Develop Reference

What is stopping this program from having an output

I made this code to brute force anagrams by printing all possible permutables of the characters in a string, however there is no output. I do not need simplifications as I am a student still learning the basics, I just need help getting it to work this way so I can better understand it.
using System;
using System.Collections.Generic;
namespace anagramSolver
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("Enter anagram:");
string anagram = Console.ReadLine();
string temp = "";
List<string> storeStr = new List<string>(0);
List<int> storeInt = new List<int>(0);
//creates factorial number for anagram
int factorial = 1;
for (int i = anagram.Length; i > 0; i--)
{
factorial *= i;
}
while (storeStr.Count != factorial)
{
Random rnd = new Random();
while(temp.Length != anagram.Length)
{
int num = rnd.Next(anagram.Length - 1);
if (storeInt.Contains(num) == true)
{
}
else
{
storeInt.Add(num);
temp += anagram[num];
}
}
if (storeStr.Contains(temp) == true)
{
temp = "";
}
else
{
storeStr.Add(temp);
Console.WriteLine(temp, storeStr.Count);
temp = "";
}
}
}
}
}
edit: added temp reset after it is deemed contained by storeStr

Two main issues causing infinite loop:
1)
As per Random.Next documentation, the parameter is the "exclusive" upper bound. This means, if you want a random number between 0 and anagram.Length - 1 included, you should use rnd.Next(anagram.Length);.
With rnd.Next(anagram.Length - 1), you'll never hit anagram.Length - 1.
2)
Even if you solve 1, only the first main iteration goes well.
storeInt is never reset. Meaning, after the first main iteration, it will have already all the numbers in it.
So, during the second iteration, you will always hit the case if (storeInt.Contains(num) == true), which does nothing and the inner loop will go on forever.

Several issues here...
The expression rnd.Next(anagram.Length - 1) generates a value between 0 and anagram.Length - 2. For an input with two characters it will always return 0, so you'll never actually generate a full string. Remove the - 1 from it and you'll be fine.
Next, you're using a list to keep track of the character indices you've used already, but you never clear the list. You'll get one output (eventually, when the random number generator covers all the values) and then enter an infinite loop on the next generation pass. Clear storeInt after the generation loop to fix this.
While not a true infinite loop, creating a new instance of the random number generator each time will give you a lot of duplication. new Random() uses the current time as a seed value and you could potentially get through a ton of loops with each seed, generating exactly the same values until the time changes enough to change your random sequence. Create the random number generator once before you start your main loop.
And finally, your code doesn't handle repeated input letters. If the input is "aa" then there is only a single distinct output, but your code won't stop until it gets two. If the input was "aab" there are three distinct permutations ("aab", "aba", "baa") which is half of the expected results. You'll never reach your exit condition in this case. You could do it by keeping track of the indices you've used instead of the generated strings, it just makes it a bit more complex.
There are a few ways to generate permutations that are less error-prone. In general you should try to avoid "keep generating random garbage until I find the result I'm looking for" solutions. Think about how you personally would go about writing down the full list of permutations for 3, 4 or 5 inputs. What steps would you take? Think about how that would work in a computer program.

this loop
while(temp.Length != anagram.Length)
{
int num = rnd.Next(anagram.Length - 1);
if (storeInt.Contains(num) == true)
{
}
else
{
storeInt.Add(num);
temp += anagram[num];
}
}
gets stuck.
once the number is in storeInt you never change storeStr, yet thats what you are testing for loop exit

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)

Calculate amount of memory needed for recursive function

I have following function:
public static long Fibon(long num)
{
if (num == 1)
{
return 1;
}
else if (num == 2)
{
return 1;
}
return fibon(num - 1) + fibon(num - 2);
}
this function uses recursion in order to calculate Fibonacci number. How can I calculate amount of required stack memory for executing this function before executing it? For example I want to execute this function in few separated threads with some big numbers, and before executing threads I want to know how much stack memory available I need to have.

Just looking at it, the code won't work because when num == 2, the method tries to find fibon(0).
Try
public static long Fibon(long num)
{
if (num == 1)
{
return 1;
}
else if (num == 2)
{
return 1;
}
return fibon(num - 1) + fibon(num - 2);
}
will give you 1, 1, 2, 3, 5, ...
Sorry this wasn't an answer, I don't have the reputation to comment.
edit: You'll also be able compute greater entries bu using ulong.

