Develop Reference

Trying to find Frequency of elements in array

I am trying to count how many times a number appears in an array 1 (a1) and then trying to print out that number by storing it in array 2 (a2) just once and then try to print array 2. But first using for loop and a function, I will check that if a number already exist in array 2 then move to next index in array 1, unfortunateley this code is not working; can someone please help me in trying to fix it, I don't need some complex solutions like dictionaries or lists athe moment, although it might be helpful too. thanks, I am not an expert in programming and I try to practise it in my free time, so please help me.
I just want this code to be fixed for my understanding and knowledge
class Program
{
static void Main(string[] args)
{
int i, j;
int[] a1 = new int[10];
int[] a2 = new int[10];
int[] a3 = new int[10];
//takes an input
for (i = 0; i < a1.Length; i++)
{
a1[i] = Convert.ToInt32(Console.ReadLine());
}
for (i = 0; i < a1.Length; i++)
{
Cn(a1, a2); //calls in function
i++; //increments is if true
int count = 0;
for (j = 0; j < a1.Length; j++)
{
//if a number matches with a number in second array
if (a1[i] == a1[j])
{
//do count ++
count++;
// store that number into second array
a2[i] = a1[i];
}
}
//store the number of counts in third array
a3[i] = count;
}
for (i = 0; i < a2.Length; i++)
{
if (a2[i] != 0)
{
Console.WriteLine(a2[i]);
}
}
Console.ReadLine();
}
//function to check if element at current index of array 1 exists in array 2 if yes than break
public static void Cn (int[] aa1, int [] aa2)
{
int k, j;
for ( k = 0; k < aa1.Length; k++)
{
for (j = 0; j < aa2.Length; j++)
{
if (aa1[k] == aa2[j])
break;
}
}
}
}

You probably want to do a group by count:
int[] a1 = new int[10];
var rnd = new Random();
//takes an input
for (int i = 0; i < a1.Length; i++)
{
a1[i] = Convert.ToInt32(rnd.Next(0, 11)); // or Console.ReadLine()
}
var grouped = a1
.GroupBy(x => x)
.Select(g => new
{
Item = g.Key,
Count = g.Count()
}).ToList(); // ToList() is optional, materializes the IEnumerable
foreach (var item in grouped)
{
Console.WriteLine($"number: {item.Item}, count: {item.Count}");
}
This uses a Hash algorithm internally.
You can solve this without a Hash or Dictionary but it wouldn't be very efficient because you need to do lots of linear searches through the arrays.
The advantage of a Hash algorithm is that your lookups or groupings are much faster than if you loop over a complete array to find / increment an item.

Hackerrank: Climbing the Leaderboard

i have a deal with a hackerrank algorithm problem.
It works at all cases, except 6-7-8-9. It gives timeout error. I had spent so much time at this level. Someone saw where is problem?
static long[] climbingLeaderboard(long[] scores, long[] alice)
{
//long[] ranks = new long[scores.Length];
long[] aliceRanks = new long[alice.Length]; // same length with alice length
long lastPoint = 0;
long lastRank;
for (long i = 0; i < alice.Length; i++)
{
lastPoint = scores[0];
lastRank = 1;
bool isIn = false; // if never drop in if statement
for (long j = 0; j < scores.Length; j++)
{
if (lastPoint != scores[j]) //if score is not same, raise the variable
{
lastPoint = scores[j];
lastRank++;
}
if (alice[i] >= scores[j])
{
aliceRanks[i] = lastRank;
isIn = true;
break;
}
aliceRanks[i] = !isIn & j + 1 == scores.Length ? ++lastRank : aliceRanks[i]; //drop in here
}
}
return aliceRanks;
}

