Finding Isogram Word












2












$begingroup$


An isogram (also known as a "nonpattern word") is a word or phrase without a repeating letter, however spaces and hyphens are allowed to appear multiple times.



The IsIsogram() method takes a string and returns boolean value.




  1. INPUT: anik OUTPUT: true

  2. INPUT: A nika OUTPUT: false


Approach 01: Using Dictionary<TKey, TValue>



static bool IsIsogram(string str)
{
//create a dictionary with 26 keys and assign false as default value
var storeOccurance = new Dictionary<int, bool>();
for (int i = 0; i <= 25; i++)
{
storeOccurance[i] = false;
}

str = Regex.Replace(str, "[ -]", ""); //ignore whitespace and dash

//iterate over each character of the string
foreach (var ch in str.ToLower())
{
//(ch - 'a') is used to calculate the key. Check if the character key has false as it's value,
//meaning they were never stored
if (!storeOccurance[ch - 'a'])
{
storeOccurance[ch - 'a'] = true; //assign true when they are stored in the dictionary
}
else //otherwise, if a character key has true as it's value then it means that it exists in the dictionary
{
return false; //return false and exits the iteration..Multiple occurance in dictionary
//means the string has repeating character so it's not Isogram
}
}
return true;
}


Approach 02: Using HashSet<T>



static bool IsIsogram(string str)
{
//TRICK: HashSet<T>.Add(T) returns false if an item already exists in HashSet<T>,
//otherwise returns true and add the character to the datastructure

var charHash = new HashSet<char>();

foreach (var ch in str.ToLower())
{
//check if the charHash.Add() returns false, if it does then terminate
//as it means the character already exists
if (!charHash.Add(ch) && char.IsLetter(ch))
return false;
}
return true; //otherwise return true
}


Approach 03: Using LINQ



static bool IsIsogram(string str)
{ //Suppose "anik a" is a given string
return str.Where(char.IsLetter) //filter only the letters ||here- only a, n, i, k, a will be taken
.GroupBy(char.ToLower) //create group by characters(and transform them to lowercase) || here, there will be a group for each character a, n, i, k
.All(g => g.Count() == 1); //for every group, count the number of it's element and check if it's 1
//|| here, 'a' group has 2 elements so it return false though other groups has only one element in their group
}


Given those above approaches:




  1. Which approach would you recommend regarding readability, clean code and performance?

  2. Is there any scope for improvement in the existing approaches?










share|improve this question











$endgroup$

















    2












    $begingroup$


    An isogram (also known as a "nonpattern word") is a word or phrase without a repeating letter, however spaces and hyphens are allowed to appear multiple times.



    The IsIsogram() method takes a string and returns boolean value.




    1. INPUT: anik OUTPUT: true

    2. INPUT: A nika OUTPUT: false


    Approach 01: Using Dictionary<TKey, TValue>



    static bool IsIsogram(string str)
    {
    //create a dictionary with 26 keys and assign false as default value
    var storeOccurance = new Dictionary<int, bool>();
    for (int i = 0; i <= 25; i++)
    {
    storeOccurance[i] = false;
    }

    str = Regex.Replace(str, "[ -]", ""); //ignore whitespace and dash

    //iterate over each character of the string
    foreach (var ch in str.ToLower())
    {
    //(ch - 'a') is used to calculate the key. Check if the character key has false as it's value,
    //meaning they were never stored
    if (!storeOccurance[ch - 'a'])
    {
    storeOccurance[ch - 'a'] = true; //assign true when they are stored in the dictionary
    }
    else //otherwise, if a character key has true as it's value then it means that it exists in the dictionary
    {
    return false; //return false and exits the iteration..Multiple occurance in dictionary
    //means the string has repeating character so it's not Isogram
    }
    }
    return true;
    }


    Approach 02: Using HashSet<T>



    static bool IsIsogram(string str)
    {
    //TRICK: HashSet<T>.Add(T) returns false if an item already exists in HashSet<T>,
    //otherwise returns true and add the character to the datastructure

    var charHash = new HashSet<char>();

    foreach (var ch in str.ToLower())
    {
    //check if the charHash.Add() returns false, if it does then terminate
    //as it means the character already exists
    if (!charHash.Add(ch) && char.IsLetter(ch))
    return false;
    }
    return true; //otherwise return true
    }


