Number of loop recursions as a parameter - c#

I'm building a list of strings that contains all permutations of 2-letter strings, for instance "aa" to "zz". This is what I have:
public List<string> SomeMethod(int NumberOfChars) {
for (var i = 0; i < 26; i++)
{
char character1 = (char)(i + 97);
var Letter1 = character1.ToString();
for (var j = 0; j < 26; j++)
{
char character2 = (char)(j + 97);
var Letter2 = character2.ToString();
string TheString = Letter1 + Letter2;
TheList.Add(TheString);
}
}
return TheList;
}
Basically, it's a loop inside a loop that combines characters from the alphabet. Now suppose I want to include NumberOfChars as a parameter that determines the length of each string. For instance, if I pass in 2 it would return all 676 2-letter strings from "aa" to "zz" and if I pass in 3 it would return all 17,576 3-letter strings from "aaa" to "zzz".
For the moment, the easiest way to do it would be to have two different methods, one that returns 2-letter strings with two nested loops and another one that returns 3-letter strings with 3 nested loops.
What's a cleaner way to do this?
Thanks.

How about a method like this:
public IEnumerable<string> SomeMethod(int NumberOfChars)
{
if (NumberOfChars == 0)
{
yield return string.Empty;
}
else
{
for (var i = 'a'; i <= 'z'; i++)
{
foreach (var s in SomeMethod(NumberOfChars - 1))
{
yield return i + s;
}
}
}
}
And just for fun, here's another solution using Linq:
public IEnumerable<string> SomeMethod(int n)
{
var r = Enumerable.Range('a', 26).Select(x => ((char)x).ToString());
return (n > 1) ? r.SelectMany(x => SomeMethod(n - 1), (x, y) => x + y) : r;
}

Here's an alternative that uses a loop instead of recursion:
public static List<string> SomeMethod(int numberOfChars)
{
IEnumerable<string> results = new List<string> { "" };
for (int i = 0; i < numberOfChars; ++i)
results = from s in results
from c in Enumerable.Range('a', 26)
select s + (char)c;
return results.ToList();
}

Related

How to find all possible strings of a certain length using only the chars out of an array [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Listing all permutations of a string/integer
For example,
aaa .. aaz .. aba .. abz .. aca .. acz .. azz .. baa .. baz .. bba .. bbz .. zzz
Basically, imagine counting binary but instead of going from 0 to 1, it goes from a to z.
I have been trying to get this working to no avail and the formula is getting quite complex. I'm not sure if there's a simpler way to do it.
Edit
I have something like this at the moment but it's not quite there and I'm not sure if there is a better way:
private IEnumerable<string> GetWordsOfLength(int length)
{
char letterA = 'a', letterZ = 'z';
StringBuilder currentLetters = new StringBuilder(new string(letterA, length));
StringBuilder endingLetters = new StringBuilder(new string(letterZ, length));
int currentIndex = length - 1;
while (currentLetters.ToString() != endingLetters.ToString())
{
yield return currentLetters.ToString();
for (int i = length - 1; i > 0; i--)
{
if (currentLetters[i] == letterZ)
{
for (int j = i; j < length; j++)
{
currentLetters[j] = letterA;
}
if (currentLetters[i - 1] != letterZ)
{
currentLetters[i - 1]++;
}
}
else
{
currentLetters[i]++;
break;
}
}
}
}
For a variable amount of letter combinations, you can do the following:
var alphabet = "abcdefghijklmnopqrstuvwxyz";
var q = alphabet.Select(x => x.ToString());
int size = 4;
for (int i = 0; i < size - 1; i++)
q = q.SelectMany(x => alphabet, (x, y) => x + y);
foreach (var item in q)
Console.WriteLine(item);
var alphabet = "abcdefghijklmnopqrstuvwxyz";
//or var alphabet = Enumerable.Range('a', 'z' - 'a' + 1).Select(i => (char)i);
var query = from a in alphabet
from b in alphabet
from c in alphabet
select "" + a + b + c;
foreach (var item in query)
{
Console.WriteLine(item);
}
__EDIT__
For a general solution, you can use the CartesianProduct here
int N = 4;
var result = Enumerable.Range(0, N).Select(_ => alphabet).CartesianProduct();
foreach (var item in result)
{
Console.WriteLine(String.Join("",item));
}
// Eric Lippert’s Blog
// Computing a Cartesian Product with LINQ
// http://blogs.msdn.com/b/ericlippert/archive/2010/06/28/computing-a-cartesian-product-with-linq.aspx
public static IEnumerable<IEnumerable<T>> CartesianProduct<T>(this IEnumerable<IEnumerable<T>> sequences)
{
// base case:
IEnumerable<IEnumerable<T>> result = new[] { Enumerable.Empty<T>() };
foreach (var sequence in sequences)
{
var s = sequence; // don't close over the loop variable
// recursive case: use SelectMany to build the new product out of the old one
result =
from seq in result
from item in s
select seq.Concat(new[] { item });
}
return result;
}
You have 26^3 counts for 3 "digits". Just iterate from 'a' to 'z' in three loops.
Here's a very simple solution:
for(char first = 'a'; first <= (int)'z'; first++)
for(char second = 'a'; second <= (int)'z'; second++)
for(char third = 'a'; third <= (int)'z'; third++)
Console.WriteLine(first.ToString() + second + third);

