Develop Reference

How to Order By or Sort an integer List and select the Nth element

I have a list, and I want to select the fifth highest element from it:
List<int> list = new List<int>();
list.Add(2);
list.Add(18);
list.Add(21);
list.Add(10);
list.Add(20);
list.Add(80);
list.Add(23);
list.Add(81);
list.Add(27);
list.Add(85);
But OrderbyDescending is not working for this int list...

int fifth = list.OrderByDescending(x => x).Skip(4).First();

Depending on the severity of the list not having more than 5 elements you have 2 options.
If the list never should be over 5 i would catch it as an exception:
int fifth;
try
{
fifth = list.OrderByDescending(x => x).ElementAt(4);
}
catch (ArgumentOutOfRangeException)
{
//Handle the exception
}
If you expect that it will be less than 5 elements then you could leave it as default and check it for that.
int fifth = list.OrderByDescending(x => x).ElementAtOrDefault(4);
if (fifth == 0)
{
//handle default
}
This is still some what flawed because you could end up having the fifth element being 0. This can be solved by typecasting the list into a list of nullable ints at before the linq:
var newList = list.Select(i => (int?)i).ToList();
int? fifth = newList.OrderByDescending(x => x).ElementAtOrDefault(4);
if (fifth == null)
{
//handle default
}

Without LINQ expressions:
int result;
if(list != null && list.Count >= 5)
{
list.Sort();
result = list[list.Count - 5];
}
else // define behavior when list is null OR has less than 5 elements
This has a better performance compared to LINQ expressions, although the LINQ solutions presented in my second answer are comfortable and reliable.
In case you need extreme performance for a huge List of integers, I'd recommend a more specialized algorithm, like in Matthew Watson's answer.
Attention: The List gets modified when the Sort() method is called. If you don't want that, you must work with a copy of your list, like this:
List<int> copy = new List<int>(original);
List<int> copy = original.ToList();

The easiest way to do this is to just sort the data and take N items from the front. This is the recommended way for small data sets - anything more complicated is just not worth it otherwise.
However, for large data sets it can be a lot quicker to do what's known as a Partial Sort.
There are two main ways to do this: Use a heap, or use a specialised quicksort.
The article I linked describes how to use a heap. I shall present a partial sort below:
public static IList<T> PartialSort<T>(IList<T> data, int k) where T : IComparable<T>
{
int start = 0;
int end = data.Count - 1;
while (end > start)
{
var index = partition(data, start, end);
var rank = index + 1;
if (rank >= k)
{
end = index - 1;
}
else if ((index - start) > (end - index))
{
quickSort(data, index + 1, end);
end = index - 1;
}
else
{
quickSort(data, start, index - 1);
start = index + 1;
}
}
return data;
}
static int partition<T>(IList<T> lst, int start, int end) where T : IComparable<T>
{
T x = lst[start];
int i = start;
for (int j = start + 1; j <= end; j++)
{
if (lst[j].CompareTo(x) < 0) // Or "> 0" to reverse sort order.
{
i = i + 1;
swap(lst, i, j);
}
}
swap(lst, start, i);
return i;
}
static void swap<T>(IList<T> lst, int p, int q)
{
T temp = lst[p];
lst[p] = lst[q];
lst[q] = temp;
}
static void quickSort<T>(IList<T> lst, int start, int end) where T : IComparable<T>
{
if (start >= end)
return;
int index = partition(lst, start, end);
quickSort(lst, start, index - 1);
quickSort(lst, index + 1, end);
}
Then to access the 5th largest element in a list you could do this:
PartialSort(list, 5);
Console.WriteLine(list[4]);
For large data sets, a partial sort can be significantly faster than a full sort.
Addendum
See here for another (probably better) solution that uses a QuickSelect algorithm.

