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  1. Prepare
  2. Mathematics
  3. Combinatorics
  4. Picking Cards
  5. Discussions

Picking Cards

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47 Discussions

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  • venom1724
     + 0 comments

    3 line Python solution, 100%

    cnt, mod = [0] * max(len(c), max(c)+1), 1000000007
for i in c: cnt[i]+=1
return math.prod(x-i for i,x in enumerate(accumulate(cnt)))%mod

    Could get it down to 1 line with using collections.Counter instead of building the counter array cnt manully in the first 2 lines of code here, but gets a bit too ugly for 1 liner.

    ¯_(ツ)_/¯

  • oliveranthony231
     + 0 comments

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  • macdonald_l_kyn
     + 0 comments

    Python Solution that runs on O(n) times only

    def solve(c):
    lyl = []
    c.sort(reverse=True)
    ii = 0
    n = len(c)
    prod = 1
    for i in c:
        if i < n:
            prod *= (n - i - ii)
        else:
            return 0
        ii += 1
    return prod % 1000000007

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     + 0 comments

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  • aniketsinharoy
     + 0 comments

    Java 8 Solution:

    public static int solve(List<Integer> c) 
    {
        TreeMap<Integer,Integer> card_count = new TreeMap<>();
        int i;
        for(i=0;i<c.size();++i)
        {
            if(card_count.containsKey(c.get(i))==false)
            {
                card_count.put(c.get(i), 1);
            }
            else
            {
                card_count.put(c.get(i), card_count.get(c.get(i))+1);
            }
        }
        if(card_count.containsKey(0)==false)
        {
            return 0;
        }
        int last_key=card_count.lastKey();
        int remaining_card=0;
        long temp;
        long ans=1;
        for(i=0;i<=last_key;++i)
        {
            if(card_count.containsKey(i)==true)
            {
                remaining_card=remaining_card+card_count.get(i);
            }
            if(remaining_card==0)
            {
                return 0;
            }
            ans=(ans*remaining_card)%1000000007;
            --remaining_card;
        }
        for(i=remaining_card;i>0;--i)
        {
            ans=(ans*i)%1000000007;
        }
        ans=ans%1000000007;
        return (int)ans;
    }