This problem can be solved in O(n) time, no binary search needed at all. First, we need to extract the most useful piece of data given in the problem statement, which is,
The existing leaderboard, scores, is in descending order.
Alice's scores, alice, are in ascending order.
An approach that makes this useful is to create two pointers, one at the start of alice array, let's call it "i", and the second is at the end of scores array, let's call it "j". We then loop until i reaches the end of alice array and at each iteration, we check for three main conditions. We increment i by one if alice[i] is less than scores[j] because the next element of alice may be also less than the current element of scores, or we decrement j if alice[i] is greater than scores[j] because we are sure that the next elements of alice are also greater than those elements discarded in scores. The last condition is that if alice[i] == scores[j], we only increment i.
I solved this question in C++, my goal here is to make you understand the algorithm, I think you can easily convert it to C# if you understand it. If there are any confusions, please tell me. Here is the code:
// Complete the climbingLeaderboard function below.
vector<int> climbingLeaderboard(vector<int> scores, vector<int> alice) {
int j = 1, i = 1;
// this is to remove duplicates from the scores vector
for(i =1; i < scores.size(); i++){
if(scores[i] != scores[i-1]){
scores[j++] = scores[i];
}
}
int size = scores.size();
for(i = 0; i < size-j; i++){
scores.pop_back();
}
vector<int> ranks;
i = 0;
j = scores.size()-1;
while(i < alice.size()){
if(j < 0){
ranks.push_back(1);
i++;
continue;
}
if(alice[i] < scores[j]){
ranks.push_back(j+2);
i++;
} else if(alice[i] > scores[j]){
j--;
} else {
ranks.push_back(j+1);
i++;
}
}
return ranks;
}
I think this may help you too:
vector is like an array list that resizes itself.
push_back() is inserting at the end of the vector.
pop_back() is removing from the end of the vector.

Here is my solution with c#
public static List<int> climbingLeaderboard(List<int> ranked, List<int> player)
{
List<int> result = new List<int>();
ranked = ranked.Distinct().ToList();
var pLength = player.Count;
var rLength = ranked.Count-1;
int j = rLength;
for (int i = 0; i < pLength; i++)
{
for (; j >= 0; j--)
{
if (player[i] == ranked[j])
{
result.Add(j + 1);
break;
}
else if(player[i] < ranked[j])
{
result.Add(j + 2);
break;
}
else if(player[i] > ranked[j]&&j==0)
{
result.Add(1);
break;
}enter code here
}
}
return result;
}

Here is a solution that utilizes BinarySearch. This method returns the index of the searched number in the array, or if the number is not found then it returns a negative number that is the bitwise complement of the index of the next element in the array. Binary search only works in sorted arrays.
public static int[] GetRanks(long[] scores, long[] person)
{
var defaultComparer = Comparer<long>.Default;
var reverseComparer = Comparer<long>.Create((x, y) => -defaultComparer.Compare(x, y));
var distinctOrderedScores = scores.Distinct().OrderBy(i => i, reverseComparer).ToArray();
return person
.Select(i => Array.BinarySearch(distinctOrderedScores, i, reverseComparer))
.Select(pos => (pos >= 0 ? pos : ~pos) + 1)
.ToArray();
}
Usage example:
var scores = new long[] { 100, 100, 50, 40, 40, 20, 10 };
var alice = new long[] { 5, 25, 50, 120 };
var ranks = GetRanks(scores, alice);
Console.WriteLine($"Ranks: {String.Join(", ", ranks)}");
Output:
Ranks: 6, 4, 2, 1

I was bored so i gave this a go with Linq and heavily commented it for you,
Given
public static IEnumerable<int> GetRanks(long[] scores, long[] person)
// Convert scores to a tuple
=> scores.Select(s => (scores: s, isPerson: false))
// convert persons score to a tuple and concat
.Concat(person.Select(s => (scores: s, isPerson: true)))
// Group by scores
.GroupBy(x => x.scores)
// order by score
.OrderBy(x => x.Key)
// select into an indexable tuple so we know everyones rank
.Select((groups, i) => (rank: i, groups))
// Filter the person
.Where(x => x.groups.Any(y => y.isPerson))
// select the rank
.Select(x => x.rank);
Usage
static void Main(string[] args)
{
var scores = new long[]{1, 34, 565, 43, 44, 56, 67};
var alice = new long[]{578, 40, 50, 67, 6};
var ranks = GetRanks(scores, alice);
foreach (var rank in ranks)
Console.WriteLine(rank);
}
Output
1
3
6
8
10