Finding unique letter combinations within a word [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Listing all permutations of a string/integer
For example,
aaa .. aaz .. aba .. abz .. aca .. acz .. azz .. baa .. baz .. bba .. bbz .. zzz
Basically, imagine counting binary but instead of going from 0 to 1, it goes from a to z.
I have been trying to get this working to no avail and the formula is getting quite complex. I'm not sure if there's a simpler way to do it.
Edit
I have something like this at the moment but it's not quite there and I'm not sure if there is a better way:
private IEnumerable<string> GetWordsOfLength(int length)
{
char letterA = 'a', letterZ = 'z';
StringBuilder currentLetters = new StringBuilder(new string(letterA, length));
StringBuilder endingLetters = new StringBuilder(new string(letterZ, length));
int currentIndex = length - 1;
while (currentLetters.ToString() != endingLetters.ToString())
{
yield return currentLetters.ToString();
for (int i = length - 1; i > 0; i--)
{
if (currentLetters[i] == letterZ)
{
for (int j = i; j < length; j++)
{
currentLetters[j] = letterA;
}
if (currentLetters[i - 1] != letterZ)
{
currentLetters[i - 1]++;
}
}
else
{
currentLetters[i]++;
break;
}
}
}
}
For a variable amount of letter combinations, you can do the following:
var alphabet = "abcdefghijklmnopqrstuvwxyz";
var q = alphabet.Select(x => x.ToString());
int size = 4;
for (int i = 0; i < size - 1; i++)
q = q.SelectMany(x => alphabet, (x, y) => x + y);
foreach (var item in q)
Console.WriteLine(item);
var alphabet = "abcdefghijklmnopqrstuvwxyz";
//or var alphabet = Enumerable.Range('a', 'z' - 'a' + 1).Select(i => (char)i);
var query = from a in alphabet
from b in alphabet
from c in alphabet
select "" + a + b + c;
foreach (var item in query)
{
Console.WriteLine(item);
}
__EDIT__
For a general solution, you can use the CartesianProduct here
int N = 4;
var result = Enumerable.Range(0, N).Select(_ => alphabet).CartesianProduct();
foreach (var item in result)
{
Console.WriteLine(String.Join("",item));
}
// Eric Lippert’s Blog
// Computing a Cartesian Product with LINQ
// http://blogs.msdn.com/b/ericlippert/archive/2010/06/28/computing-a-cartesian-product-with-linq.aspx
public static IEnumerable<IEnumerable<T>> CartesianProduct<T>(this IEnumerable<IEnumerable<T>> sequences)
{
// base case:
IEnumerable<IEnumerable<T>> result = new[] { Enumerable.Empty<T>() };
foreach (var sequence in sequences)
{
var s = sequence; // don't close over the loop variable
// recursive case: use SelectMany to build the new product out of the old one
result =
from seq in result
from item in s
select seq.Concat(new[] { item });
}
return result;
}
You have 26^3 counts for 3 "digits". Just iterate from 'a' to 'z' in three loops.
Here's a very simple solution:
for(char first = 'a'; first <= (int)'z'; first++)
for(char second = 'a'; second <= (int)'z'; second++)
for(char third = 'a'; third <= (int)'z'; third++)
Console.WriteLine(first.ToString() + second + third);

Which iterator(other than for, foreach) can be used to count the number of character in a string?