This LINQ approach retrieves the 5th biggest element OR throws an exception WHEN the list is null or contains less than 5 elements:
int fifth = list?.Count >= 5 ?
list.OrderByDescending(x => x).Take(5).Last() :
throw new Exception("list is null OR has not enough elements");
This one retrieves the 5th biggest element OR null WHEN the list is null or contains less than 5 elements:
int? fifth = list?.Count >= 5 ?
list.OrderByDescending(x => x).Take(5).Last() :
default(int?);
if(fifth == null) // define behavior
This one retrieves the 5th biggest element OR the smallest element WHEN the list contains less than 5 elements:
if(list == null || list.Count <= 0)
throw new Exception("Unable to retrieve Nth biggest element");
int fifth = list.OrderByDescending(x => x).Take(5).Last();
All these solutions are reliable, they should NEVER throw "unexpected" exceptions.
PS: I'm using .NET 4.7 in this answer.

Here there is a C# implementation of the QuickSelect algorithm to select the nth element in an unordered IList<>.
You have to put all the code contained in that page in a static class, like:
public static class QuickHelpers
{
// Put the code here
}
Given that "library" (in truth a big fat block of code), then you can:
int resA = list.QuickSelect(2, (x, y) => Comparer<int>.Default.Compare(y, x));
int resB = list.QuickSelect(list.Count - 1 - 2);
Now... Normally the QuickSelect would select the nth lowest element. We reverse it in two ways:
For resA we create a reverse comparer based on the default int comparer. We do this by reversing the parameters of the Compare method. Note that the index is 0 based. So there is a 0th, 1th, 2th and so on.
For resB we use the fact that the 0th element is the list-1 th element in the reverse order. So we count from the back. The highest element would be the list.Count - 1 in an ordered list, the next one list.Count - 1 - 1, then list.Count - 1 - 2 and so on
Theorically using Quicksort should be better than ordering the list and then picking the nth element, because ordering a list is on average a O(NlogN) operation and picking the nth element is then a O(1) operation, so the composite is O(NlogN) operation, while QuickSelect is on average a O(N) operation. Clearly there is a but. The O notation doesn't show the k factor... So a O(k1 * NlogN) with a small k1 could be better than a O(k2 * N) with a big k2. Only multiple real life benchmarks can tell us (you) what is better, and it depends on the size of the collection.
A small note about the algorithm:
As with quicksort, quickselect is generally implemented as an in-place algorithm, and beyond selecting the k'th element, it also partially sorts the data. See selection algorithm for further discussion of the connection with sorting.
So it modifies the ordering of the original list.

Shuffling a string so that no two adjacent letters are the same

I've been trying to solve this interview problem which asks to shuffle a string so that no two adjacent letters are identical
For example,
ABCC -> ACBC
The approach I'm thinking of is to
1) Iterate over the input string and store the (letter, frequency)
pairs in some collection
2) Now build a result string by pulling the highest frequency (that is > 0) letter that we didn't just pull
3) Update (decrement) the frequency whenever we pull a letter
4) return the result string if all letters have zero frequency
5) return error if we're left with only one letter with frequency greater than 1
With this approach we can save the more precious (less frequent) letters for last. But for this to work, we need a collection that lets us efficiently query a key and at the same time efficiently sort it by values. Something like this would work except we need to keep the collection sorted after every letter retrieval.
I'm assuming Unicode characters.
Any ideas on what collection to use? Or an alternative approach?

You can sort the letters by frequency, split the sorted list in half, and construct the output by taking letters from the two halves in turn. This takes a single sort.
Example:
Initial string: ACABBACAB
Sort: AAAABBBCC
Split: AAAA+BBBCC
Combine: ABABABCAC
If the number of letters of highest frequency exceeds half the length of the string, the problem has no solution.

Why not use two Data Structures: One for sorting (Like a Heap) and one for key retrieval, like a Dictionary?