Based on the given constraint brute-force solution will not be efficient for the problem.
you have to optimize your code and the key part here is to look up for exact place which can be effectively done by using binary search.
Here is the solution using binary search:-
static int[] climbingLeaderboard(int[] scores, int[] alice) {
int n = scores.length;
int m = alice.length;
int res[] = new int[m];
int[] rank = new int[n];
rank[0] = 1;
for (int i = 1; i < n; i++) {
if (scores[i] == scores[i - 1]) {
rank[i] = rank[i - 1];
} else {
rank[i] = rank[i - 1] + 1;
}
}
for (int i = 0; i < m; i++) {
int aliceScore = alice[i];
if (aliceScore > scores[0]) {
res[i] = 1;
} else if (aliceScore < scores[n - 1]) {
res[i] = rank[n - 1] + 1;
} else {
int index = binarySearch(scores, aliceScore);
res[i] = rank[index];
}
}
return res;
}
private static int binarySearch(int[] a, int key) {
int lo = 0;
int hi = a.length - 1;
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
if (a[mid] == key) {
return mid;
} else if (a[mid] < key && key < a[mid - 1]) {
return mid;
} else if (a[mid] > key && key >= a[mid + 1]) {
return mid + 1;
} else if (a[mid] < key) {
hi = mid - 1;
} else if (a[mid] > key) {
lo = mid + 1;
}
}
return -1;
}
You can refer to this link for a more detailed video explanation.

static int[] climbingLeaderboard(int[] scores, int[] alice) {
int[] uniqueScores = IntStream.of(scores).distinct().toArray();
int [] rank = new int [alice.length];
int startIndex=0;
for(int j=alice.length-1; j>=0;j--) {
for(int i=startIndex; i<=uniqueScores.length-1;i++) {
if (alice[j]<uniqueScores[uniqueScores.length-1]){
rank[j]=uniqueScores.length+1;
break;
}
else if(alice[j]>=uniqueScores[i]) {
rank[j]=i+1;
startIndex=i;
break;
}
else{continue;}
}
}
return rank;
}

My solution in javascript for climbing the Leaderboard Hackerrank problem. The time complexity of the problem can be O(i+j), i is the length of scores and j is the length of alice. The space complexity is O(1).
// Complete the climbingLeaderboard function below.
function climbingLeaderboard(scores, alice) {
const ans = [];
let count = 0;
// the alice array is arranged in ascending order
let j = alice.length - 1;
for (let i = 0 ; i < scores.length ; i++) {
const score = scores[i];
for (; j >= 0 ; j--) {
if (alice[j] >= score) {
// if higher than score
ans.unshift(count+1);
} else if (i === scores.length - 1) {
// if smallest
ans.unshift(count+2);
} else {
break;
}
}
// actual rank of the score in leaderboard
if (score !== scores[i-1]) {
count++;
}
}
return ans;
}

Here is my solution
List<int> distinct = null;
List<int> rank = new List<int>();
foreach (int item in player)
{
ranked.Add(item);
ranked.Sort();
ranked.Reverse();
distinct = ranked.Distinct().ToList();
for (int i = 0; i < distinct.Count; i++)
{
if (item == distinct[i])
{
rank.Add(i + 1);
break;
}
}
}
return rank;
This can be modified by removing the inner for loop also
List<int> distinct = null;
List<int> rank = new List<int>();
foreach (int item in player)
{
ranked.Add(item);
ranked.Sort();
ranked.Reverse();
distinct = ranked.Distinct().ToList();
var index = ranked.FindIndex(x => x == item);
rank.Add(index + 1);
}
return rank;

This is my solution in c# for Hackerrank Climbing the Leaderboard based on C++ answer here.
public static List<int> climbingLeaderboard(List<int> ranked, List<int> player)
{
List<int> _ranked = new List<int>();
_ranked.Add(ranked[0]);
for(int a=1; a < ranked.Count(); a++)
if(_ranked[_ranked.Count()-1] != ranked[a])
_ranked.Add(ranked[a]);
int j = _ranked.Count()-1;
int i = 0;
while(i < player.Count())
{
if(j < 0)
{
player[i] = 1;
i++;
continue;
}
if(player[i] < _ranked[j])
{
player[i] = j+2;
i++;
}
else
if(player[i] == _ranked[j])
{
player[i] = j+1;
i++;
}
else
{
j--;
}
}
return player;
}

