Comments on Cumulative Counts
Parent
Cumulative Counts
Challenge
Given an array of numbers return the cumulative count of each item.
This is the number of times an item has occurred so far.
Examples
[1,1,2,2,2,1,1,1,3,3] -> [1,2,1,2,3,3,4,5,1,2]
[3,7,5,4,9,2,3,2,6,6] -> [1,1,1,1,1,1,2,2,1,2]
Brownie points for beating my 7 byte APL answer.
BQN, 5 bytes ``` 1+⊒ ``` T …
3y ago
[APL (Dyalog Unicode)], 11 byt …
3y ago
[Jelly], 4 bytes ¹Ƥċ" …
3y ago
Japt, 14 11 8 bytes £¯Y …
3y ago
[APL (Dyalog Unicode)], 11 7 b …
3y ago
[Ruby], 36 bytes ```ruby - …
3y ago
[JavaScript (Node.js)], 34 29 …
2y ago
[Husk], 4 bytes Sz#ḣ …
3y ago
Scala 3, 50 44 bytes ```scala …
3y ago
[Jelly], 7 bytes =þ`ÄŒD …
3y ago
Haskell + hgl, 10 bytes ``` …
1y ago
Python 3, 70 bytes ```pytho …
2y ago
Factor, 122 bytes ``` USIN …
2y ago
Vyxal, 4 bytes ``` KƛtO ``` …
2y ago
[Python 3], 74 bytes …
2y ago
[Python 3.8 (pre-release)], 69 …
2y ago
J, 9 bytes ```J 1#.]=&><\ …
2y ago
J, 24 char ```([: +/ [: ( …
2y ago
Ruby, 31 bytes ```ruby ->a …
2y ago
Post
APL (Dyalog Unicode), 11 7 bytes (SBCS)
Razetime and rak1507 came up with 7 byte equivalents of my original dfn (this one's rak1507's). See their solutions below.
+/¨⊢=,\
+/¨⊢=,\
,\ ⍝ Prefixes of the list
= ⍝ Compare every prefix
⊢ ⍝ to the corresponding element in the original list
+/ ⍝ Sum each to get a count of how many elements in each prefix match
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