(arr, idx);
}
}

Simple solution:
using System.Linq;
...
public static int[] Distinct(int[] handles)
{
return handles.ToList().Distinct().ToArray();
}

Maybe hashset which do not store duplicate elements and silently ignore requests to add
duplicates.
static void Main()
{
string textWithDuplicates = "aaabbcccggg";
Console.WriteLine(textWithDuplicates.Count());
var letters = new HashSet<char>(textWithDuplicates);
Console.WriteLine(letters.Count());
foreach (char c in letters) Console.Write(c);
Console.WriteLine("");
int[] array = new int[] { 12, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2 };
Console.WriteLine(array.Count());
var distinctArray = new HashSet<int>(array);
Console.WriteLine(distinctArray.Count());
foreach (int i in distinctArray) Console.Write(i + ",");
}

NOTE : NOT tested!
string[] test(string[] myStringArray)
{
List<String> myStringList = new List<string>();
foreach (string s in myStringArray)
{
if (!myStringList.Contains(s))
{
myStringList.Add(s);
}
}
return myStringList.ToString();
}
Might do what you need...
EDIT Argh!!! beaten to it by rob by under a minute!

Tested the below & it works. What's cool is that it does a culture sensitive search too
class RemoveDuplicatesInString
{
public static String RemoveDups(String origString)
{
String outString = null;
int readIndex = 0;
CompareInfo ci = CultureInfo.CurrentCulture.CompareInfo;
if(String.IsNullOrEmpty(origString))
{
return outString;
}
foreach (var ch in origString)
{
if (readIndex == 0)
{
outString = String.Concat(ch);
readIndex++;
continue;
}
if (ci.IndexOf(origString, ch.ToString().ToLower(), 0, readIndex) == -1)
{
//Unique char as this char wasn't found earlier.
outString = String.Concat(outString, ch);
}
readIndex++;
}
return outString;
}
static void Main(string[] args)
{
String inputString = "aAbcefc";
String outputString;
outputString = RemoveDups(inputString);
Console.WriteLine(outputString);
}
}
--AptSenSDET

This code 100% remove duplicate values from an array[as I used a[i]].....You can convert it in any OO language..... :)
for(int i=0;i<size;i++)
{
for(int j=i+1;j<size;j++)
{
if(a[i] == a[j])
{
for(int k=j;k<size;k++)
{
a[k]=a[k+1];
}
j--;
size--;
}
}
}

Generic Extension method :
public static IEnumerable<TSource> Distinct<TSource>(this IEnumerable<TSource> source, IEqualityComparer<TSource> comparer)
{
if (source == null)
throw new ArgumentNullException(nameof(source));
HashSet<TSource> set = new HashSet<TSource>(comparer);
foreach (TSource item in source)
{
if (set.Add(item))
{
yield return item;
}
}
}

you can using This code when work with an ArrayList
ArrayList arrayList;
//Add some Members :)
arrayList.Add("ali");
arrayList.Add("hadi");
arrayList.Add("ali");
//Remove duplicates from array
for (int i = 0; i < arrayList.Count; i++)
{
for (int j = i + 1; j < arrayList.Count ; j++)
if (arrayList[i].ToString() == arrayList[j].ToString())
arrayList.Remove(arrayList[j]);

Below is an simple logic in java you traverse elements of array twice and if you see any same element you assign zero to it plus you don't touch the index of element you are comparing.
import java.util.*;
class removeDuplicate{
int [] y ;
public removeDuplicate(int[] array){
y=array;
for(int b=0;b<y.length;b++){
int temp = y[b];
for(int v=0;v<y.length;v++){
if( b!=v && temp==y[v]){
y[v]=0;
}
}
}
}

public static int RemoveDuplicates(ref int[] array)
{
int size = array.Length;
// if 0 or 1, return 0 or 1:
if (size < 2) {
return size;
}
int current = 0;
for (int candidate = 1; candidate < size; ++candidate) {
if (array[current] != array[candidate]) {
array[++current] = array[candidate];
}
}
// index to count conversion:
return ++current;
}

The best way? Hard to say, the HashSet approach looks fast,
but (depending on the data) using a sort algorithm (CountSort ?)
can be much faster.
using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
static void Main()
{
Random r = new Random(0); int[] a, b = new int[1000000];
for (int i = b.Length - 1; i >= 0; i--) b[i] = r.Next(b.Length);
a = new int[b.Length]; Array.Copy(b, a, b.Length);
a = dedup0(a); Console.WriteLine(a.Length);
a = new int[b.Length]; Array.Copy(b, a, b.Length);
var w = System.Diagnostics.Stopwatch.StartNew();
a = dedup0(a); Console.WriteLine(w.Elapsed); Console.Read();
}
static int[] dedup0(int[] a) // 48 ms
{
return new HashSet<int>(a).ToArray();
}
static int[] dedup1(int[] a) // 68 ms
{
Array.Sort(a); int i = 0, j = 1, k = a.Length; if (k < 2) return a;
while (j < k) if (a[i] == a[j]) j++; else a[++i] = a[j++];
Array.Resize(ref a, i + 1); return a;
}
static int[] dedup2(int[] a) // 8 ms
{
var b = new byte[a.Length]; int c = 0;
for (int i = 0; i < a.Length; i++)
if (b[a[i]] == 0) { b[a[i]] = 1; c++; }
a = new int[c];
for (int j = 0, i = 0; i < b.Length; i++) if (b[i] > 0) a[j++] = i;
return a;
}
}
Almost branch free. How? Debug mode, Step Into (F11) with a small array: {1,3,1,1,0}
static int[] dedupf(int[] a) // 4 ms
{
if (a.Length < 2) return a;
var b = new byte[a.Length]; int c = 0, bi, ai, i, j;
for (i = 0; i < a.Length; i++)
{ ai = a[i]; bi = 1 ^ b[ai]; b[ai] |= (byte)bi; c += bi; }
a = new int[c]; i = 0; while (b[i] == 0) i++; a[0] = i++;
for (j = 0; i < b.Length; i++) a[j += bi = b[i]] += bi * i; return a;
}
A solution with two nested loops might take some time,
especially for larger arrays.
static int[] dedup(int[] a)
{
int i, j, k = a.Length - 1;
for (i = 0; i < k; i++)
for (j = i + 1; j <= k; j++) if (a[i] == a[j]) a[j--] = a[k--];
Array.Resize(ref a, k + 1); return a;
}