My solution in Java for climbing the Leaderboard Hackerrank problem.
// Complete the climbingLeaderboard function below.
static int[] climbingLeaderboard(int[] scores, int[] alice) {
Arrays.sort(scores);
HashSet<Integer> set = new HashSet<Integer>();
int[] ar = new int[alice.length];
int sc = 0;
for(int i=0; i<alice.length; i++){
sc = 1;
set.clear();
for(int j=0; j<scores.length; j++){
if(alice[i] < scores[j] && !set.contains(scores[j])){
sc++;
set.add(scores[j]);
}
}
ar[i] = sc;
}return ar;
}

Remove Nulls from string[,]

I have a string array defined in c# as
string[,] options = new string[100,3];
Throughout the code it gets populated with data but not always filled.
So if I have 80 parts of it filled and 20 parts of it not filled. The 20 parts have nulls in them or 60 nulls at the end. Is there an easy way to resize the array so that after filling it the array is the same as
String[,] options = new string[80,3];
It would have to be resized based on the position of the first set of 3 nulls it found.
If this was a jagged array I would have done
options = options.Where(x => x != null).ToArray();

The method is quite long, because it has to check every row twice...
public static string[,] RemoveEmptyRows(string[,] strs)
{
int length1 = strs.GetLength(0);
int length2 = strs.GetLength(1);
// First we count the non-emtpy rows
int nonEmpty = 0;
for (int i = 0; i < length1; i++)
{
for (int j = 0; j < length2; j++)
{
if (strs[i, j] != null)
{
nonEmpty++;
break;
}
}
}
// Then we create an array of the right size
string[,] strs2 = new string[nonEmpty, length2];
for (int i1 = 0, i2 = 0; i2 < nonEmpty; i1++)
{
for (int j = 0; j < length2; j++)
{
if (strs[i1, j] != null)
{
// If the i1 row is not empty, we copy it
for (int k = 0; k < length2; k++)
{
strs2[i2, k] = strs[i1, k];
}
i2++;
break;
}
}
}
return strs2;
}
Use it like:
string[,] options = new string[100, 3];
options[1, 0] = "Foo";
options[3, 1] = "Bar";
options[90, 2] = "fiz";
options = RemoveEmptyRows(options);
As suggested by Alexei, there is another way of doing this:
public static string[,] RemoveEmptyRows2(string[,] strs)
{
int length1 = strs.GetLength(0);
int length2 = strs.GetLength(1);
// First we put somewhere a list of the indexes of the non-emtpy rows
var nonEmpty = new List<int>();
for (int i = 0; i < length1; i++)
{
for (int j = 0; j < length2; j++)
{
if (strs[i, j] != null)
{
nonEmpty.Add(i);
break;
}
}
}
// Then we create an array of the right size
string[,] strs2 = new string[nonEmpty.Count, length2];
// And we copy the rows from strs to strs2, using the nonEmpty
// list of indexes
for (int i1 = 0; i1 < nonEmpty.Count; i1++)
{
int i2 = nonEmpty[i1];
for (int j = 0; j < length2; j++)
{
strs2[i1, j] = strs[i2, j];
}
}
return strs2;
}
This one, in the tradeoff memory vs time, chooses time. It is probably faster, because it doesn't have to check every row twice, but it uses more memory, because it puts somewhere a list of the non-empty indexes.