The accepted answer may produce a correct result, but is likely not the 'correct' answer to this interview brain teaser, nor the most efficient algorithm.
The simple answer is to take the premise of a basic sorting algorithm and alter the looping predicate to check for adjacency rather than magnitude. This ensures that the 'sorting' operation is the only step required, and (like all good sorting algorithms) does the least amount of work possible.
Below is a c# example akin to insertion sort for simplicity (though many sorting algorithm could be similarly adjusted):
string NonAdjacencySort(string stringInput)
{
var input = stringInput.ToCharArray();
for(var i = 0; i < input.Length; i++)
{
var j = i;
while(j > 0 && j < input.Length - 1 &&
(input[j+1] == input[j] || input[j-1] == input[j]))
{
var tmp = input[j];
input[j] = input[j-1];
input[j-1] = tmp;
j--;
}
if(input[1] == input[0])
{
var tmp = input[0];
input[0] = input[input.Length-1];
input[input.Length-1] = tmp;
}
}
return new string(input);
}
The major change to standard insertion sort is that the function has to both look ahead and behind, and therefore needs to wrap around to the last index.
A final point is that this type of algorithm fails gracefully, providing a result with the fewest consecutive characters (grouped at the front).

Since I somehow got convinced to expand an off-hand comment into a full algorithm, I'll write it out as an answer, which must be more readable than a series of uneditable comments.
The algorithm is pretty simple, actually. It's based on the observation that if we sort the string and then divide it into two equal-length halves, plus the middle character if the string has odd length, then corresponding positions in the two halves must differ from each other, unless there is no solution. That's easy to see: if the two characters are the same, then so are all the characters between them, which totals ⌈n/2⌉+1 characters. But a solution is only possible if there are no more than ⌈n/2⌉ instances of any single character.
So we can proceed as follows:
Sort the string.
If the string's length is odd, output the middle character.
Divide the string (minus its middle character if the length is odd) into two equal-length halves, and interleave the two halves.
At each point in the interleaving, since the pair of characters differ from each other (see above), at least one of them must differ from the last character output. So we first output that character and then the corresponding one from the other half.
The sample code below is in C++, since I don't have a C# environment handy to test with. It's also simplified in two ways, both of which would be easy enough to fix at the cost of obscuring the algorithm:
If at some point in the interleaving, the algorithm encounters a pair of identical characters, it should stop and report failure. But in the sample implementation below, which has an overly simple interface, there's no way to report failure. If there is no solution, the function below returns an incorrect solution.
The OP suggests that the algorithm should work with Unicode characters, but the complexity of correctly handling multibyte encodings didn't seem to add anything useful to explain the algorithm. So I just used single-byte characters. (In C# and certain implementations of C++, there is no character type wide enough to hold a Unicode code point, so astral plane characters must be represented with a surrogate pair.)
#include <algorithm>
#include <iostream>
#include <string>
// If possible, rearranges 'in' so that there are no two consecutive
// instances of the same character.
std::string rearrange(std::string in) {
// Sort the input. The function is call-by-value,
// so the argument itself isn't changed.
std::string out;
size_t len = in.size();
if (in.size()) {
out.reserve(len);
std::sort(in.begin(), in.end());
size_t mid = len / 2;
size_t tail = len - mid;
char prev = in[mid];
// For odd-length strings, start with the middle character.
if (len & 1) out.push_back(prev);
for (size_t head = 0; head < mid; ++head, ++tail)
// See explanatory text
if (in[tail] != prev) {
out.push_back(in[tail]);
out.push_back(prev = in[head]);
}
else {
out.push_back(in[head]);
out.push_back(prev = in[tail]);
}
}
}
return out;
}

you can do that by using a priority queue.
Please find the below explanation.
https://iq.opengenus.org/rearrange-string-no-same-adjacent-characters/