private static string[] distinct(string[] inputArray)
{
bool alreadyExists;
string[] outputArray = new string[] {};
for (int i = 0; i < inputArray.Length; i++)
{
alreadyExists = false;
for (int j = 0; j < outputArray.Length; j++)
{
if (inputArray[i] == outputArray[j])
alreadyExists = true;
}
if (alreadyExists==false)
{
Array.Resize<string>(ref outputArray, outputArray.Length + 1);
outputArray[outputArray.Length-1] = inputArray[i];
}
}
return outputArray;
}

int size = a.Length;
for (int i = 0; i < size; i++)
{
for (int j = i + 1; j < size; j++)
{
if (a[i] == a[j])
{
for (int k = j; k < size; k++)
{
if (k != size - 1)
{
int temp = a[k];
a[k] = a[k + 1];
a[k + 1] = temp;
}
}
j--;
size--;
}
}
}

So I was doing an interview session and got the same question to sort and distinct
static void Sort()
{
try
{
int[] number = new int[Convert.ToInt32(Console.ReadLine())];
for (int i = 0; i < number.Length; i++)
{
number[i] = Convert.ToInt32(Console.ReadLine());
}
Array.Sort(number);
int[] num = number.Distinct().ToArray();
for (int i = 0; i < num.Length; i++)
{
Console.WriteLine(num[i]);
}
}
catch (Exception ex)
{
Console.WriteLine(ex);
}
Console.Read();
}

using System;
using System.Collections.Generic;
using System.Linq;
namespace Rextester
{
public class Program
{
public static void Main(string[] args)
{
List<int> listofint1 = new List<int> { 4, 8, 4, 1, 1, 4, 8 };
List<int> updatedlist= removeduplicate(listofint1);
foreach(int num in updatedlist)
Console.WriteLine(num);
}
public static List<int> removeduplicate(List<int> listofint)
{
List<int> listofintwithoutduplicate= new List<int>();
foreach(var num in listofint)
{
if(!listofintwithoutduplicate.Any(p=>p==num))
{
listofintwithoutduplicate.Add(num);
}
}
return listofintwithoutduplicate;
}
}
}

strINvalues = "1,1,2,2,3,3,4,4";
strINvalues = string.Join(",", strINvalues .Split(',').Distinct().ToArray());
Debug.Writeline(strINvalues);
Kkk Not sure if this is witchcraft or just beautiful code
1 strINvalues .Split(',').Distinct().ToArray()
2 string.Join(",", XXX);
1 Splitting the array and using Distinct [LINQ] to remove duplicates
2 Joining it back without the duplicates.
Sorry I never read the text on StackOverFlow just the code. it make more sense than the text ;)

Removing duplicate and ignore case sensitive using Distinct & StringComparer.InvariantCultureIgnoreCase
string[] array = new string[] { "A", "a", "b", "B", "a", "C", "c", "C", "A", "1" };
var r = array.Distinct(StringComparer.InvariantCultureIgnoreCase).ToList();
Console.WriteLine(r.Count); // return 4 items

Find answer below.
class Program
{
static void Main(string[] args)
{
var nums = new int[] { 1, 4, 3, 3, 3, 5, 5, 7, 7, 7, 7, 9, 9, 9 };
var result = removeDuplicates(nums);
foreach (var item in result)
{
Console.WriteLine(item);
}
}
static int[] removeDuplicates(int[] nums)
{
nums = nums.ToList().OrderBy(c => c).ToArray();
int j = 1;
int i = 0;
int stop = 0;
while (j < nums.Length)
{
if (nums[i] != nums[j])
{
nums[i + 1] = nums[j];
stop = i + 2;
i++;
}
j++;
}
nums = nums.Take(stop).ToArray();
return nums;
}
}
Just a bit of contribution based on a test i just solved, maybe helpful and open to improvement by other top contributors here.
Here are the things i did:
I used OrderBy which allows me order or sort the items from smallest to the highest using LINQ
I then convert it to back to an array and then re-assign it back to the primary datasource
So i then initialize j which is my right hand side of the array to be 1 and i which is my left hand side of the array to be 0, i also initialize where i would i to stop to be 0.
I used a while loop to increment through the array by going from one position to the other left to right, for each increment the stop position is the current value of i + 2 which i will use later to truncate the duplicates from the array.
I then increment by moving from left to right from the if statement and from right to right outside of the if statement until i iterate through the entire values of the array.
I then pick from the first element to the stop position which becomes the last i index plus 2. that way i am able to remove all the duplicate items from the int array. which is then reassigned.