I went for all rows until you find an row with all null values:
Needs some clean up and will obviously remove non-null rows that occur after the first all null row. The requirement wasn't too clear here
EDIT: Just seen the comment clarifying requirement to remove all null rows - I've tweaked the below to avoid downvotes but a more comprehensive answer is already accepted (and is more efficient) :)
void Main()
{
string[,] options = new string[100,3];
options[0,0] = "bleb";
options[1,1] = "bleb";
options[2,0] = "bleb";
options[2,1] = "bleb";
options[3,2] = "bleb";
options[4,1] = "bleb";
string[,] trimmed = TrimNullRows(options);
Console.WriteLine(trimmed);
}
public string[,] TrimNullRows(string[,] options)
{
IList<string[]> nonNullRows = new List<string[]>();
for (int x = 0; x < options.GetLength(0); x++)
{
bool allNull = true;
var row = new string[options.GetLength(1)];
for (int y = 0; y < options.GetLength(1); y++)
{
row[y] = options[x,y];
allNull &= options[x,y] == null;
}
if (!allNull)
{
nonNullRows.Add(row);
}
}
var optionsTrimmed = new string[nonNullRows.Count, options.GetLength(1)];
for (int i=0;i<nonNullRows.Count;i++)
{
for (int j=0;j<options.GetLength(1);j++)
{
optionsTrimmed[i, j] = nonNullRows[i][j];
}
}
return optionsTrimmed;
}

You can also get yourself some helpers to convert between jagged and multi-dimensional representations. This is pretty silly, of course, but for arrays as small as the ones you're showing (and also, very sparse arrays), it'll be fine.
void Main()
{
string[,] options = new string[100,3];
options[3, 1] = "Hi";
options[5, 0] = "Dan";
var results =
options
.JagIt()
.Where(i => i.Any(j => j != null))
.UnjagIt();
results.Dump();
}
static class Extensions
{
public static IEnumerable<IEnumerable<T>> JagIt<T>(this T[,] array)
{
for (var i = 0; i < array.GetLength(0); i++)
yield return GetRow(array, i);
}
public static IEnumerable<T> GetRow<T>(this T[,] array, int rowIndex)
{
for (var j = 0; j < array.GetLength(1); j++)
yield return array[rowIndex, j];
}
public static T[,] UnjagIt<T>(this IEnumerable<IEnumerable<T>> jagged)
{
var rows = jagged.Count();
if (rows == 0) return new T[0, 0];
var columns = jagged.Max(i => i.Count());
var array = new T[rows, columns];
var row = 0;
var column = 0;
foreach (var r in jagged)
{
column = 0;
foreach (var c in r)
{
array[row, column++] = c;
}
row++;
}
return array;
}
}
The JagIt method is pretty simple of course - we'll just iterate over the rows, and yield the individual items. This gives us an enumerable of enumerables, which we can use in LINQ quite easily. If desired, you could transform those into arrays, of course (say, Select(i => i.ToArray()).ToArray()).
The UnjagIt method is a bit more talkative, because we need to create the target array with the correct dimensions first. And there's no unyield instruction to simplify that :D
This is pretty inefficient, of course, but that isn't necessarily a problem. You could save yourself some of the iterations by keeping the inner enumerable an array, for example - that will save us having to iterate over all the inner items.
I'm mostly keeping this as the memory-cheap, CPU-intensive alternative to #xanatos' memory-intensive, CPU-cheap (relatively).
Of course, the main bonus is that it can be used to treat any multi-dimensional arrays as jagged arrays, and convert them back again. General solutions usually aren't the most efficient :D

Yet another variant with linq
static string[,] RemoveNotNullRow(string[,] o)
{
var rowLen = o.GetLength(1);
var notNullRowIndex = (from oo in o.Cast<string>().Select((x, idx) => new { idx, x })
group oo.x by oo.idx / rowLen into g
where g.Any(f => f != null)
select g.Key).ToArray();
var res = new string[notNullRowIndex.Length, rowLen];
for (int i = 0; i < notNullRowIndex.Length; i++)
{
Array.Copy(o, notNullRowIndex[i] * rowLen, res, i * rowLen, rowLen);
}
return res;
}

Order an array in a specific order

I have this array of integers:-
int[] numbers = new int[] { 10, 20, 30, 40 };
I am trying to create an array which will have first element, last element, second element, second-last element and so on..
So, my resulting output will be:-
int[] result = {10,40,20,30};
This was my approach, in one loop start from first and go till the middle & in second loop start from last and get to the middle and select items accordingly, but I totally messed it up. Here is my attempted code:-
private static IEnumerable<int> OrderedArray(int[] numbers)
{
bool takeFirst = true;
if (takeFirst)
{
takeFirst = false;
for (int i = 0; i < numbers.Length / 2; i++)
{
yield return numbers[i];
}
}
else
{
takeFirst = true;
for (int j = numbers.Length; j < numbers.Length / 2; j--)
{
yield return numbers[j];
}
}
}
Need Help.

