List indexes of duplicate values in a list with Python

Question

I am trying to modify this definition that lists duplicate items so that it lists indexes of duplicate values. Also, I would like it to list ALL of the duplicates that means the resultant for a = [1,2,3,2,1,5,6,5,5,5] would be duplicate_indexes = [3,4,7,8,9] Here's the definition:

def list_duplicates(seq):
    seen = set()
    seen_add = seen.add
    # adds all elements it doesn't know yet to seen and all other to seen_twice
    seen_twice = set( x for x in seq if x in seen or seen_add(x) )
    # turn the set into a list (as requested)
    return list( seen_twice )

a = [1,2,3,2,1,5,6,5,5,5]
list_duplicates(a) # yields [1, 2, 5]

Answer 1

List comprehension to print the index of duplicates. It slices the list till the selected index and return the index value if the item is already present in the sliced list

a= [1, 2, 3, 2, 1, 5, 6, 5, 5, 5]
result=[idx for idx, item in enumerate(a) if item in a[:idx]]
print result #[3, 4, 7, 8, 9]

Answer 2

def list_duplicates(seq):
    d = {}
    for i in seq:
        if i in d:
            d[i] += 1
        else:
            d[i] = 1
    dups = []
    for i in d:
        if d[i] > 1:
            dups.append(i)
    lst = []
    for i in dups:
        l = []
        for index in range(len(seq)):
            if seq[index] == i:
                l.append(index)
        lst.append(l[1:])
    new = []
    for i in lst:
        for index in i:
            new.append(index)   
    return new