    Approach 03: Using LINQ



    static bool IsIsogram(string str)
    { //Suppose "anik a" is a given string
    return str.Where(char.IsLetter) //filter only the letters ||here- only a, n, i, k, a will be taken
    .GroupBy(char.ToLower) //create group by characters(and transform them to lowercase) || here, there will be a group for each character a, n, i, k
    .All(g => g.Count() == 1); //for every group, count the number of it's element and check if it's 1
    //|| here, 'a' group has 2 elements so it return false though other groups has only one element in their group
    }


    Given those above approaches:




    1. Which approach would you recommend regarding readability, clean code and performance?

    2. Is there any scope for improvement in the existing approaches?










    share|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      An isogram (also known as a "nonpattern word") is a word or phrase without a repeating letter, however spaces and hyphens are allowed to appear multiple times.



      The IsIsogram() method takes a string and returns boolean value.




      1. INPUT: anik OUTPUT: true

      2. INPUT: A nika OUTPUT: false


      Approach 01: Using Dictionary<TKey, TValue>



      static bool IsIsogram(string str)
      {
      //create a dictionary with 26 keys and assign false as default value
      var storeOccurance = new Dictionary<int, bool>();
      for (int i = 0; i <= 25; i++)
      {
      storeOccurance[i] = false;
      }

      str = Regex.Replace(str, "[ -]", ""); //ignore whitespace and dash

      //iterate over each character of the string
      foreach (var ch in str.ToLower())
      {
      //(ch - 'a') is used to calculate the key. Check if the character key has false as it's value,
      //meaning they were never stored
      if (!storeOccurance[ch - 'a'])
      {
      storeOccurance[ch - 'a'] = true; //assign true when they are stored in the dictionary
      }
      else //otherwise, if a character key has true as it's value then it means that it exists in the dictionary
      {
      return false; //return false and exits the iteration..Multiple occurance in dictionary
      //means the string has repeating character so it's not Isogram
      }
      }
      return true;
      }


      Approach 02: Using HashSet<T>



      static bool IsIsogram(string str)
      {
      //TRICK: HashSet<T>.Add(T) returns false if an item already exists in HashSet<T>,
      //otherwise returns true and add the character to the datastructure

      var charHash = new HashSet<char>();

      foreach (var ch in str.ToLower())
      {
      //check if the charHash.Add() returns false, if it does then terminate
      //as it means the character already exists
      if (!charHash.Add(ch) && char.IsLetter(ch))
      return false;
      }
      return true; //otherwise return true
      }


      Approach 03: Using LINQ



      static bool IsIsogram(string str)
      { //Suppose "anik a" is a given string
      return str.Where(char.IsLetter) //filter only the letters ||here- only a, n, i, k, a will be taken
      .GroupBy(char.ToLower) //create group by characters(and transform them to lowercase) || here, there will be a group for each character a, n, i, k
      .All(g => g.Count() == 1); //for every group, count the number of it's element and check if it's 1
      //|| here, 'a' group has 2 elements so it return false though other groups has only one element in their group
      }


      Given those above approaches:




      1. Which approach would you recommend regarding readability, clean code and performance?

      2. Is there any scope for improvement in the existing approaches?










      share|improve this question











      $endgroup$




      An isogram (also known as a "nonpattern word") is a word or phrase without a repeating letter, however spaces and hyphens are allowed to appear multiple times.



      The IsIsogram() method takes a string and returns boolean value.




