Develop Reference

Optimal solution for "Bitwise AND" problem in C#

Problem statement:
Given an array of non-negative integers, count the number of unordered pairs of array elements, such that their bitwise AND is a power of 2.
Example:
arr = [10, 7, 2, 8, 3]
Answer: 6 (10&7, 10&2, 10&8, 10&3, 7&2, 2&3)
Constraints:
1 <= arr.Count <= 2*10^5
0 <= arr[i] <= 2^12
Here's my brute-force solution that I've come up with:
private static Dictionary<int, bool> _dictionary = new Dictionary<int, bool>();
public static long CountPairs(List<int> arr)
{
long result = 0;
for (var i = 0; i < arr.Count - 1; ++i)
{
for (var j = i + 1; j < arr.Count; ++j)
{
if (IsPowerOfTwo(arr[i] & arr[j])) ++result;
}
}
return result;
}
public static bool IsPowerOfTwo(int number)
{
if (_dictionary.TryGetValue(number, out bool value)) return value;
var result = (number != 0) && ((number & (number - 1)) == 0);
_dictionary[number] = result;
return result;
}
For small inputs this works fine, but for big inputs this works slow.
My question is: what is the optimal (or at least more optimal) solution for the problem? Please provide a graceful solution in C#. 😊

One way to accelerate your approach is to compute the histogram of your data values before counting.
This will reduce the number of computations for long arrays because there are fewer options for value (4096) than the length of your array (200000).
Be careful when counting bins that are powers of 2 to make sure you do not overcount the number of pairs by including cases when you are comparing a number with itself.

We can adapt the bit-subset dynamic programming idea to have a solution with O(2^N * N^2 + n * N) complexity, where N is the number of bits in the range, and n is the number of elements in the list. (So if the integers were restricted to [1, 4096] or 2^12, with n at 100,000, we would have on the order of 2^12 * 12^2 + 100000*12 = 1,789,824 iterations.)
The idea is that we want to count instances for which we have overlapping bit subsets, with the twist of adding a fixed set bit. Given Ai -- for simplicity, take 6 = b110 -- if we were to find all partners that AND to zero, we'd take Ai's negation,
110 -> ~110 -> 001
Now we can build a dynamic program that takes a diminishing mask, starting with the full number and diminishing the mask towards the left
001
^^^
001
^^
001
^
Each set bit on the negation of Ai represents a zero, which can be ANDed with either 1 or 0 to the same effect. Each unset bit on the negation of Ai represents a set bit in Ai, which we'd like to pair only with zeros, except for a single set bit.
We construct this set bit by examining each possibility separately. So where to count pairs that would AND with Ai to zero, we'd do something like
001 ->
001
000
we now want to enumerate
011 ->
011
010
101 ->
101
100
fixing a single bit each time.
We can achieve this by adding a dimension to the inner iteration. When the mask does have a set bit at the end, we "fix" the relevant bit by counting only the result for the previous DP cell that would have the bit set, and not the usual union of subsets that could either have that bit set or not.
Here is some JavaScript code (sorry, I do not know C#) to demonstrate with testing at the end comparing to the brute-force solution.
var debug = 0;
function bruteForce(a){
let answer = 0;
for (let i = 0; i < a.length; i++) {
for (let j = i + 1; j < a.length; j++) {
let and = a[i] & a[j];
if ((and & (and - 1)) == 0 && and != 0){
answer++;
if (debug)
console.log(a[i], a[j], a[i].toString(2), a[j].toString(2))
}
}
}
return answer;
}
function f(A, N){
const n = A.length;
const hash = {};
const dp = new Array(1 << N);
for (let i=0; i<1<<N; i++){
dp[i] = new Array(N + 1);
for (let j=0; j<N+1; j++)
dp[i][j] = new Array(N + 1).fill(0);
}
for (let i=0; i<n; i++){
if (hash.hasOwnProperty(A[i]))
hash[A[i]] = hash[A[i]] + 1;
else
hash[A[i]] = 1;
}
for (let mask=0; mask<1<<N; mask++){
// j is an index where we fix a 1
for (let j=0; j<=N; j++){
if (mask & 1){
if (j == 0)
dp[mask][j][0] = hash[mask] || 0;
else
dp[mask][j][0] = (hash[mask] || 0) + (hash[mask ^ 1] || 0);
} else {
dp[mask][j][0] = hash[mask] || 0;
}
for (let i=1; i<=N; i++){
if (mask & (1 << i)){
if (j == i)
dp[mask][j][i] = dp[mask][j][i-1];
else
dp[mask][j][i] = dp[mask][j][i-1] + dp[mask ^ (1 << i)][j][i - 1];
} else {
dp[mask][j][i] = dp[mask][j][i-1];
}
}
}
}
let answer = 0;
for (let i=0; i<n; i++){
for (let j=0; j<N; j++)
if (A[i] & (1 << j))
answer += dp[((1 << N) - 1) ^ A[i] | (1 << j)][j][N];
}
for (let i=0; i<N + 1; i++)
if (hash[1 << i])
answer = answer - hash[1 << i];
return answer / 2;
}
var As = [
[10, 7, 2, 8, 3] // 6
];
for (let A of As){
console.log(JSON.stringify(A));
console.log(`DP, brute force: ${ f(A, 4) }, ${ bruteForce(A) }`);
console.log('');
}
var numTests = 1000;
for (let i=0; i<numTests; i++){
const N = 6;
const A = [];
const n = 10;
for (let j=0; j<n; j++){
const num = Math.floor(Math.random() * (1 << N));
A.push(num);
}
const fA = f(A, N);
const brute = bruteForce(A);
if (fA != brute){
console.log('Mismatch:');
console.log(A);
console.log(fA, brute);
console.log('');
}
}
console.log("Done testing.");

