How to obtain O(log(N)) complexity of the counting std::map of specific value

Question

I am thinking about the problem of counting elements of specific values of map and how to achieve O(log(N)) complexity in C++. It is would be only possible when it is sorted by the values I guess, but then it would not be sorted by keys and the get from map would be then O(N).

In the first version I use std::count_if. Here is my code:

std::count_if( std::begin(_map), std::end(_map),
    [&val]( const std::pair<const std::string,  std::string> &p )
    {
        return p.second == val;
    } );

It has complexity the O(N). Do you know any other containers which would be useful for this problem? I was thinking about tree map but there might be a similar problem of sorting by keys.

score 2 · Accepted Answer · answered Apr 30 '21 at 08:40

2

Boost's bimap allows lookup by either key or value.

auto range = map.right.equal_range(val);
std::distance(range.first, range.second);

answered Apr 30 '21 at 08:40

Caleth

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Wow, that's looking really good, I think that is giving me te most capabilities so far – Przemekeke Apr 30 '21 at 08:46

score 1 · Answer 2 · answered Apr 30 '21 at 08:19

1

You could maintain two maps so items would be indexed by by both key and value. You would need a multimap to allow for duplicate values.

answered Apr 30 '21 at 08:19

John Kugelman

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I was considering this as well but it doubling the memory usage and it would be problem with big maps. I am thinking about this solution as a trade off – Przemekeke Apr 30 '21 at 08:22

score 1 · Answer 3 · answered Apr 30 '21 at 08:24

1

Depending one your other typical uses of the container you can consider to use a std::vector<std::pair<key_t,mapped_t>> instead of the map. You have to do the sorting yourself, but you can sort for keys or values, and then do a binary search.

answered Apr 30 '21 at 08:24

463035818_is_not_an_ai

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Thanks, how do you suggest to sort in order to meet time complexity limits? – Przemekeke Apr 30 '21 at 08:28
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@PPM sorting according to values is `O(N logN)` though you need to sort only once and can then use `O(log N)` binary search (`std::binary_search`) – 463035818_is_not_an_ai Apr 30 '21 at 08:32

score 0 · Answer 4 · answered Sep 14 '21 at 15:18

0

I've got an idea that it can be achieved by creating a hash_map hash function which increments bucket value by one if the value from map equals the bucket key. Then, it is only a matter of reading bucket values.

answered Sep 14 '21 at 15:18

Przemekeke

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Add this to the description of the problem. – Rishabh Deep Singh Sep 14 '21 at 16:33
As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-ask). – Community Sep 14 '21 at 16:33

How to obtain O(log(N)) complexity of the counting std::map of specific value

4 Answers4