You might try this:
int[] result = numbers.Zip(numbers.Reverse(), (n1,n2) => new[] {n1, n2})
.SelectMany(x =>x)
.Take(numbers.Length)
.ToArray();
Explanation: This approach basically pairs up the elements of the original collection with the elements of its reverse ordered collection (using Zip). So you get a collection of pairs like [first, last], [second, second from last], etc.
It then flattens those collection of pairs into a single collection (using SelectMany). So the collection becomes [first, last, second, second from last,...].
Finally, we limit the number of elements to the length of the original array (n). Since we are iterating through twice as many elements (normal and reverse), it works out that iterating through n elements allow us to stop in the middle of the collection.
As a different approach, this is a modification on your existing method:
private static IEnumerable<int> OrderedArray(int[] numbers)
{
var count = (numbers.Length + 1) / 2;
for (int i = 0; i < count; i++)
{
yield return numbers[i];
int reverseIdx = numbers.Length - 1 - i;
if(i != reverseIdx)
yield return numbers[reverseIdx];
}
}

ok,
public static class Extensions
{
public static IEnumerable<T> EndToEnd<T>(this IReadOnlyList<T> source)
{
var length = source.Count;
var limit = length / 2;
for (var i = 0; i < limit; i++)
{
yield return source[i];
yield return source[length - i - 1];
}
if (length % 2 > 0)
{
yield return source[limit];
}
}
}
Which you could use like this,
var result = numbers.EndToEnd().ToArray();
more optimally,
public static class Extensions
{
public static IEnumerable<T> EndToEnd<T>(this IReadOnlyList<T> source)
{
var c = source.Count;
for (int i = 0, f = 0, l = c - 1; i < c; i++, f++, l--)
{
yield return source[f];
if (++i == c)
{
break;
}
yield return source[l];
}
}
}
no divide or modulus required.

With a simple for;
int len = numbers.Length;
int[] result = new int[len];
for (int i = 0, f = 0, l = len - 1; i < len; f++, l--)
{
result[i++] = numbers[f];
if (f != l)
result[i++] = numbers[l];
}

Based on Selman22's now deleted answer:
int[] numbers = new int[] { 10, 20, 30, 40 };
int[] result = numbers
.Select((x,idx) => idx % 2 == 0
? numbers[idx/2]
: numbers[numbers.Length - 1 -idx/2])
.ToArray();
result.Dump();
(The last line is LinqPad's way of outputting the results)
Or in less LINQy form as suggested by Jeppe Stig Nielsen
var result = new int[numbers.Length];
for (var idx = 0; idx < result.Length; idx++) {
result[idx] = idx % 2 == 0 ? numbers[idx/2] : numbers[numbers.Length - 1 -idx/2];
}
The principle is that you have two sequences, one for even elements (in the result) and one for odd. The even numbers count the first half of the array and the odds count the second half from the back.
The only modification to Selman's code is adding the /2 to the indexes to keep it counting one by one in the right half while the output index (which is what idx basically is in this case) counts on.