      1. INPUT: anik OUTPUT: true

      2. INPUT: A nika OUTPUT: false


      Approach 01: Using Dictionary<TKey, TValue>



      static bool IsIsogram(string str)
      {
      //create a dictionary with 26 keys and assign false as default value
      var storeOccurance = new Dictionary<int, bool>();
      for (int i = 0; i <= 25; i++)
      {
      storeOccurance[i] = false;
      }

      str = Regex.Replace(str, "[ -]", ""); //ignore whitespace and dash

      //iterate over each character of the string
      foreach (var ch in str.ToLower())
      {
      //(ch - 'a') is used to calculate the key. Check if the character key has false as it's value,
      //meaning they were never stored
      if (!storeOccurance[ch - 'a'])
      {
      storeOccurance[ch - 'a'] = true; //assign true when they are stored in the dictionary
      }
      else //otherwise, if a character key has true as it's value then it means that it exists in the dictionary
      {
      return false; //return false and exits the iteration..Multiple occurance in dictionary
      //means the string has repeating character so it's not Isogram
      }
      }
      return true;
      }


      Approach 02: Using HashSet<T>



      static bool IsIsogram(string str)
      {
      //TRICK: HashSet<T>.Add(T) returns false if an item already exists in HashSet<T>,
      //otherwise returns true and add the character to the datastructure

      var charHash = new HashSet<char>();

      foreach (var ch in str.ToLower())
      {
      //check if the charHash.Add() returns false, if it does then terminate
      //as it means the character already exists
      if (!charHash.Add(ch) && char.IsLetter(ch))
      return false;
      }
      return true; //otherwise return true
      }


      Approach 03: Using LINQ



      static bool IsIsogram(string str)
      { //Suppose "anik a" is a given string
      return str.Where(char.IsLetter) //filter only the letters ||here- only a, n, i, k, a will be taken
      .GroupBy(char.ToLower) //create group by characters(and transform them to lowercase) || here, there will be a group for each character a, n, i, k
      .All(g => g.Count() == 1); //for every group, count the number of it's element and check if it's 1
      //|| here, 'a' group has 2 elements so it return false though other groups has only one element in their group
      }


      Given those above approaches:




      1. Which approach would you recommend regarding readability, clean code and performance?

      2. Is there any scope for improvement in the existing approaches?







      c# algorithm programming-challenge linq dictionary






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 23 mins ago







      AKdeBerg

















      asked 10 hours ago









      AKdeBergAKdeBerg

      935




      935






















          4 Answers
          4






          active

          oldest

          votes


















          4












          $begingroup$

          Choice of Map Key



          Ultimately, as you are querying your map based upon the character, you should really make your key as the character. This makes your regular subtraction of 'a' unnecessary.



              var storeOccurance = new Dictionary<char, bool>();
          for (char c = 'a'; c <= 'z'; c++)
          {
          storeOccurance[c] = false;
          }


          Inconsistent behaviour for non-letter characters



          Given a string such as anik00, the first approach will produce a false response, as the duplicate 0s are detected like any other letter. The other two approaches will produce true, as 0s are ignored.



          Comments



          I don't think these comments are necessary, as the code is clear enough on its own. If you feel the need to add comments to explain what something is doing, you should extract methods instead to make these intentions clear.



          Which approach would I choose?



          The litmus test is how easy it is to understand. I've not seen any serious LINQ for a number of years, and despite that I can understand your LINQ query without difficulty. If you can create understandable code in fewer lines, it's a good thing.






          share|improve this answer









          $endgroup$





















            4












            $begingroup$

            I like the second approach since it allows you to exit as soon as a false condition is reached and is simpler than the first. I would offer one minor optimization, keep the check for isletter separate and only check for duplicate if isletter is true:



            static bool IsIsogram(string str)
            {
            //TRICK: HashSet<T>.Add(T) returns false if an item already exists in HashSet<T>,
            //otherwise returns true and add the character to the datastructure

            var charHash = new HashSet<char>();

            foreach (var ch in str.ToLower())
            {
            //check if the charHash.Add() returns false, if it does then terminate
            //as it means the character already exists
            if(char.IsLetter(ch))
            {
            if (!charHash.Add(ch))
            return false;
            }

            }
            return true; //otherwise return true
            }





            share|improve this answer









            $endgroup$





















              1












              $begingroup$

              I suggest using an array of boolean, so instead of



              var storeOccurance = new Dictionary<int, bool>();


              use simply



              var storeOccurance = new bool[26];


              Although if you want to allow a large character set instead of just a-z, say the whole of unicode, the HashSet approach may be appropriate, although in that case you would have to consider surrogates ( the situation where a character is represented by two chars ).