Here is a probabilistic approach. The algorithm is:
10) Select a random char from the input string.
20) Try to insert the selected char in a random position in the output string.
30) If it can't be inserted because of proximity with the same char, go to 10.
40) Remove the selected char from the input string and go to 10.
50) Continue until there are no more chars in the input string, or the failed attempts are too many.
public static string ShuffleNoSameAdjacent(string input, Random random = null)
{
if (input == null) return null;
if (random == null) random = new Random();
string output = "";
int maxAttempts = input.Length * input.Length * 2;
int attempts = 0;
while (input.Length > 0)
{
while (attempts < maxAttempts)
{
int inputPos = random.Next(0, input.Length);
var outputPos = random.Next(0, output.Length + 1);
var c = input[inputPos];
if (outputPos > 0 && output[outputPos - 1] == c)
{
attempts++; continue;
}
if (outputPos < output.Length && output[outputPos] == c)
{
attempts++; continue;
}
input = input.Remove(inputPos, 1);
output = output.Insert(outputPos, c.ToString());
break;
}
if (attempts >= maxAttempts) throw new InvalidOperationException(
$"Shuffle failed to complete after {attempts} attempts.");
}
return output;
}
Not suitable for strings longer than 1,000 chars!
Update: And here is a more complicated deterministic approach. The algorithm is:
Group the elements and sort the groups by length.
Create three empty piles of elements.
Insert each group to a separate pile, inserting always the largest group to the smallest pile, so that the piles differ in length as little as possible.
Check that there is no pile with more than half the total elements, in which case satisfying the condition of not having same adjacent elements is impossible.
Shuffle the piles.
Start yielding elements from the piles, selecting a different pile each time.
When the piles that are eligible for selection are more than one, select randomly, weighting by the size of each pile. Piles containing near half of the remaining elements should be much preferred. For example if the remaining elements are 100 and the two eligible piles have 49 and 40 elements respectively, then the first pile should be 10 times more preferable than the second (because 50 - 49 = 1 and 50 - 40 = 10).
public static IEnumerable<T> ShuffleNoSameAdjacent<T>(IEnumerable<T> source,
Random random = null, IEqualityComparer<T> comparer = null)
{
if (source == null) yield break;
if (random == null) random = new Random();
if (comparer == null) comparer = EqualityComparer<T>.Default;
var grouped = source
.GroupBy(i => i, comparer)
.OrderByDescending(g => g.Count());
var piles = Enumerable.Range(0, 3).Select(i => new Pile<T>()).ToArray();
foreach (var group in grouped)
{
GetSmallestPile().AddRange(group);
}
int totalCount = piles.Select(e => e.Count).Sum();
if (piles.Any(pile => pile.Count > (totalCount + 1) / 2))
{
throw new InvalidOperationException("Shuffle is impossible.");
}
piles.ForEach(pile => Shuffle(pile));
Pile<T> previouslySelectedPile = null;
while (totalCount > 0)
{
var selectedPile = GetRandomPile_WeightedByLength();
yield return selectedPile[selectedPile.Count - 1];
selectedPile.RemoveAt(selectedPile.Count - 1);
totalCount--;
previouslySelectedPile = selectedPile;
}
List<T> GetSmallestPile()
{
List<T> smallestPile = null;
int smallestCount = Int32.MaxValue;
foreach (var pile in piles)
{
if (pile.Count < smallestCount)
{
smallestPile = pile;
smallestCount = pile.Count;
}
}
return smallestPile;
}
void Shuffle(List<T> pile)
{
for (int i = 0; i < pile.Count; i++)
{
int j = random.Next(i, pile.Count);
if (i == j) continue;
var temp = pile[i];
pile[i] = pile[j];
pile[j] = temp;
}
}
Pile<T> GetRandomPile_WeightedByLength()
{
var eligiblePiles = piles
.Where(pile => pile.Count > 0 && pile != previouslySelectedPile)
.ToArray();
Debug.Assert(eligiblePiles.Length > 0, "No eligible pile.");
eligiblePiles.ForEach(pile =>
{
pile.Proximity = ((totalCount + 1) / 2) - pile.Count;
pile.Score = 1;
});
Debug.Assert(eligiblePiles.All(pile => pile.Proximity >= 0),
"A pile has negative proximity.");
foreach (var pile in eligiblePiles)
{
foreach (var otherPile in eligiblePiles)
{
if (otherPile == pile) continue;
pile.Score *= otherPile.Proximity;
}
}
var sumScore = eligiblePiles.Select(p => p.Score).Sum();
while (sumScore > Int32.MaxValue)
{
eligiblePiles.ForEach(pile => pile.Score /= 100);
sumScore = eligiblePiles.Select(p => p.Score).Sum();
}
if (sumScore == 0)
{
return eligiblePiles[random.Next(0, eligiblePiles.Length)];
}
var randomScore = random.Next(0, (int)sumScore);
int accumulatedScore = 0;
foreach (var pile in eligiblePiles)
{
accumulatedScore += (int)pile.Score;
if (randomScore < accumulatedScore) return pile;
}
Debug.Fail("Could not select a pile randomly by weight.");
return null;
}
}
private class Pile<T> : List<T>
{
public int Proximity { get; set; }
public long Score { get; set; }
}
This implementation can suffle millions of elements. I am not completely convinced that the quality of the suffling is as perfect as the previous probabilistic implementation, but should be close.

