The editorial has the most inefficient solution to this problem. Three for loops--are you sure? This was my linear python3 solution (I tried to make it as readable as possible and in line with the vars defined in the problem's description):

def solve(n, y):
    V = 0
    y.sort()
    for i in range(n):
        if i == 0 or (i > 0 and y[i] > y[i-1]): 
            k = n-i
            v_i = (n+1)/(k+1) 
        V += v_i
    return ["%.2f" % V]

This also exercises Linearity of Expectation, but all it's doing is sorting the input array y then walking through it to sum up v_i. If there's a repeating value, it'll use the first instance of v_i for that value (since v_i will continue to decrease as i increases).