int[] numbers = new[] { 10, 7, 2, 8, 3 };
static bool IsPowerOfTwo(int n) => (n != 0) && ((n & (n - 1)) == 0);
long result = numbers.AsParallel()
.Select((a, i) => numbers
.Skip(i + 1)
.Select(b => a & b)
.Count(IsPowerOfTwo))
.Sum();
If I understand the problem correctly, this should work and should be faster.
First, for each number in the array we grab all elements in the array after it to get a collection of numbers to pair with.
Then we transform each pair number with a bitwise AND, then counting the number that satisfy our 'IsPowerOfTwo;' predicate (implementation here).
Finally we simply get the sum of all the counts - our output from this case is 6.
I think this should be more performant than your dictionary based solution - it avoids having to perform a lookup each time you wish to check power of 2.
I think also given the numerical constraints of your inputs it is fine to use int data types.

Cycling the elements of one C# List within another List [duplicate]

Lists say I have a list List<int> {1,2,3,4,5}
Rotate means:
=> {2,3,4,5,1} => {3,4,5,1,2} => {4,5,1,2,3}
Maybe rotate is not the best word for this, but hope you understand what I means
My question, whats the easiest way (in short code, c# 4 Linq ready), and will not be hit by performance (reasonable performance)
Thanks.

List<T>
The simplest way (for a List<T>) is to use:
int first = list[0];
list.RemoveAt(0);
list.Add(first);
Performance is nasty though - O(n).
Array
This is basically equivalent to the List<T> version, but more manual:
int first = array[0];
Array.Copy(array, 1, array, 0, array.Length - 1);
array[array.Length - 1] = first;
LinkedList<T>
If you could use a LinkedList<T> instead, that would be much simpler:
int first = linkedList.First;
linkedList.RemoveFirst();
linkedList.AddLast(first);
This is O(1) as each operation is constant time.
Queue<T>
cadrell0's solution of using a queue is a single statement, as Dequeue removes the element and returns it:
queue.Enqueue(queue.Dequeue());
While I can't find any documentation of the performance characteristic of this, I'd expect Queue<T> to be implemented using an array and an index as the "virtual starting point" - in which case this is another O(1) solution.
Note that in all of these cases you'd want to check for the list being empty first. (You could deem that to be an error, or a no-op.)

You could implement it as a queue. Dequeue and Enqueue the same value.
**I wasn't sure about performance in converting a List to a Queue, but people upvoted my comment, so I'm posting this as an answer.