func shuffle(str:String)-> String{
var shuffleArray = [Character](str)
//Sorting
shuffleArray.sort()
var shuffle1 = [Character]()
var shuffle2 = [Character]()
var adjacentStr = ""
//Split
for i in 0..<shuffleArray.count{
if i > shuffleArray.count/2 {
shuffle2.append(shuffleArray[i])
}else{
shuffle1.append(shuffleArray[i])
}
}
let count = shuffle1.count > shuffle2.count ? shuffle1.count:shuffle2.count
//Merge with adjacent element
for i in 0..<count {
if i < shuffle1.count{
adjacentStr.append(shuffle1[i])
}
if i < shuffle2.count{
adjacentStr.append(shuffle2[i])
}
}
return adjacentStr
}
let s = shuffle(str: "AABC")
print(s)

Getting all combinations of K and less elements in List of N elements with big K

I want to have all combination of elements in a list for a result like this:
List: {1,2,3}
1
2
3
1,2
1,3
2,3
My problem is that I have 180 elements, and I want to have all combinations up to 5 elements. With my tests with 4 elements, it took a long time (2 minutes) but all went well. But with 5 elements, I get a run out of memory exception.
My code presently is this:
public IEnumerable<IEnumerable<Rondin>> getPossibilites(List<Rondin> rondins)
{
var combin5 = rondins.Combinations(5);
var combin4 = rondins.Combinations(4);
var combin3 = rondins.Combinations(3);
var combin2 = rondins.Combinations(2);
var combin1 = rondins.Combinations(1);
return combin5.Concat(combin4).Concat(combin3).Concat(combin2).Concat(combin1).ToList();
}
With the fonction: (taken from this question: Algorithm to return all combinations of k elements from n)
public static IEnumerable<IEnumerable<T>> Combinations<T>(this IEnumerable<T> elements, int k)
{
return k == 0 ? new[] { new T[0] } :
elements.SelectMany((e, i) =>
elements.Skip(i + 1).Combinations(k - 1).Select(c => (new[] { e }).Concat(c)));
}
I need to search in the list for a combination where each element added up is near (with a certain precision) to a value, this for each element in an other list. There is all my code for this part:
var possibilites = getPossibilites(opt.rondins);
possibilites = possibilites.Where(p => p.Sum(r => r.longueur + traitScie) < 144);
foreach(BilleOptimisee b in opt.billesOptimisees)
{
var proches = possibilites.Where(p => p.Sum(r => (r.longueur + traitScie)) < b.chute && Math.Abs(b.chute - p.Sum(r => r.longueur)) - (p.Count() * 0.22) < 0.01).OrderByDescending(p => p.Sum(r => r.longueur)).ElementAt(0);
if(proches != null)
{
foreach (Rondin r in proches)
{
opt.rondins.Remove(r);
b.rondins.Add(r);
possibilites = possibilites.Where(p => !p.Contains(r));
}
}
}
With the code I have, how can I limit the memory taken by my list ? Or is there a better solution to search in a very big set of combinations ?
Please, if my question is not good, tell me why and I will do my best to learn and ask better questions next time ;)

