Given a list and a number, create a new list that contains each number of list at most N times, without reordering.

For example if the input number is 2, and the input list is [1,2,3,1,2,1,2,3], you take [1,2,3,1,2], drop the next [1,2] since this would lead to 1 and 2 being in the result 3 times, and then take 3, which leads to [1,2,3,1,2,3].

With list [20,37,20,21] and number 1, the result would be [20,37,21].

def delete_nth(order,max_e):
    s = []
    for i in order:
        if s.count(i) < max_e:
            s.append(i)
    return s
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