Came up with this
static void Main(string[] args)
{
List<int> numbers = new List<int>() { 10, 20, 30, 40, 50, 60, 70};
List<int> numbers2 = new List<int>();
int counter1 = 0;
int counter2 = numbers.Count - 1;
int remainder = numbers.Count % 2 == 0 ? 1: 0;
while (counter1-1 < counter2)
{
if (counter1 + counter2 % 2 == remainder)
{
numbers2.Add(numbers[counter1]);
counter1++;
}
else
{
numbers2.Add(numbers[counter2]);
counter2--;
}
}
string s = "";
for(int a = 0; a< numbers2.Count;a++)
s+=numbers2[a] + " ";
Console.Write(s);
Console.ReadLine();
}

This late answer steals a lot from the existing answers!
The idea is to allocate the entire result array at once (since its length is known). Then fill out all even-indexed members first, from one end of source. And finally fill out odd-numbered entries from the back end of source.
public static TElement[] EndToEnd<TElement>(this IReadOnlyList<TElement> source)
{
var count = source.Count;
var result = new TElement[count];
for (var i = 0; i < (count + 1) / 2; i++)
result[2 * i] = source[i];
for (var i = 1; i <= count / 2; i++)
result[2 * i - 1] = source[count - i];
return result;
}

Came up with this
public int[] OrderedArray(int[] numbers)
{
int[] final = new int[numbers.Length];
var limit=numbers.Length;
int last = numbers.Length - 1;
var finalCounter = 0;
for (int i = 0; finalCounter < numbers.Length; i++)
{
final[finalCounter] = numbers[i];
final[((finalCounter + 1) >= limit ? limit - 1 : (finalCounter + 1))] = numbers[last];
finalCounter += 2;
last--;
}
return final;
}

Implementation of an anagram function in C#

Possible Duplicate:
What is an easy way to tell if a list of words are anagrams of each other?
What is the best way (performance wide) to write a function in C# that takes two strings and returns true when the strings are anagrams of each other and otherwise returns false. Example of anagrams are:
abet beat beta bate
abides biased
anagrams link
In implementing this, is it possible that there is space in each string?
Any idea would be very much appreciated!

A simple (naïve?) way, using LINQ:
"abides".OrderBy(c=>c).SequenceEqual("biased".OrderBy(c=>c))

An easy solution would be to sort the characters alphabetically and compare them to one another.
public static class AnagramExtensions
{
public static bool IsAnagramOf(this string word1, string word2)
{
return word1.OrderBy(x => x).SequenceEqual(word2.OrderBy(x => x));
}
}
Then, to use it:
static void Main()
{
string word1 = "cat";
string word2 = "tac";
Console.WriteLine(word1.IsAnagramOf(word2));
string word3 = "cat";
string word4 = "dog";
Console.WriteLine(word3.IsAnagramOf(word4));
}
The output in this case would be
True
False

How not to do this: Remove all whitespace from each of the strings. Use one of the algorithms at Algorithm to generate anagrams to generate all possible permutations of the first string. Finally, search the list of permuations for a match; if there is one, then the two are anagrams, otherwise, not.

I have a solution for a List of Strings (Not just two Strings).
If you are interested in it or any other one, you can use it.
Given an array of strings, remove each string that is an anagram of an earlier string, then return the remaining array in stored order.
private static List<string> GencoAnagrams(List<string> textList)
{
var listCount = textList.Count;
if (listCount == 1) return textList;
for (var j = 1; j < listCount; j++)
{
for (var i = 0; i < j; i++)
{
if (string.Concat(textList[j].OrderBy(x => x)).Equals(string.Concat(textList[i].OrderBy(y => y))))
{
textList.RemoveAt(j);
--listCount;
--j;
if (listCount == 1) break;
}
}
if (listCount == 1) break;
}
return textList;
}

and count all sceniro (n*n+1)/2
public static int sherlockAndAnagrams(string s)
{
int count = 0;
string[] stringList = new string[(s.Length * (s.Length + 1)) / 2];
int index = 0;
Dictionary<string, int> hash = new Dictionary<string, int>();
for (int i = 0; i < s.Length; i++)
{
for (int j = i; j < s.Length; j++)
{
stringList[index] = s.Substring(i, j - i + 1);
index++;
}
}
foreach (var str in stringList)
{
var keyString = string.Concat(str.OrderBy(x => x));
if (hash.ContainsKey(keyString))
{
hash[keyString]++;
}
else
{
hash[keyString] = 1;
}
}
foreach (var key in hash.Keys)
{
if (hash[key] <= 1)
{
continue;
}
else
{
count = count + ((hash[key] * (hash[key] - 1) )/ 2);
}
}
return count;
}