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] ``` 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 ```

# 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; ``` <>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.

# 29 bytes ``` lambda a,b:min(a,key=b.count) ``` [Try it online!][TIO-k7dy5gfl] [TIO-k7dy5gfl]: https://tio.run/##bY7BCoMwEETvfsUetSylica2gl9iPcRWMVSjlEjx69NNsCVIYQ@782aYnVfTT5rbrrzZQY7NQ4LEphiVjiU@27Vsjvdp0Sax714NLbBifiltoIsPSs@LiZPEVgyBI6QIGYJAyGuEii7mBea1ax0FNmcIQ/v9n/k3/DvpxqOKYhdi1MNyQk7lLkZI7LvcgwhnV3EKGN25h5k3ksQcFRv9AA "Python 2 – Try It Online" Conveniently, Python chooses the earliest element in case of a tie.

# 6 bytes Anonymous tacit infix function taking `A` as left argument and `B` as right argument. ``` ⊃⍧⍋⍛⊇⊣ ``` [Try it online!][TIO-k7cy1b6e] `⊃` the first of `⍧` the **C**ount-`A`-**I**n-`B` `⍋` ascending ordered `⍛` then `⊇` used to reorder `⊣` `A` [TIO-k7cy1b6e]: https://tio.run/##SyzI0U2pTMzJT9dNrShJzUtJTfn//1FX86Pe5Y96ux/1zn7U1f6oa/H/tEdtEx719oFl1jzq3XJovfGjtomP@qYGBzkDyRAPz@D/jkA1hgpGCsYKJgqmCmZcTkC@sYIhkGcI5FtyOSqkKThxcSEpAyuBaUGicSuFQCMwNAaLQ5U@6p2rAFJuomChYGiqYGimYASUNVJQAEmU5xdlFyvkp6XlZOal6igklZYohHj6K2QkAgVzUhTKUouKM/PzwGY4gT26BmIewhU6pihuBXpPwRzZlTqGBggFyBJmQLUmQD1GCoZgBaYIBQA "APL (Dyalog Extended) – Try It Online"

# 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 ``` `minBy` always appears to take the earlier item in the event of a tie. [Try it online!][TIO-k845wpk5] [TIO-k845wpk5]: https://tio.run/##Xc5LCsIwFIXheVdxhgmkhfoCxURwJgiCO0jEQrBNpb0tSMjaY6xQxDs6g4@f@2ipti6OukZlu56uurv3tMPZ9rQ/OVIFmydHrpDGATJ6k6vGuuPLm@LWDo78mKtRSkshhFgNDo22jnH4DOk@fQ2JOrUu1VRjpcBCYCmwElgLbPgszZ9MppxYOcntVz4764j9fM20MJxnIb4B "Kotlin – Try It Online"

# 5 bytes Input should be taken in the order `B, A` instead. s¢Wkè [Try it online!][TIO-k7ddy8ki] [TIO-k7ddy8ki]: https://tio.run/##yy9OTMpM/f@/@NCi8OzDK/7/jzbWUTDUUTABk6Y6CpaxXNFAlpGOgjFYFChkFgsA "05AB1E – 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 ```