frequency of largest number group

[DeaD_EyE]

a = [112211121112121122111] is a list with only one element.
The answer to your question is effortless because there is only one element in your list.

I guess you mean a sequence like this:

a = [1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 2, 1, 2, 1, 1, 2, 2, 1, 1, 1]

The items in the list are Integers, but they could also be str or something else.

Make a function, which saves the last_value, start_index and current index.
If there is a difference between last and current value, you've to add the last value with start_index, end_index to a list. Then you assign the current_value to last_value and the start_index. After your loop is done, you've to add the last remaining group.

Then you have a list with items where one value is the value itself, start_index and end_index.
You can get the shortest and longest group with a key function.

So if you save the groups like this pattern:

(value, start_index, end_index)

You can get the longest and shortest distance by calculating it.

def by_distance(element):
    value, start, end = element
    return end - start


result = [(42, 4, 22), (33, 22, 25)]
print("Shortest")
print(min(result, key=by_distance))
print("Longest")
print(max(result, key=by_distance))