I use this one:
public static List<T> Rotate<T>(this List<T> list, int offset)
{
return list.Skip(offset).Concat(list.Take(offset)).ToList();
}

It seems like some answerers have treated this as a chance to explore data structures. While those answers are informative and useful, they are not very Linq'ish.
The Linq'ish approach is: You get an extension method which returns a lazy IEnumerable that knows how to build what you want. This method doesn't modify the source and should only allocate a copy of the source if necessary.
public static IEnumerable<IEnumerable<T>> Rotate<T>(this List<T> source)
{
for(int i = 0; i < source.Count; i++)
{
yield return source.TakeFrom(i).Concat(source.TakeUntil(i));
}
}
//similar to list.Skip(i-1), but using list's indexer access to reduce iterations
public static IEnumerable<T> TakeFrom<T>(this List<T> source, int index)
{
for(int i = index; i < source.Count; i++)
{
yield return source[i];
}
}
//similar to list.Take(i), but using list's indexer access to reduce iterations
public static IEnumerable<T> TakeUntil<T>(this List<T> source, int index)
{
for(int i = 0; i < index; i++)
{
yield return source[i];
}
}
Used as:
List<int> myList = new List<int>(){1, 2, 3, 4, 5};
foreach(IEnumerable<int> rotation in myList.Rotate())
{
//do something with that rotation
}

How about this:
var output = input.Skip(rot)
.Take(input.Count - rot)
.Concat(input.Take(rot))
.ToList();
Where rot is the number of spots to rotate - which must be less than the number of elements in the input list.
As #cadrell0 answer shows if this is all you do with your list, you should use a queue instead of a list.

My solution maybe too basic (I wouldn't like to say it's lame...) and not LINQ'ish.
However, it has a pretty good performance.
int max = 5; //the fixed size of your array.
int[] inArray = new int[5] {0,0,0,0,0}; //initial values only.
void putValueToArray(int thisData)
{
//let's do the magic here...
Array.Copy(inArray, 1, inArray, 0, max-1);
inArray[max-1] = thisData;
}

Try
List<int> nums = new List<int> {1,2,3,4,5};
var newNums = nums.Skip(1).Take(nums.Count() - 1).ToList();
newNums.Add(nums[0]);
Although, I like Jon Skeet's answer better.

My solution for Arrays:
public static void ArrayRotate(Array data, int index)
{
if (index > data.Length)
throw new ArgumentException("Invalid index");
else if (index == data.Length || index == 0)
return;
var copy = (Array)data.Clone();
int part1Length = data.Length - index;
//Part1
Array.Copy(copy, 0, data, index, part1Length);
//Part2
Array.Copy(copy, part1Length, data, 0, index);
}

I've used the following extensions for this:
static class Extensions
{
public static IEnumerable<T> RotateLeft<T>(this IEnumerable<T> e, int n) =>
n >= 0 ? e.Skip(n).Concat(e.Take(n)) : e.RotateRight(-n);
public static IEnumerable<T> RotateRight<T>(this IEnumerable<T> e, int n) =>
e.Reverse().RotateLeft(n).Reverse();
}
They're certainly easy (OP title request), and they've got reasonable performance (OP write-up request). Here's a little demo I ran in LINQPad 5 on an above-average-powered laptop:
void Main()
{
const int n = 1000000;
const int r = n / 10;
var a = Enumerable.Range(0, n);
var t = Stopwatch.StartNew();
Console.WriteLine(a.RotateLeft(r).ToArray().First());
Console.WriteLine(a.RotateLeft(-r).ToArray().First());
Console.WriteLine(a.RotateRight(r).ToArray().First());
Console.WriteLine(a.RotateRight(-r).ToArray().First());
Console.WriteLine(t.ElapsedMilliseconds); // e.g. 236
}

You can use below code for left Rotation.
List<int> backUpArray = array.ToList();
for (int i = 0; i < array.Length; i++)
{
int newLocation = (i + (array.Length - rotationNumber)) % n;
array[newLocation] = backUpArray[i];
}

You can play nice in .net framework.
I understand that what you want to do is more up to be an iteration behavior than a new collection type; so I would suggest you to try this extension method based on IEnumerable, which will work with Collections, Lists and so on...
class Program
{
static void Main(string[] args)
{
int[] numbers = { 1, 2, 3, 4, 5, 6, 7 };
IEnumerable<int> circularNumbers = numbers.AsCircular();
IEnumerable<int> firstFourNumbers = circularNumbers
.Take(4); // 1 2 3 4
IEnumerable<int> nextSevenNumbersfromfourth = circularNumbers
.Skip(4).Take(7); // 4 5 6 7 1 2 3
}
}
public static class CircularEnumerable
{
public static IEnumerable<T> AsCircular<T>(this IEnumerable<T> source)
{
if (source == null)
yield break; // be a gentleman
IEnumerator<T> enumerator = source.GetEnumerator();
iterateAllAndBackToStart:
while (enumerator.MoveNext())
yield return enumerator.Current;
enumerator.Reset();
if(!enumerator.MoveNext())
yield break;
else
yield return enumerator.Current;
goto iterateAllAndBackToStart;
}
}
Reasonable performance
Flexible
If you want go further, make a CircularList and hold the same enumerator to skip the Skip() when rotating like in your sample.

