Challenge #59: Is it an Anagram?

[JoeM]

The solution to last week's challenge can be found HERE.

This week, let's take a break from all of those business uses cases and have some fun! In this challenge, create a field that indicates whether two values in each row are anagrams for each other. An anagram is a word formed by re-arranging the letters of another word. In our case, all anagrams are one word and are not split to multiple. No letter can be used more than once, and all letters must be used.

[MarqueeCrew]

Good morning @JoeM,

In my response I also added a TEST to determine if my results are correct.  It turns out that I learned something new here that might be useful to others.  It concerns Yes and No.  This might be obvious to everyone, but I am a binary in nature and like boolean variables.  Your results were Yes and No, so I had to reformat my boolean (True/False) results.  After writing the workflow, I found that if you SUMMARIZE data and look for the Minimum value of a field, that No is the minimum of {Yes, No}.

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Happy Mardi Gras!

Mark

[alex]

Here is my take on it. 

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[lbansal]

Here's my solution to the above.

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[Max06270]

Hi guys,

My solution is below, it is not the very optimized but it does the work ;)

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Have a great week!

Max

[AdamR_AYX]

@alex Mine was very similar to yours but I used one more summarise than I needed to

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Now to think if I can do it in any less tools :)

[alex]

Simple and quick!  Nice work.

[Hartwick96]

Here is my go at it!

[AdamR_AYX]

OK I have a one tool solution :) (Technically two if the RecordId is required)

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iif(REGEX_CountMatches([word1], "a")==REGEX_CountMatches([word2], "a")
&& REGEX_CountMatches([word1], "b")==REGEX_CountMatches([word2], "b")
&& REGEX_CountMatches([word1], "c")==REGEX_CountMatches([word2], "c")
&& REGEX_CountMatches([word1], "d")==REGEX_CountMatches([word2], "d")
&& REGEX_CountMatches([word1], "e")==REGEX_CountMatches([word2], "e")
&& REGEX_CountMatches([word1], "f")==REGEX_CountMatches([word2], "f")
&& REGEX_CountMatches([word1], "g")==REGEX_CountMatches([word2], "g")
&& REGEX_CountMatches([word1], "h")==REGEX_CountMatches([word2], "h")
&& REGEX_CountMatches([word1], "i")==REGEX_CountMatches([word2], "i")
&& REGEX_CountMatches([word1], "j")==REGEX_CountMatches([word2], "j")
&& REGEX_CountMatches([word1], "k")==REGEX_CountMatches([word2], "k")
&& REGEX_CountMatches([word1], "l")==REGEX_CountMatches([word2], "l")
&& REGEX_CountMatches([word1], "m")==REGEX_CountMatches([word2], "m")
&& REGEX_CountMatches([word1], "n")==REGEX_CountMatches([word2], "n")
&& REGEX_CountMatches([word1], "o")==REGEX_CountMatches([word2], "o")
&& REGEX_CountMatches([word1], "p")==REGEX_CountMatches([word2], "p")
&& REGEX_CountMatches([word1], "q")==REGEX_CountMatches([word2], "q")
&& REGEX_CountMatches([word1], "r")==REGEX_CountMatches([word2], "r")
&& REGEX_CountMatches([word1], "s")==REGEX_CountMatches([word2], "s")
&& REGEX_CountMatches([word1], "t")==REGEX_CountMatches([word2], "t")
&& REGEX_CountMatches([word1], "u")==REGEX_CountMatches([word2], "u")
&& REGEX_CountMatches([word1], "v")==REGEX_CountMatches([word2], "v")
&& REGEX_CountMatches([word1], "w")==REGEX_CountMatches([word2], "w")
&& REGEX_CountMatches([word1], "x")==REGEX_CountMatches([word2], "x")
&& REGEX_CountMatches([word1], "y")==REGEX_CountMatches([word2], "y")
&& REGEX_CountMatches([word1], "z")==REGEX_CountMatches([word2], "z"),
"Yes", "No")

[MarqueeCrew]

@AdamR_AYX,

You're my hero