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First rarest element

Code Golf

counting add tag

DJMcMayhem

Given two lists, A and B, return the first element of A that appears the fewest times in B.

For example,

A = [1, 2, 3, 4, 5, 6]
B = [3, 1, 4, 1, 5, 9]

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A = [1, 2, 3, 4, 5, 6]B = [3, 1, 4, 1, 5, 9]

The elements ‘2’ and ‘6’ both appear 0 times in B. Because ‘2’ occurs first, your program/function should return 2. (note that ‘9’ appears in B but not in A.)

B may be empty, but A will always have at least 1 element.

More examples:

A = [1, 2, 3, 4]
B = [1, 2, 3, 4, 1, 2, 3, 4, 1, 2]
-> 3

A = [1, 2, 3, 4]
B = [1, 1, 1, 1, 2, 2, 2, 3, 3, 4]
-> 4

A = [4, 8, 15, 16, 23, 42]
B = []
-> 4

A = [5]
B = [1, 2, 3, 4, 5, 6, 7]
-> 5

A = [10]
B = [1, 2, 3]
-> 10

A = [6, 5, 4, 3, 2, 1]
B = [5, 2, 3]
-> 6

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A = [1, 2, 3, 4]B = [1, 2, 3, 4, 1, 2, 3, 4, 1, 2]-> 3​A = [1, 2, 3, 4]B = [1, 1, 1, 1, 2, 2, 2, 3, 3, 4]-> 4​A = [4, 8, 15, 16, 23, 42]B = []-> 4​A = [5]B = [1, 2, 3, 4, 5, 6, 7]-> 5​A = [10]B = [1, 2, 3]-> 10​A = [6, 5, 4, 3, 2, 1]B = [5, 2, 3]-> 6
show all 23 lines

Python 2 29 bytes

xnor

SQL (Postgres) 148 bytes

Jack Douglas

APL (Dyalog Extended) 6 bytes

Adám

Kotlin 29 bytes

ajc2

05AB1E 5 bytes

liek

29 bytes

lambda a,b:min(a,key=b.count)

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lambda a,b:min(a,key=b.count)

Try it online!

Conveniently, Python chooses the earliest element in case of a tie.

Answer #2 SQL (Postgres)

Jack Douglas

148 bytes

select v from(select*from unnest((select v from a))with ordinality a(v,o))w natural left join b group by v order by count(b.v) desc,min(o) limit 1;

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select v from(select*from unnest((select v from a))with ordinality a(v,o))w natural left join b group by v order by count(b.v) desc,min(o) limit 1;

<>https://dbfiddle.uk/?rdbms=postgres_12&fiddle=6c90524b802f73aacb2f6d372a310e18&hide=15

Explanations etc.

The order of elements in B is immaterial so it is represented as a regular table but A is an array as the order matters in that input.

Answer #3 APL (Dyalog Extended)

Adám

6 bytesAnonymous tacit infix function taking A as left argument and B as right argument.
⊃⍧⍋⍛⊇⊣
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⊃⍧⍋⍛⊇⊣
Try it online!
⊃ the first of
⍧ the Count-A-In-B
⍋ ascending ordered
⍛ then
⊇ used to reorder
⊣ A

Answer #4 Kotlin

ajc2

29 bytes

-6 by changing sortedBy to minBy.

{b->minBy{b.count{v->v==it}}}
    
{b->                           // A is this, B is arg
    minBy{                     // select minimum value of A by
          b.count{v->v==it}}}  // # of occurrences in B

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{b->minBy{b.count{v->v==it}}}    {b->                           // A is this, B is arg    minBy{                     // select minimum value of A by          b.count{v->v==it}}}  // # of occurrences in B

minBy always appears to take the earlier item in the event of a tie.

Try it online!

Answer #5 05AB1E

liek

5 bytes

Input should be taken in the order B, A instead.

s¢Wkè

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s¢Wkè

Try it online!

Explanation

s      Swap the arguments so that the order becomes A, B
 ¢     Count how many times the items in A occur in B
  W    Push the minimum of these occurences
       without popping the original list
   k   Index into the original list, yielding the first
       index satisfying that
    è  Index into the same index of the original list

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s      Swap the arguments so that the order becomes A, B ¢     Count how many times the items in A occur in B  W    Push the minimum of these occurences       without popping the original list   k   Index into the original list, yielding the first       index satisfying that    è  Index into the same index of the original list

5 Answers

29 bytes

148 bytes

6 bytes

29 bytes

5 bytes

Explanation