              share|improve this answer









              $endgroup$





















                1












                $begingroup$

                I think i would go for LINQ approach. For this simple operation, it is short and easy to understand. But just for fun i try different approach :



                public static bool IsIsogram( string input){
                if (string.IsNullOrEmpty(input))
                {
                throw new ArgumentNullException(nameof(input));
                //or we can return true , no character, no repeated character :)
                }
                var preprocessedInput = Regex.Replace(input.ToLower(), "[ -]", "");
                if (input.Length==1)
                {
                return true;
                }
                var firstHalf = preprocessedInput.Substring(0,preprocessedInput.Length/2);
                var secondHalf = preprocessedInput.Remove(0,preprocessedInput.Length/2);

                if (firstHalf.Intersect(secondHalf).Any())
                {
                return false;

                return IsIsogram(firstHalf) && IsIsogram(secondHalf);
                }


                The input string is divided into two string and checked for any intersected characters. If there is not any intersected character then, each string divided and called the method recursively. As i mentioned, this was just for fun.






                share|improve this answer








                New contributor




                Renas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                $endgroup$









                • 1




                  $begingroup$
                  You have presented an alternative solution but don't explain why it is better. Can you edit your answer to add that please?
                  $endgroup$
                  – bruglesco
                  5 hours ago











                Your Answer





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                4 Answers
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                active

                oldest

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                active

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                4












                $begingroup$

                Choice of Map Key



                Ultimately, as you are querying your map based upon the character, you should really make your key as the character. This makes your regular subtraction of 'a' unnecessary.



                    var storeOccurance = new Dictionary<char, bool>();
                for (char c = 'a'; c <= 'z'; c++)
                {
                storeOccurance[c] = false;
                }


                Inconsistent behaviour for non-letter characters



                Given a string such as anik00, the first approach will produce a false response, as the duplicate 0s are detected like any other letter. The other two approaches will produce true, as 0s are ignored.



                Comments



                I don't think these comments are necessary, as the code is clear enough on its own. If you feel the need to add comments to explain what something is doing, you should extract methods instead to make these intentions clear.



                Which approach would I choose?



                The litmus test is how easy it is to understand. I've not seen any serious LINQ for a number of years, and despite that I can understand your LINQ query without difficulty. If you can create understandable code in fewer lines, it's a good thing.






                share|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  Choice of Map Key



                  Ultimately, as you are querying your map based upon the character, you should really make your key as the character. This makes your regular subtraction of 'a' unnecessary.



                      var storeOccurance = new Dictionary<char, bool>();
                  for (char c = 'a'; c <= 'z'; c++)
                  {
                  storeOccurance[c] = false;
                  }


                  Inconsistent behaviour for non-letter characters



                  Given a string such as anik00, the first approach will produce a false response, as the duplicate 0s are detected like any other letter. The other two approaches will produce true, as 0s are ignored.



                  Comments



                  I don't think these comments are necessary, as the code is clear enough on its own. If you feel the need to add comments to explain what something is doing, you should extract methods instead to make these intentions clear.



                  Which approach would I choose?



                  The litmus test is how easy it is to understand. I've not seen any serious LINQ for a number of years, and despite that I can understand your LINQ query without difficulty. If you can create understandable code in fewer lines, it's a good thing.






                  share|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    Choice of Map Key



                    Ultimately, as you are querying your map based upon the character, you should really make your key as the character. This makes your regular subtraction of 'a' unnecessary.



                        var storeOccurance = new Dictionary<char, bool>();
                    for (char c = 'a'; c <= 'z'; c++)
                    {
                    storeOccurance[c] = false;
                    }


                    Inconsistent behaviour for non-letter characters



                    Given a string such as anik00, the first approach will produce a false response, as the duplicate 0s are detected like any other letter. The other two approaches will produce true, as 0s are ignored.



                    Comments



                    I don't think these comments are necessary, as the code is clear enough on its own. If you feel the need to add comments to explain what something is doing, you should extract methods instead to make these intentions clear.



                    Which approach would I choose?



