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ABC81 C - Not so Diverse を解いた

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abc81_1.png
abc81_2.png
abc81_3.png
abc81_4.png

各要素を数えてみる。
K 種類以下になるまで、要素数の少ない順に足し合わせたら答えになると思った。

NotSoDiverse.py
N,K = map(int,input().split())
A = list(map(int,input().split()))

dic = {}
for n in range(N):
    if A[n] not in dic:
        dic[A[n]] = 0
    dic[A[n]] += 1

target = len(dic)-K
dic = sorted(dic.items(), key=lambda t:t[1])
#print(dic)
ans = 0
for i in range(target):
    ans += dic[i][1]
print(ans)

辞書の sort がいっつも忘れる。

神サイトに感謝

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