below is my approach. Thank you
public static int[] RotationOfArray(int[] A, int k)
{
if (A == null || A.Length==0)
return null;
int[] result =new int[A.Length];
int arrayLength=A.Length;
int moveBy = k % arrayLength;
for (int i = 0; i < arrayLength; i++)
{
int tmp = i + moveBy;
if (tmp > arrayLength-1)
{
tmp = + (tmp - arrayLength);
}
result[tmp] = A[i];
}
return result;
}

public static int[] RightShiftRotation(int[] a, int times) {
int[] demo = new int[a.Length];
int d = times,i=0;
while(d>0) {
demo[d-1] = a[a.Length - 1 - i]; d = d - 1; i = i + 1;
}
for(int j=a.Length-1-times;j>=0;j--) { demo[j + times] = a[j]; }
return demo;
}

Using Linq,
List<int> temp = new List<int>();
public int[] solution(int[] array, int range)
{
int tempLength = array.Length - range;
temp = array.Skip(tempLength).ToList();
temp.AddRange(array.Take(array.Length - range).ToList());
return temp.ToArray();
}

If you're working with a string you can do this quite efficiently using ReadOnlySpans:
ReadOnlySpan<char> apiKeySchema = "12345";
const int apiKeyLength = 5;
for (int i = 0; i < apiKeyLength; i++)
{
ReadOnlySpan<char> left = apiKeySchema.Slice(start: i, length: apiKeyLength - i);
ReadOnlySpan<char> right = apiKeySchema.Slice(start: 0, length: i);
Console.WriteLine(string.Concat(left, right));
}
Output:
12345
23451
34512
45123
51234

I was asked to reverse a character array with minimal memory usage.
char[] charArray = new char[]{'C','o','w','b','o','y'};
Method:
static void Reverse(ref char[] s)
{
for (int i=0; i < (s.Length-i); i++)
{
char leftMost = s[i];
char rightMost = s[s.Length - i - 1];
s[i] = rightMost;
s[s.Length - i - 1] = leftMost;
}
}

How about using modular arithmetic :
public void UsingModularArithmetic()
{
string[] tokens_n = Console.ReadLine().Split(' ');
int n = Convert.ToInt32(tokens_n[0]);
int k = Convert.ToInt32(tokens_n[1]);
int[] a = new int[n];
for(int i = 0; i < n; i++)
{
int newLocation = (i + (n - k)) % n;
a[newLocation] = Convert.ToInt32(Console.ReadLine());
}
foreach (int i in a)
Console.Write("{0} ", i);
}
So basically adding the values to the array when I am reading from console.

Reverse binary representation of int (only significant bits)

I'm trying to write a program for reversing numbers in binary. For instance, the binary representation of 13 is 1101, and reversing it gives 1011, which corresponds to number 11 right?
Here's my code:
static void Main(string[] args)
{
Console.WriteLine("Enter a Number");
int numb = int.Parse(Console.ReadLine());
int reverse = 0;
while (numb > 0)
{
int rem = numb % 10;
reverse = (reverse * 10) + rem;
numb = numb / 10;
}
Console.WriteLine("Reverse number={0}", reverse);
Console.ReadLine();
}
By this code I only get the numbers to reverse (13 -> 31)...
The input should contain a single line with an integer N, 1≤N≤1000000000 and I want my output in one line with one integer, the number I want to get by reversing the binary representation of N.

