Efficient way to get middle (median) of an std::set?

Question

std::set is a sorted tree. It provides begin and end methods so I can get minimum and maximum and lower_bound and upper_bound for binary search. But what if I want to get iterator pointing to the middle element (or one of them if there are even number of elements there)?

Is there an efficient way (O(log(size)) not O(size)) to do that?

{1} => 1
{1,2} => 1 or 2
{1,2,3} => 2
{1,2,3,4} => 2 or 3 (but in the same direction from middle as for {1,2})
{1,312,10000,14000,152333} => 10000

PS: Same question in Russian.

Answer 1

Depending on how often you insert/remove items versus look up the middle/median, a possibly more efficient solution than the obvious one is to keep a persistent iterator to the middle element and update it whenever you insert/delete items from the set. There are a bunch of edge cases which will need handling (odd vs even number of items, removing the middle item, empty set, etc.), but the basic idea would be that when you insert an item that's smaller than the current middle item, your middle iterator may need decrementing, whereas if you insert a larger one, you need to increment. It's the other way around for removals.

At lookup time, this is of course O(1), but it also has an essentially O(1) cost at each insertion/deletion, i.e. O(N) after N insertions, which needs to be amortised across a sufficient number of lookups to make it more efficient than brute forcing.

Answer 2

This suggestion is pure magic and will fail if there are some duplicated items

Depending on how often you insert/remove items versus look up the middle/median, a possibly more efficient solution than the obvious one is to keep a persistent iterator to the middle element and update it whenever you insert/delete items from the set. There are a bunch of edge cases which will need handling (odd vs even number of items, removing the middle item, empty set, etc.), but the basic idea would be that when you insert an item that's smaller than the current middle item, your middle iterator may need decrementing, whereas if you insert a larger one, you need to increment. It's the other way around for removals.

Suggestions

1. first suggestion is to use a std::multiset instead of std::set, so that it can work well when items could be duplicated
2. my suggestion is to use 2 multisets to track the smaller potion and the bigger potion and balance the size between them

Algorithm

1. keep the sets balanced, so that size_of_small==size_of_big or size_of_small + 1 == size_of_big

void balance(multiset<int> &small, multiset<int> &big)
{
    while (true)
    {
        int ssmall = small.size();
        int sbig = big.size();

        if (ssmall == sbig || ssmall + 1 == sbig) break; // OK

        if (ssmall < sbig)
        {
            // big to small
            auto v = big.begin();
            small.emplace(*v);
            big.erase(v);
        }
        else 
        {
            // small to big
            auto v = small.end();
            --v;
            big.emplace(*v);
            small.erase(v);
        }
    }
}

2. if the sets are balanced, the medium item is always the first item in the big set

auto medium = big.begin();
cout << *medium << endl;

3. take caution when add a new item

auto v = big.begin();
if (v != big.end() && new_item > *v)
    big.emplace(new_item );
else
    small.emplace(new_item );

balance(small, big);

complexity explained

• it is O(1) to find the medium value
• add a new item takes O(log n)
• you can still search a item in O(log n), but you need to search 2 sets

Answer 3

It's going to be O(size) to get the middle of a binary search tree. You can get it with std::advance() as follows:

std::set<int>::iterator it = s.begin();
std::advance(it, s.size() / 2);

Answer 4

Be aware that the std::set does NOT store duplicate values. If you insert the following values {1, 2, 3, 3, 3, 3, 3, 3, 3}, the median you will retrieve is 2.

std::set<int>::iterator it = s.begin();
std::advance(it, s.size() / 2);
int median = *it;

If you want to include duplicates when considering the median you can use std::multiset ({1, 2, 3, 3, 3, 3, 3, 3, 3} median's would be 3) :

std::multiset<int>::iterator it = s.begin();
std::advance(it, s.size() / 2);
int median = *it;

If the only reason you want the data sorted is to get the median, you are better off with a plain old std::vector + std::sort in my opinion.

With a large test sample and multiple iterations, I completed the test in 5s with std::vector and std::sort and 13 to 15s with either std::set or std::multiset. Your milage may vary depending on the size and number of duplicate values you have.

Answer 5

As told by @pmdj we use the iterator to keep the track of the middle element. The below is the code implementation of the following :

class RollingMedian {
public:
multiset<int> order;
multiset<int>::iterator it;
RollingMedian() {
}

void add(int val) {
    order.insert(val);
    if (order.size() == 1) {
        it = order.begin();
    } else {
        if (val < *it and order.size() % 2 == 0) {
            --it;
        }
        if (val >= *it and order.size() % 2 != 0) {
            ++it;
        }
    }
}

double median() {
    if (order.size() % 2 != 0) {
        return double(*it);
    } else {
        auto one = *it, two = *next(it);
        return double(one + two) / 2.0;
    }
}  };

Feel free to copy and use any part of this code. Also, you can use set instead of multiset if the repetition is not there.

Answer 6

If your data is static, those you could precalcate it and do not insert new elements - it’s simplier to use vector , sort it , and access median just by index in O(1)

vector<int> data;
// fill data
std::sort(data.begin(), data.end());
auto median = data[data.size() / 2];