Your output list for combinations of 5 elements will have ~1.5*10^9 (that's billion with b) sublists of size 5. If you use 32bit integers, even neglecting lists overhead and assuming you have a perfect list with 0b overhead - that will be ~200GB!
You should reconsider if you actually need to generate the list like you do, some alternative might be: streaming the list of elements - i.e. generating them on the fly.
That can be done by creating a function, which gets the last combination as an argument - and outputs the next. (to think how it is done, think about increasing by one a number. you go from last to first, remembering a "carry over" until you are done)
A streaming example for choosing 2 out of 4:
start: {4,3}
curr = start {4, 3}
curr = next(curr) {4, 2} // reduce last by one
curr = next(curr) {4, 1} // reduce last by one
curr = next(curr) {3, 2} // cannot reduce more, reduce the first by one, and set the follower to maximal possible value
curr = next(curr) {3, 1} // reduce last by one
curr = next(curr) {2, 1} // similar to {3,2}
done.
Now, you need to figure how to do it for lists of size 2, then generalize it for arbitrary size - and program your streaming combination generator.
Good Luck!

Let your precision be defined in the imaginary spectrum.
Use a real index to access the leaf and then traverse the leaf with the required precision.
See PrecisLise # http://net7mma.codeplex.com/SourceControl/latest#Common/Collections/Generic/PrecicseList.cs
While the implementation is not 100% complete as linked you can find where I used a similar concept here:
http://net7mma.codeplex.com/SourceControl/latest#RtspServer/MediaTypes/RFC6184Media.cs
Using this concept I was able to re-order h.264 Access Units and their underlying Network Access Layer Components in what I consider a very interesting way... outside of interesting it also has the potential to be more efficient using close the same amount of memory.
et al, e.g, 0 can be proceeded by 0.1 or 0.01 or 0.001, depending on the type of the key in the list (double, float, Vector, inter alia) you may have the added benefit of using the FPU or even possibly Intrinsics if supported by your processor, thus making sorting and indexing much faster than would be possible on normal sets regardless of the underlying storage mechanism.
Using this concept allows for very interesting ordering... especially if you provide a mechanism to filter the precision.
I was also able to find several bugs in the bit-stream parser of quite a few well known media libraries using this methodology...

I found my solution, I'm writing it here so that other people that has a similar problem than me can have something to work with...
I made a recursive fonction that check for a fixed amount of possibilities that fit the conditions. When the amount of possibilities is found, I return the list of possibilities, do some calculations with the results, and I can restart the process. I added a timer to stop the research when it takes too long. Since my condition is based on the sum of the elements, I do every possibilities with distinct values, and search for a small amount of possibilities each time (like 1).
So the fonction return a possibility with a very high precision, I do what I need to do with this possibility, I remove the elements of the original list, and recall the fontion with the same precision, until there is nothing returned, so I can continue with an other precision. When many precisions are done, there is only about 30 elements in my list, so I can call for all the possibilities (that still fits the maximum sum), and this part is much easier than the beginning.
There is my code:
public List<IEnumerable<Rondin>> getPossibilites(IEnumerable<Rondin> rondins, int nbElements, double minimum, double maximum, int instance = 0, double longueur = 0)
{
if(instance == 0)
timer = DateTime.Now;
List<IEnumerable<Rondin>> liste = new List<IEnumerable<Rondin>>();
//Get all distinct rondins that can fit into the maximal length
foreach (Rondin r in rondins.Where(r => r.longueur < (maximum - longueur)).DistinctBy(r => r.longueur).OrderBy(r => r.longueur))
{
//Check the current length
double longueur2 = longueur + r.longueur + traitScie;
//If the current length is under the maximal length
if (longueur2 < maximum)
{
//Get all the possibilities with all rondins except the current one, and add them to the list
foreach (IEnumerable<Rondin> poss in getPossibilites(rondins.Where(rondin => rondin.id != r.id), nbElements - liste.Count, minimum, maximum, instance + 1, longueur2).Select(possibilite => possibilite.Concat(new Rondin[] { r })))
{
liste.Add(poss);
if (liste.Count >= nbElements && nbElements > 0)
break;
}
//If this the current length in higher than the minimum, add it to the list
if (longueur2 >= minimum)
liste.Add(new Rondin[] { r });
}
//If we have enough possibilities, we stop the research
if (liste.Count >= nbElements && nbElements > 0)
break;
//If the research is taking too long, stop the research and return the list;
if (DateTime.Now.Subtract(timer).TotalSeconds > 30)
break;
}
return liste;
}

Randomly select a specific quantity of indices from an array?

