From f4496a8648e8c2a7c7b60b6b2ef3c90ecaba20bf Mon Sep 17 00:00:00 2001
From: Joseph Ferano <joseph@ferano.io>
Date: Tue, 21 Nov 2023 15:21:21 +0700
Subject: [PATCH] Haskell: Exercise #9

---
 Haskell.org | 44 ++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 44 insertions(+)

diff --git a/Haskell.org b/Haskell.org
index f07bcee..694ba53 100644
--- a/Haskell.org
+++ b/Haskell.org
@@ -195,3 +195,47 @@ import Data.List
 compress :: Eq a => [a] -> [a]
 compress = map head . group
 #+end_src
+*** #9 Pack consecutive duplicates of list elements into sublists
+
+If a list contains repeated elements they should be placed in separate sublists.
+
+Example:
+
+#+begin_src haskell
+,* (pack '(a a a a b c c a a d e e e e))
+((A A A A) (B) (C C) (A A) (D) (E E E E))
+#+end_src
+
+Example in Haskell:
+
+#+begin_src haskell
+λ> pack ['a', 'a', 'a', 'a', 'b', 'c', 'c', 'a', 'a', 'd', 'e', 'e', 'e', 'e']
+["aaaa","b","cc","aa","d","eeee"]
+#+end_src
+**** Solution
+In python you can use ~Counter~
+#+begin_src python
+from collections import Counter
+def pack(arr):
+    return [[key] * val for key,val in Counter(arr).items()]
+#+end_src
+
+In Haskell, this is just ~group~
+
+#+begin_src haskell
+pack :: [a] -> [[a]]
+pack = group
+#+end_src
+
+Here's my naive implementation
+
+#+begin_src haskell
+pack :: Eq a => [a] -> [[a]]
+pack xs = reverse $ go [[]] xs
+  where
+    go acc [] = acc
+    go ([]:acc) (x:xs) = go ([x]:acc) xs
+    go ((a:as):acc) (x:xs)
+       | a == x = go ((x:a:as):acc) xs
+       | otherwise = go ([x]:(a:as):acc) xs
+#+end_src