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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)_/¯

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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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