Since you only have to remember the previous two terms to calculate the current one, you will not face any memory problem if using a non-recursive procedure :
public static long Fibon(long num)
{
long result ;
if (num == 1) { return 1; }
else if (num=2) { return 1; }
long grandfather = 1 ;
long father = 1 ;
for (in i=2;i<=num;i++)
{
result = father + grandFather;
grandfather = father ;
father = result ;
}
return result ;
}

For nth Fibonacci term the amount of memory needed by your function is O(n), i.e., linear in the index of the term in the Fibonacci sequence. More precisely, it will be n-1 times the amount of memory needed for each recursive call, which is implementation-dependent (plus some constant).
The amount of memory needed is equal to the amount of memory in each recursive call plus the "depth" of the "execution tree". In each recursive call you either terminate or make two new calls, one on the argument n-1 and one on the argument n-2; it is obvious this has to stop after n-1 calls.
If you imagine the whole process as a binary tree with nodes labeled f(k), where the node f(k) has a left child labeled f(k-1) and a right child labeled f(k-2), then the space complexity of f corresponds to the depth of the execution tree.

I believe the number of longs needed is actually equal to the returned long.
To return 2, you need to add 2 longs. To return 3, you need to add the number of longs needed to return 2 (which is 2 longs) to 1 which == 3. The pattern continues.
Since a long is 64 bits, the memory needed is equal to the fibonacci value * 64 bits.

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.

How do I turn a recursive algorithm into a tail-recursive algorithm?

As a first attempt to get into merge sort i produced the following code which works on strings because they are easier than lists to deal with.
class Program
{
static int iterations = 0;
static void Main(string[] args)
{
string test = "zvutsrqponmlihgfedcba";
test = MergeSort(test);
// test is sorted after 41 iterations
}
static string MergeSort(string input)
{
iterations++;
if (input.Length < 2)
return input;
int pivot = 0;
foreach (char c in input)
pivot += c;
pivot /= input.Length;
string left = "";
string right = "";
foreach (char c in input)
if (c <= (char)pivot)
left += c;
else
right += c;
return string.Concat(new string[] { MergeSort(left), MergeSort(right) });
}
}
Reading on Wikipedia about possible optimizations i found the following hint "To make sure at most O(log N) space is used, recurse first into the smaller half of the array, and use a tail call to recurse into the other." but honestly i have no idea how to apply this to my case.
I have some vague memories of tail calls from my IT class when we were taught about recursion and factorials but i really can't understand how to apply Wikipedia's advice to my piece of code.
Any help would be much appreciated.

There are numerous problems with this question, starting with the fact that you've implemented a very slow version of QuickSort but asked a question about MergeSort. MergeSort is not typically implemented as a tail recursive algorithm.
Let me ask a better question on your behalf:
How do I turn a recursive algorithm into a tail-recursive algorithm?
Let me sketch out a simpler tail-recursive transformation and then you can work out how to apply that to your sort, should you decide that doing so is a good idea.
Suppose you have the following recursive algorithm:
static int Count(Tree tree)
{
if (tree.IsEmpty)
return 0;
return 1 + Count(tree.Left) + Count(tree.Right);
}
Let's break that down into more steps using the following somewhat bizarre transformation:
static int Count(Tree tree)
{
int total = 0;
Tree current = tree;
if (current.IsEmpty)
return 0;
total += 1;
int countLeft = Count(current.Left);
total += countLeft;
current = current.Right;
int countRight = Count(current);
total += countRight;
return total;
}
Notice that this is exactly the same program as before, just more verbose. Of course you would not write the program in such a verbose manner, but it will help us make it tail recursive.
The point of tail recursion is to turn a recursive call into a goto. We can do that like this:
static int Count(Tree tree)
{
int total = 0;
Tree current = tree;
Restart:
if (current.IsEmpty)
return total;
int countLeft = Count(current.Left);
total += 1;
total += countLeft;
current = current.Right;
goto Restart;
}
See what we've done there? Instead of recursing, we reset the current reference to the thing that would have been recursed on, and go back to the start, while maintaining the state of the accumulator.
Now is it clear how to do the same thing to the QuickSort algorithm?

This looks like a less-than-optimal variant of QuickSort, not a MergeSort. You are missing a C# equivalent of this part:
function merge(left, right)
var list result
while length(left) > 0 or length(right) > 0
if length(left) > 0 and length(right) > 0
if first(left) <= first(right)
append first(left) to result
left = rest(left)
else
append first(right) to result
right = rest(right)
else if length(left) > 0
append first(left) to result
left = rest(left)
else if length(right) > 0
append first(right) to result
right = rest(right)
end while
return result