                    The litmus test is how easy it is to understand. I've not seen any serious LINQ for a number of years, and despite that I can understand your LINQ query without difficulty. If you can create understandable code in fewer lines, it's a good thing.






                    share|improve this answer









                    $endgroup$



                    Choice of Map Key



                    Ultimately, as you are querying your map based upon the character, you should really make your key as the character. This makes your regular subtraction of 'a' unnecessary.



                        var storeOccurance = new Dictionary<char, bool>();
                    for (char c = 'a'; c <= 'z'; c++)
                    {
                    storeOccurance[c] = false;
                    }


                    Inconsistent behaviour for non-letter characters



                    Given a string such as anik00, the first approach will produce a false response, as the duplicate 0s are detected like any other letter. The other two approaches will produce true, as 0s are ignored.



                    Comments



                    I don't think these comments are necessary, as the code is clear enough on its own. If you feel the need to add comments to explain what something is doing, you should extract methods instead to make these intentions clear.



                    Which approach would I choose?



                    The litmus test is how easy it is to understand. I've not seen any serious LINQ for a number of years, and despite that I can understand your LINQ query without difficulty. If you can create understandable code in fewer lines, it's a good thing.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 9 hours ago









                    Joe CJoe C

                    805211




                    805211

























                        4












                        $begingroup$

                        I like the second approach since it allows you to exit as soon as a false condition is reached and is simpler than the first. I would offer one minor optimization, keep the check for isletter separate and only check for duplicate if isletter is true:



                        static bool IsIsogram(string str)
                        {
                        //TRICK: HashSet<T>.Add(T) returns false if an item already exists in HashSet<T>,
                        //otherwise returns true and add the character to the datastructure

                        var charHash = new HashSet<char>();

                        foreach (var ch in str.ToLower())
                        {
                        //check if the charHash.Add() returns false, if it does then terminate
                        //as it means the character already exists
                        if(char.IsLetter(ch))
                        {
                        if (!charHash.Add(ch))
                        return false;
                        }

                        }
                        return true; //otherwise return true
                        }





                        share|improve this answer









                        $endgroup$


















                          4












                          $begingroup$

                          I like the second approach since it allows you to exit as soon as a false condition is reached and is simpler than the first. I would offer one minor optimization, keep the check for isletter separate and only check for duplicate if isletter is true:



                          static bool IsIsogram(string str)
                          {
                          //TRICK: HashSet<T>.Add(T) returns false if an item already exists in HashSet<T>,
                          //otherwise returns true and add the character to the datastructure

                          var charHash = new HashSet<char>();

                          foreach (var ch in str.ToLower())
                          {
                          //check if the charHash.Add() returns false, if it does then terminate
                          //as it means the character already exists
                          if(char.IsLetter(ch))
                          {
                          if (!charHash.Add(ch))
                          return false;
                          }

                          }
                          return true; //otherwise return true
                          }





                          share|improve this answer









                          $endgroup$
















                            4












                            4








                            4





                            $begingroup$

                            I like the second approach since it allows you to exit as soon as a false condition is reached and is simpler than the first. I would offer one minor optimization, keep the check for isletter separate and only check for duplicate if isletter is true:



                            static bool IsIsogram(string str)
                            {
                            //TRICK: HashSet<T>.Add(T) returns false if an item already exists in HashSet<T>,
                            //otherwise returns true and add the character to the datastructure

                            var charHash = new HashSet<char>();

                            foreach (var ch in str.ToLower())
                            {
                            //check if the charHash.Add() returns false, if it does then terminate
                            //as it means the character already exists
                            if(char.IsLetter(ch))
                            {
                            if (!charHash.Add(ch))
                            return false;
                            }

                            }
                            return true; //otherwise return true
                            }





                            share|improve this answer









                            $endgroup$



                            I like the second approach since it allows you to exit as soon as a false condition is reached and is simpler than the first. I would offer one minor optimization, keep the check for isletter separate and only check for duplicate if isletter is true:



                            static bool IsIsogram(string str)
                            {
                            //TRICK: HashSet<T>.Add(T) returns false if an item already exists in HashSet<T>,
                            //otherwise returns true and add the character to the datastructure

                            var charHash = new HashSet<char>();