I have an array of boolean values and need to randomly select a specific quantity of indices for values which are true.
What is the most efficient way to generate the array of indices?
For instance,
BitArray mask = GenerateSomeMask(length: 100000);
int[] randomIndices = RandomIndicesForTrue(mask, quantity: 10);
In this case the length of randomIndices would be 10.

There's a faster way to do this that requires only a single scan of the list.
Consider picking a line at random from a text file when you don't know how many lines are in the file, and the file is too large to fit in memory. The obvious solution is to read the file once to count the lines, pick a random number in the range of 0 to Count-1, and then read the file again up to the chosen line number. That works, but requires you to read the file twice.
A faster solution is to read the first line and save it as the selected line. You replace the selected line with the next line with probability 1/2. When you read the third line, you replace with probability 1/3, etc. When you've read the entire file, you have selected a line at random, and every line had equal probability of being selected. The code looks something like this:
string selectedLine = null;
int numLines = 0;
Random rnd = new Random();
foreach (var line in File.ReadLines(filename))
{
++numLines;
double prob = 1.0/numLines;
if (rnd.Next() >= prob)
selectedLine = line;
}
Now, what if you want to select 2 lines? You select the first two. Then, as each line is read the probability that it will replace one of the two lines is 2/n, where n is the number of lines already read. If you determine that you need to replace a line, you randomly select the line to be replaced. You can follow that same basic idea to select any number of lines at random. For example:
string[] selectedLines = new int[M];
int numLines = 0;
Random rnd = new Random();
foreach (var line in File.ReadLines(filename))
{
++numLines;
if (numLines <= M)
{
selectedLines[numLines-1] = line;
}
else
{
double prob = (double)M/numLines;
if (rnd.Next() >= prob)
{
int ix = rnd.Next(M);
selectedLines[ix] = line;
}
}
}
You can apply that to your BitArray quite easily:
int[] selected = new int[quantity];
int num = 0; // number of True items seen
Random rnd = new Random();
for (int i = 0; i < items.Length; ++i)
{
if (items[i])
{
++num;
if (num <= quantity)
{
selected[num-1] = i;
}
else
{
double prob = (double)quantity/num;
if (rnd.Next() > prob)
{
int ix = rnd.Next(quantity);
selected[ix] = i;
}
}
}
}
You'll need some special case code at the end to handle the case where there aren't quantity set bits in the array, but you'll need that with any solution.
This makes a single pass over the BitArray, and the only extra memory it uses is for the list of selected indexes. I'd be surprised if it wasn't significantly faster than the LINQ version.
Note that I used the probability calculation to illustrate the math. You can change the inner loop code in the first example to:
if (rnd.Next(numLines+1) == numLines)
{
selectedLine = line;
}
++numLines;
You can make a similar change to the other examples. That does the same thing as the probability calculation, and should execute a little faster because it eliminates a floating point divide for each item.

There are two families of approaches you can use: deterministic and non-deterministic. The first one involves finding all the eligible elements in the collection and then picking N at random; the second involves randomly reaching into the collection until you have found N eligible items.
Since the size of your collection is not negligible at 100K and you only want to pick a few out of those, at first sight non-deterministic sounds like it should be considered because it can give very good results in practice. However, since there is no guarantee that N true values even exist in the collection, going non-deterministic could put your program into an infinite loop (less catastrophically, it could just take a very long time to produce results).
Therefore I am going to suggest going for a deterministic approach, even though you are going to pay for the guarantees you need through the nose with resource usage. In particular, the operation will involve in-place sorting of an auxiliary collection; this will practically undo the nice space savings you got by using BitArray.
Theory aside, let's get to work. The standard way to handle this is:
Filter all eligible indices into an auxiliary collection.
Randomly shuffle the collection with Fisher-Yates (there's a convenient implementation on StackOverflow).
Pick the N first items of the shuffled collection. If there are less than N then your input cannot satisfy your requirements.
Translated into LINQ:
var results = mask
.Select((i, f) => Tuple.Create) // project into index/bool pairs
.Where(t => t.Item2) // keep only those where bool == true
.Select(t => t.Item1) // extract indices
.ToList() // prerequisite for next step
.Shuffle() // Fisher-Yates
.Take(quantity) // pick N
.ToArray(); // into an int[]
if (results.Length < quantity)
{
// not enough true values in input
}

