PAST4F
from 第四回 アルゴリズム実技検定
PAST4F
上または下と比べて出現回数が同じならAMBIGUOUS
pythondef solve(N, K, SS):
from collections import Counter
c = Counter(SS).most_common()
K -= 1
ck, ckv = c[K]
if K > 0:
_, pv = c[K - 1]
if pv == ckv:
return "AMBIGUOUS"
if K < len(c) - 1:
_, pv = c[K + 1]
if pv == ckv:
return "AMBIGUOUS"
return ckRelated Pages
"Engineer's way of creating knowledge" the English version of my book is now available on [Engineer's way of creating knowledge]
(C)NISHIO Hirokazu / Converted from [Scrapbox] at [Edit]