                            foreach (var ch in str.ToLower())
                            {
                            //check if the charHash.Add() returns false, if it does then terminate
                            //as it means the character already exists
                            if(char.IsLetter(ch))
                            {
                            if (!charHash.Add(ch))
                            return false;
                            }

                            }
                            return true; //otherwise return true
                            }






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 9 hours ago









                            tinstaafltinstaafl

                            6,6511827




                            6,6511827























                                1












                                $begingroup$

                                I suggest using an array of boolean, so instead of



                                var storeOccurance = new Dictionary<int, bool>();


                                use simply



                                var storeOccurance = new bool[26];


                                Although if you want to allow a large character set instead of just a-z, say the whole of unicode, the HashSet approach may be appropriate, although in that case you would have to consider surrogates ( the situation where a character is represented by two chars ).






                                share|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  I suggest using an array of boolean, so instead of



                                  var storeOccurance = new Dictionary<int, bool>();


                                  use simply



                                  var storeOccurance = new bool[26];


                                  Although if you want to allow a large character set instead of just a-z, say the whole of unicode, the HashSet approach may be appropriate, although in that case you would have to consider surrogates ( the situation where a character is represented by two chars ).






                                  share|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    I suggest using an array of boolean, so instead of



                                    var storeOccurance = new Dictionary<int, bool>();


                                    use simply



                                    var storeOccurance = new bool[26];


                                    Although if you want to allow a large character set instead of just a-z, say the whole of unicode, the HashSet approach may be appropriate, although in that case you would have to consider surrogates ( the situation where a character is represented by two chars ).






                                    share|improve this answer









                                    $endgroup$



                                    I suggest using an array of boolean, so instead of



                                    var storeOccurance = new Dictionary<int, bool>();


                                    use simply



                                    var storeOccurance = new bool[26];


                                    Although if you want to allow a large character set instead of just a-z, say the whole of unicode, the HashSet approach may be appropriate, although in that case you would have to consider surrogates ( the situation where a character is represented by two chars ).







                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered 7 hours ago









                                    George BarwoodGeorge Barwood

                                    1127




                                    1127























                                        1












                                        $begingroup$

                                        I think i would go for LINQ approach. For this simple operation, it is short and easy to understand. But just for fun i try different approach :



                                        public static bool IsIsogram( string input){
                                        if (string.IsNullOrEmpty(input))
                                        {
                                        throw new ArgumentNullException(nameof(input));
                                        //or we can return true , no character, no repeated character :)
                                        }
                                        var preprocessedInput = Regex.Replace(input.ToLower(), "[ -]", "");
                                        if (input.Length==1)
                                        {
                                        return true;
                                        }
                                        var firstHalf = preprocessedInput.Substring(0,preprocessedInput.Length/2);
                                        var secondHalf = preprocessedInput.Remove(0,preprocessedInput.Length/2);

                                        if (firstHalf.Intersect(secondHalf).Any())
                                        {
                                        return false;

                                        return IsIsogram(firstHalf) && IsIsogram(secondHalf);
                                        }


                                        The input string is divided into two string and checked for any intersected characters. If there is not any intersected character then, each string divided and called the method recursively. As i mentioned, this was just for fun.






                                        share|improve this answer








                                        New contributor




                                        Renas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.






                                        $endgroup$









                                        • 1




                                          $begingroup$
                                          You have presented an alternative solution but don't explain why it is better. Can you edit your answer to add that please?
                                          $endgroup$
                                          – bruglesco
                                          5 hours ago
















                                        1












                                        $begingroup$

                                        I think i would go for LINQ approach. For this simple operation, it is short and easy to understand. But just for fun i try different approach :



                                        public static bool IsIsogram( string input){
                                        if (string.IsNullOrEmpty(input))
                                        {
                                        throw new ArgumentNullException(nameof(input));
                                        //or we can return true , no character, no repeated character :)
                                        }
                                        var preprocessedInput = Regex.Replace(input.ToLower(), "[ -]", "");
                                        if (input.Length==1)
                                        {
                                        return true;
                                        }
                                        var firstHalf = preprocessedInput.Substring(0,preprocessedInput.Length/2);
                                        var secondHalf = preprocessedInput.Remove(0,preprocessedInput.Length/2);

                                        if (firstHalf.Intersect(secondHalf).Any())
                                        {
                                        return false;

                                        return IsIsogram(firstHalf) && IsIsogram(secondHalf);
                                        }


                                        The input string is divided into two string and checked for any intersected characters. If there is not any intersected character then, each string divided and called the method recursively. As i mentioned, this was just for fun.