I dont know what iteration method to be used for more efficiency, Here i have listed my solution which i have tried. is there any other way to iterate, i mean any special methods or ways?
Method One :
Here i have used two for loops so the iteration goes for 2N times
public void CountChar()
{
String s = Ipstring();
int[] counts = new int[256];
char[] c = s.ToCharArray();
for (int i = 0; i < c.Length; ++i)
{
counts[c[i]]++;
}
for (int i = 0; i < c.Length; i++)
{
Console.WriteLine(c[i].ToString() + " " + counts[c[i]]);
Console.WriteLine();
}
}
Method 2 :
public void CountChar()
{
_inputWord = Ipstring();
char[] test = _inputWord.ToCharArray();
char temp;
int count = 0, tcount = 0;
Array.Sort(test);
int length = test.Length;
temp = test[0];
while (length > 0)
{
for (int i = 0; i < test.Length; i++)
{
if (temp == test[i])
{
count++;
}
}
Console.WriteLine(temp + " " + count);
tcount = tcount + count;
length = length - count;
count = 0;
if (tcount != test.Length)
temp = test[tcount];
//atchutharam. aaachhmrttu
}
}
Method three:
public void CountChar()
{
int indexcount = 0;
s = Ipstring();
int[] count = new int[s.Length];
foreach (char c in s)
{
Console.Write(c);
count[s.IndexOf(c)]++;
}
foreach (char c in s)
{
if (indexcount <= s.IndexOf(c))
{
Console.WriteLine(c);
Console.WriteLine(count[s.IndexOf(c)]);
Console.WriteLine("");
}
indexcount++;
////atchutharam
}
}
You can use LINQ methods to group the characters and count them:
public void CountChar() {
String s = Ipstring();
foreach (var g in s.GroupBy(c => c)) {
Console.WriteLine("{0} : {1}", g.Key, g.Count());
}
}
Your loops are not nested so your complexity is not N*N (O(n^2)) but 2*N which gives O(N) because you can always ignore constants :
for(){}
for(){} // O(2N) = O(N)
for()
{
for(){}
} // O(N*N) = O(N^2)
If you really want to know which one of these 3 solutions have the fastest execution time in a specific environment, do a benchmark.
If you want the one that is the most clean and readable (And you should almost always aim for that), just use LINQ :
String s = Ipstring();
int count = s.Count();
It will execute in O(N) too.
If you need the results in arrays:
var groups = s.GroupBy(i => i ).OrderBy( g => g.Key );
var chars = groups.Select(g => g.Key).ToArray();
var counts = groups.Select(g => g.Count()).ToArray();
Otherwise:
var dict = s.GroupBy(i => i).ToDictionary(g => g.Key, g => g.Count());
foreach (var g in dict)
{
Console.WriteLine( "{0}: {1}", g.Key, g.Value );
}

Counting how many times a certain char appears in a string before any other char appears

I have many strings. Each string is prepended with at least 1 $. What is the best way to loop through the chars of each string to count how many $'s there are per string.
eg:
"$hello" - 1
"$$hello" - 2
"$$h$ello" - 2
You could use the Count method
var count = mystring.Count(x => x == '$')
int count = myString.TakeWhile(c => c == '$').Count();
And without LINQ
int count = 0;
while(count < myString.Length && myString[count] == '$') count++;
The simplest approach would be to use LINQ:
var count = text.TakeWhile(c => c == '$').Count();
There are certainly more efficient approaches, but that's probably the simplest.
You could do this, it doesn't require LINQ, but it's not the best way to do it(since you make split the whole string and put it in an array and just pick the length of it, you could better just do a while loop and check every character), but it works.
int count = test.Split('$').Length - 1;
var str ="hello";
str.Where(c => c == 'l').Count() // 2
int count = yourText.Length - yourText.TrimStart('$').Length;
int count = Regex.Matches(myString,"$").Count;
public static int GetHowManyTimeOccurenceCharInString(string text, char c)
{
int count = 0;
foreach(char ch in text)
{
if(ch.Equals(c))
{
count++;
}
}
return count;
}
just a simple answer:
public static int CountChars(string myString, char myChar)
{
int count = 0;
for (int i = 0; i < myString.Length; i++)
{
if (myString[i] == myChar) ++count;
}
return count;
}
Cheers! - Rick
One approach you could take is the following method:
// Counts how many of a certain character occurs in the given string
public static int CharCountInString(char chr, string str)
{
return str.Split(chr).Length-1;
}
As per the parameters this method returns the count of a specific character within a specific string.
This method works by splitting the string into an array by the specified character and then returning the length of that array -1.
//This code worked for me
class CountOfLettersOfString
{
static void Main()
{
Console.WriteLine("Enter string to check count of letters");
string name = Console.ReadLine();
//Method1
char[] testedalphabets = new char[26];
int[] letterCount = new int[26];
int countTestesd = 0;
Console.WriteLine($"Given String is:{name}");
for (int i = 0; i < name.Length - 1; i++)
{
int countChar = 1;
bool isCharTested = false;
for (int j = 0; j < testedalphabets.Length - 1; j++)
{
if (name[i] == testedalphabets[j])
{
isCharTested = true;
break;
}
}
if (!isCharTested)
{
testedalphabets[countTestesd] = name[i];
for (int k = i + 1; k < name.Length - 1; k++)
{
if (name[i] == name[k])
{
countChar++;
}
}
letterCount[countTestesd] = countChar;
countTestesd++;
}
else
{
continue;
}
}
for (int i = 0; i < testedalphabets.Length - 1; i++)
{
if (!char.IsLetter(testedalphabets[i]))
{
continue;
}
Console.WriteLine($"{testedalphabets[i]}-{letterCount[i]}");
}
//Method2
var g = from c in name.ToLower().ToCharArray() // make sure that L and l are the same eg
group c by c into m
select new { Key = m.Key, Count = m.Count() };
foreach (var item in g)
{
Console.WriteLine(string.Format("Character:{0} Appears {1} times", item.Key.ToString(), item.Count));
}
Console.ReadLine();
}
}
This is a similar Solution to find how many email addresses included in a string. This way is more efficient`
int count = 0;
foreach (char c in email.Trim())
if (c == '#') count++;

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.

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