If you have 10 indices to choose from, you could generate a random number from 0 to 2^10 - 1, and use that as you mask.

Most efficient sorting algorithm for sorted sub-sequences

I have several sorted sequences of numbers of type long (ascending order) and want to generate one master sequence that contains all elements in the same order. I look for the most efficient sorting algorithm to solve this problem. I target C#, .Net 4.0 and thus also welcome ideas targeting parallelism.
Here is an example:
s1 = 1,2,3,5,7,13
s2 = 2,3,6
s3 = 4,5,6,7,8
resulting Sequence = 1,2,2,3,3,4,5,5,6,6,7,7,8,13
Edit: When there are two (or more) identical values then the order of those two (or more) does not matter.

Just merge the sequences. You do not have to sort them again.

There is no .NET Framework method that I know of to do a K-way merge. Typically, it's done with a priority queue (often a heap). It's not difficult to do, and it's quite efficient. Given K sorted lists, together holding N items, the complexity is O(N log K).
I show a simple binary heap class in my article A Generic Binary Heap Class. In Sorting a Large Text File, I walk through the creation of multiple sorted sub-files and using the heap to do the K-way merge. Given an hour (perhaps less) of study, and you can probably adapt that to use in your program.

You just have to merge your sequences like in a merge sort.
And this is parallelizable:
merge sequences (1 and 2 in 1/2), (3 and 4 in 3/4), …
merge sequences (1/2 and 3/4 in 1/2/3/4), (5/6 and 7/8 in 5/6/7/8), …
…
Here is the merge function :
int j = 0;
int k = 0;
for(int i = 0; i < size_merged_seq; i++)
{
if (j < size_seq1 && seq1[j] < seq2[k])
{
merged_seq[i] = seq1[j];
j++;
}
else
{
merged_seq[i] = seq2[k];
k++;
}
}

Easy way is to merge them with each other one by one. However, this will require O(n*k^2) time, where k is number of sequences and n is the average number of items in sequences. However, using divide and conquer approach you can lower this time to O(n*k*log k). The algorithm is as follows:
Divide k sequences to k/2 groups, each of 2 elements (and 1 groups of 1 element if k is odd).
Merge sequences in each group. Thus you will get k/2 new groups.
Repeat until you get single sequence.

UPDATE:
Turns out that with all the algorithms... It's still faster the simple way:
private static List<T> MergeSorted<T>(IEnumerable<IEnumerable<T>> sortedBunches)
{
var list = sortedBunches.SelectMany(bunch => bunch).ToList();
list.Sort();
return list;
}
And for legacy purposes...
Here is the final version by prioritizing:
private static IEnumerable<T> MergeSorted<T>(IEnumerable<IEnumerable<T>> sortedInts) where T : IComparable<T>
{
var enumerators = new List<IEnumerator<T>>(sortedInts.Select(ints => ints.GetEnumerator()).Where(e => e.MoveNext()));
enumerators.Sort((e1, e2) => e1.Current.CompareTo(e2.Current));
while (enumerators.Count > 1)
{
yield return enumerators[0].Current;
if (enumerators[0].MoveNext())
{
if (enumerators[0].Current.CompareTo(enumerators[1].Current) == 1)
{
var tmp = enumerators[0];
enumerators[0] = enumerators[1];
enumerators[1] = tmp;
}
}
else
{
enumerators.RemoveAt(0);
}
}
do
{
yield return enumerators[0].Current;
} while (enumerators[0].MoveNext());
}