                                        share|improve this answer








                                        New contributor




                                        Renas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.






                                        $endgroup$









                                        • 1




                                          $begingroup$
                                          You have presented an alternative solution but don't explain why it is better. Can you edit your answer to add that please?
                                          $endgroup$
                                          – bruglesco
                                          5 hours ago














                                        1












                                        1








                                        1





                                        $begingroup$

                                        I think i would go for LINQ approach. For this simple operation, it is short and easy to understand. But just for fun i try different approach :



                                        public static bool IsIsogram( string input){
                                        if (string.IsNullOrEmpty(input))
                                        {
                                        throw new ArgumentNullException(nameof(input));
                                        //or we can return true , no character, no repeated character :)
                                        }
                                        var preprocessedInput = Regex.Replace(input.ToLower(), "[ -]", "");
                                        if (input.Length==1)
                                        {
                                        return true;
                                        }
                                        var firstHalf = preprocessedInput.Substring(0,preprocessedInput.Length/2);
                                        var secondHalf = preprocessedInput.Remove(0,preprocessedInput.Length/2);

                                        if (firstHalf.Intersect(secondHalf).Any())
                                        {
                                        return false;

                                        return IsIsogram(firstHalf) && IsIsogram(secondHalf);
                                        }


                                        The input string is divided into two string and checked for any intersected characters. If there is not any intersected character then, each string divided and called the method recursively. As i mentioned, this was just for fun.






                                        share|improve this answer








                                        New contributor




                                        Renas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.






                                        $endgroup$



                                        I think i would go for LINQ approach. For this simple operation, it is short and easy to understand. But just for fun i try different approach :



                                        public static bool IsIsogram( string input){
                                        if (string.IsNullOrEmpty(input))
                                        {
                                        throw new ArgumentNullException(nameof(input));
                                        //or we can return true , no character, no repeated character :)
                                        }
                                        var preprocessedInput = Regex.Replace(input.ToLower(), "[ -]", "");
                                        if (input.Length==1)
                                        {
                                        return true;
                                        }
                                        var firstHalf = preprocessedInput.Substring(0,preprocessedInput.Length/2);
                                        var secondHalf = preprocessedInput.Remove(0,preprocessedInput.Length/2);

                                        if (firstHalf.Intersect(secondHalf).Any())
                                        {
                                        return false;

                                        return IsIsogram(firstHalf) && IsIsogram(secondHalf);
                                        }


                                        The input string is divided into two string and checked for any intersected characters. If there is not any intersected character then, each string divided and called the method recursively. As i mentioned, this was just for fun.







                                        share|improve this answer








                                        New contributor




                                        Renas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.









                                        share|improve this answer



                                        share|improve this answer






                                        New contributor




                                        Renas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.









                                        answered 7 hours ago









                                        RenasRenas

                                        132




                                        132




                                        New contributor




                                        Renas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.





                                        New contributor





                                        Renas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.






                                        Renas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.








                                        • 1




                                          $begingroup$
                                          You have presented an alternative solution but don't explain why it is better. Can you edit your answer to add that please?
                                          $endgroup$
                                          – bruglesco
                                          5 hours ago














                                        • 1




                                          $begingroup$
                                          You have presented an alternative solution but don't explain why it is better. Can you edit your answer to add that please?
                                          $endgroup$
                                          – bruglesco
                                          5 hours ago








                                        1




                                        1




                                        $begingroup$
                                        You have presented an alternative solution but don't explain why it is better. Can you edit your answer to add that please?
                                        $endgroup$
                                        – bruglesco
                                        5 hours ago




                                        $begingroup$
                                        You have presented an alternative solution but don't explain why it is better. Can you edit your answer to add that please?
                                        $endgroup$
                                        – bruglesco
                                        5 hours ago


















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