Something like that
// 13 = 1101b
int value = 13;
// 11 = 1011b
int result = Convert.ToInt32(new String(
Convert.ToString(value, 2)
.Reverse()
.ToArray()), 2);
Explanation:
Convert.ToString(value, 2) returns value in binary representation ("1101")
Reverse().ToArray() - reverse the string ('1','0','1','1') as sequence of characters and converts to array char[].
new String(...) constructs string "1011" from array of char
finally, Convert.ToInt32(..., 2) convert binary representation back to int

You can use Convert.ToString and Convert.ToInt32 methods, where 2 means binary:
int numb = int.Parse(Console.ReadLine());
var reversedString = Convert.ToString(numb, 2).ReverseString();
var result = Convert.ToInt32(reversedString, 2);
...
public static string ReverseString(this string s)
{
char[] arr = s.ToCharArray();
Array.Reverse(arr);
return new string(arr);
}

A fun excercise would be doing this without using the string conversion.
I have very little experience with bit twiddling so there is probably a faster and better way of doing this, but this seems to work:
public static IEnumerable<bool> ToBinary(this int n)
{
for (int i = 0; i < 32; i++)
{
yield return (n & (1 << i)) != 0;
}
}
public static int ToInt(this IEnumerable<bool> b)
{
var n = 0;
var counter = 0;
foreach (var i in b.Trim().Take(32))
{
n = n | (i ? 1 : 0) << counter;
counter++
}
return n;
}
private static IEnumerable<bool> Trim(this IEnumerable<bool> list)
{
bool trim = true;
foreach (var i in list)
{
if (i)
{
trim = false;
}
if (!trim)
{
yield return i;
}
}
}
And now you'd call it like this:
var reversed = n.ToBinary().Reverse().ToInt();

Incrementing characters in a particular pattern in C#

The program should take in a parameter N and print out N + 1 lines.
I have to output something like this.
This is the output which I must get at N = 5
A //Increment by 0
AB //Increment by 1
ACE //Increment by 2
ADGJ //Increment by 3
AEIMQ //Increment by 4
The algorithm uses N as the number of characters to skip in between each add. So at N=3, it's A skip 3 to D, skip 3 to G, skip three to J.
And when the program runs out of upper case characters(i.e. When N is too big), it should start with lower case characters and if it runs out of lower case then it should again start with upper case and so on.
I am a novice to programming. And I dont really know where to start. I've been working around the loops for a while and still have no clue what-so-ever.

Here's another approach using a Char[], modulo, StringBuilder and a for-loop which increments by n for efficiency:
readonly static Char[] letters =
"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz".ToCharArray();
static String appendChars(int n)
{
int length = n + 1;
StringBuilder sBuilder = new StringBuilder("A", length);
for (int i = n; sBuilder.Length < length; i += n)
{
Char nextChar = letters[i % letters.Length];
sBuilder.Append(nextChar);
}
return sBuilder.ToString();
}
test your sample data:
int n = 5;
IEnumerable<String> allWords = Enumerable.Range(0, n).Select(i => appendChars(i));
Console.Write(string.Join(Environment.NewLine, allWords));
outputs:
A
AB
ACE
ADGJ
AEIMQ
Here's the demo: http://ideone.com/0sspY

Try This :
public string GetOutPut(int increment)
{
string alphabets = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string output = string.Empty;
for(int i=0; i<=increment; i++)
{
int index = i*increment;
if(index>alphabets.Length)
index = index % alphabets.Length;
output+= alphabets[index];
}
return output;
}

It's not clear how you get the number of lines that you want - because you said N+1 but your example gives only N lines.
The following can be used to generate each one of those individual lines, and the Algo method can be modified to generate n+1 lines by sticking the code in a while loop, decrementing n and len and using AppendLine on the StringBuilder:
char[] allowedChars = Enumerable.Range('A', 26).Concat(Enumerable.Range('a', 26))
.Select(i => (char)i).ToArray();
[TestMethod]
public void Test()
{
Assert.AreEqual("A", Algo(0, 1));
Assert.AreEqual("AB", Algo(1, 2));
Assert.AreEqual("ACE", Algo(2, 3));
Assert.AreEqual("ADGJ", Algo(3, 4));
Assert.AreEqual("AEIMQ", Algo(4, 5));
}
public string Algo(int n, int len)
{
StringBuilder sb = new StringBuilder();
int nextCharIndex = 0;
for (int f = 0; f < len; f++)
{
sb.Append(allowedChars[nextCharIndex]);
//the `%`, or mod, here wraps around the next character back to upper case
nextCharIndex = (nextCharIndex + n) % allowedChars.Length;
}
return sb.ToString();
}

Most efficient way to reverse the order of a BitArray?

