New answers tagged sorting
0
SAP ABAP, 174 bytes
FORM f TABLES t.DATA r TYPE int_tab1.LOOP AT t.DATA(i) = 0.DATA(x) = sy-tabix.LOOP AT t.CHECK:sy-tabix < x,t = t[ x ].i = i + 1.ENDLOOP.APPEND i TO r.ENDLOOP.t[] = r.ENDFORM.
Subroutine which takes a table of integers and replaces all values with zero-based indices. ABAP has no array type, so a table of integers is the closest thing ...
1
K (ngn/k), 18 bytes
(,/.!'#'=x)@<,/.=x
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OLD APPROACH
K (ngn/k), 27 23 22 bytes
{x[,/.=x]:,/.!'#'=x;x}
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this is not pretty... quick and dirty solution, i will be refining this later when i get the chance to think of a better approach
explanation:
=x returns a dict where keys are items of x and values are their indices (...
0
Pyth, 7 bytes
.e/<Qkb
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.e Q # enumerated map over (implicit) Q (input): lambda k,b: (k=index, b=element)
/ b # number of occurences of b in
<Qk # Q[:k]
Alternative 1-based solution (7 bytes)
m/ded._
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m ._ # map over all prefixes of (implicit) Q (input): lambda d:
/ ed # Number of occurences of d[-1]...
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Oracle SQL, 115 bytes (Version 12c+)
select x from t match_recognize(order by i all rows per match pattern((p|{-y-})+)define p as x>=nvl(last(p.x,1),x))
It works with an assumption that input data is stored in a table t(i,x) (i for order), e.g.
with t(i,x) as (select rownum,value(v) from table(sys.odcinumberlist(1,2,5,4,3,7))v)
Test in SQL*Plus.
SQL&...
1
Japt -h, 9 bytes
à ñÊf_eZñ
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1
VBA, 173 133 bytes
Sub s(c)
d=UBound(c)
ReDim r(d)
r(i)=c(i)
For i=1To d
If r(j)<=c(i)Then j=j+1:r(j)=c(i)
Next
ReDim Preserve r(j)
c=r
End Sub
(after edits) The above code is shorter and expects the parameter c to be passed as a variant array of numbers. My original Test_Cases sub below uses the Split() function which returns an array of strings. You ...
0
Wolfram Language (Mathematica), 37 bytes
Select[Reverse@Subsets@#,OrderedQ,1]&
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The elements of Subsets[ ] are ordered from shortest to longest, so reversing it puts longer subsets first. Then, select the first one which is ordered.
0
Ruby, 63 bytes
->a{s=a.size;s-=1until r=a.combination(s).find{|e|e==e.sort};r}
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->a{ # Anonymous Proc taking array input
s=a.size; # Get array size
until # Loop until the following condition is non-nil:
r=a.combination(s) ...
0
Stax, 9 bytes
éÑ<┬Θ╡Lφ0
Run and debug it
Naive approach: powerset, filter for sorted, sort on length, get tail.
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Python 2, 89 83 bytes
f=lambda A:A>sorted(A)and max([f(A[:i]+A[i+1:])for i in range(len(A))],key=len)or A
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3
Jelly, 6 bytes
ŒPṢƑƇṪ
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How?
Pretty much the same approach as Grimy's 05AB1E entry
ŒPṢƑƇṪ - Link: list of integers
ŒP - power-set (all subsequences, from shortest to longest)
Ƈ - filter keep those which are:
Ƒ - invariant when:
Ṣ - sorted
Ṫ - tail
1
Python 3, 58 56 bytes
f=lambda x:[e for i,e in enumerate(x)if max(x[:i+1])==e]
Not sure if the input would be a string or a list, my code only would work in the second case.
3
05AB1E, 8 bytes
æʒD{Q}éθ
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æ # list of subsequences
ʒD{Q} # filter, keep only the sorted subsequences
é # sort by length
θ # get the last (longest) one
0
MathGolf, 10 bytes
╓\ô`╙┼=╛o╘
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This solution relies on stack manipulation. It could have been 2 bytes shorter if negative numbers are not included (skip the first two bytes). Basically a port of the 05AB1E answer, except there is more of a hassle to get the stack right.
Explanation
╓ minimum of two elements, min of list, minimum ...
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