I've been wondering what the most efficient way to reverse the order of a BitArray in C#. To be clear, I don't want to inverse the Bitarray by calling .Not(), I want to reverse the order of the bits in the array.
Cheers,
Chris

public void Reverse(BitArray array)
{
int length = array.Length;
int mid = (length / 2);
for (int i = 0; i < mid; i++)
{
bool bit = array[i];
array[i] = array[length - i - 1];
array[length - i - 1] = bit;
}
}

For a long array and relative few uses, just wrap it:
class BitArrayReverse
{
private BitArray _ba;
public BitArrayReverse(BitArray ba) { _ba = ba; }
public bool this[int index]
{
get { return _ba[_ba.Length - 1 - index]; }
set { _ba[_ba.Length - 1 - index] = value; }
}
}

This will be the best way
to reverse MSB <-> LSB of any length using XOR in the for loop
public static BitArray BitsReverse(BitArray bits)
{
int len = bits.Count;
BitArray a = new BitArray(bits);
BitArray b = new BitArray(bits);
for (int i = 0, j = len-1; i < len; ++i, --j)
{
a[i] = a[i] ^ b[j];
b[j] = a[i] ^ b[j];
a[i] = a[i] ^ b[j];
}
return a;
}
// in 010000011010000011100b
// out 001110000010110000010b

Dim myBA As New BitArray(4)
myBA(0) = True
myBA(1) = False
myBA(2) = True
myBA(3) = True
Dim myBoolArray1(3) As Boolean
myBA.CopyTo(myBoolArray1, 0)
Array.Reverse(myBoolArray1)
myBA = New BitArray(myBoolArray1)

For a short but inefficient answer:
using System.Linq;
var reversedBa = new BitArray(myBa.Cast<bool>().Reverse().ToArray())

Because the size if fixed at 8-bits just the "table" lookup from below is sufficient -- when dealing with a plain byte a look-up is likely the quickest way. The extra overhead of BitSet to get/set the data may, however, nullify the look-up benefit. Also the initial build cost and persistent overhead need to be considered (but the values could be coded into an array literal ... ick!)
On the other hand, if the data is only 8 bit (ever), and "performance is important", why use a BitArray at all? A BitArray could always be used for the nice features, such as "exploding" to an Enumerable while C# already has decent byte bit manipulation built-in.
Assuming a more general case that the data is 8-bit aligned... but of some undetermined length
Is this actually better (faster, more efficient, etc) than just doing it "per item" in the BitArray? I have no idea but suspect not. I would definitely start with the "simple" methods -- this is here as just a proof-of-concept and may (or may not be) interesting to compare in a benchmark. Anyway, write for clarity first ... and the below is not it! (There is at least one bug in it -- I blame the extra complexity ;-)
byte reverse (byte b) {
byte o = 0;
for (var i = 0; i < 8; i++) {
o <<= 1;
o |= (byte)(b & 1);
b >>= 1;
}
return o;
}
byte[] table;
BitArray reverse8 (BitArray ar) {
if (ar.Count % 8 != 0) {
throw new Exception("no!");
}
byte[] d = new byte[ar.Count / 8];
ar.CopyTo(d, 0);
// this only works if the bit array is
// a multiple of 8. we swap bytes and
// then reverse bits in each byte
int mid = d.Length / 2;
for (int i = 0, j = d.Length - 1; i < mid; i++, j--) {
byte t = d[i];
d[i] = table[d[j]];
d[j] = table[t];
}
return new BitArray(d);
}
string tostr (BitArray x) {
return string.Join("",
x.OfType<bool>().Select(i => i ? "1" : "0").ToArray());
}
void Main()
{
table = Enumerable.Range(0,256).Select(v => reverse((byte)v)).ToArray();
{
byte[] s = new byte[] { 1, 0xff };
BitArray ar = new BitArray(s);
// linqpad :)
tostr(ar).Dump();
tostr(reverse8(ar)).Dump();
}
"--".Dump();
{
byte[] s = new byte[] { 3, 42, 19 };
BitArray ar = new BitArray(s);
// linqpad :)
tostr(ar).Dump();
tostr(reverse8(ar)).Dump();
}
}
Output:
1000000011111111
1111111100000001
--
110000000101010011001000
000100110101010000000011
The expr.Dump() is a LINQPad feature.

Adapted the answer from #TimLoyd and turned it into an extension for easier use.
public static BitArray Reverse(this BitArray array)
{
int length = array.Length;
int mid = (length / 2);
for (int i = 0; i < mid; i++)
{
bool bit = array[i];
array[i] = array[length - i - 1];
array[length - i - 1] = bit;
}
return new BitArray(array);
}
Usage:
var bits = new BitArray(some_bytes).Reverse();