Small-To-Large Merging

Merging Data Structures

Obviously linked lists can be merged in $\mathcal{O}(1)$ time. But what about sets or vectors?

Let's consider a tree rooted at node $1$, where each node has a color.

For each node, let's store a set containing only that node, and we want to merge the sets in the nodes subtree together such that each node has a set consisting of all colors in the nodes subtree. Doing this allows us to solve a variety of problems, such as query the number of distinct colors in each subtree.

Naive Solution

Suppose that we want merge two sets $a$ and $b$ of sizes $n$ and $m$, respectively. One possiblility is the following:

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for (int x: b) a.insert(x);

which runs in $\mathcal{O}(m\log (n+m))$ time, yielding a runtime of $\mathcal{O}(N^2\log N)$ in the worst case. If we instead maintain $a$ and $b$ as sorted vectors, we can merge them in $\mathcal{O}(n+m)$ time, but $\mathcal{O}(N^2)$ is also too slow.

Better Solution

With just one additional line of code, we can significantly speed this up.

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if (a.size() < b.size()) swap(a,b);
for (int x: b) a.insert(x);

Note that swap exchanges two sets in $\mathcal{O}(1)$ time. Thus, merging a smaller set of size $m$ into the larger one of size $n$ takes $\mathcal{O}(m\log n)$ time.

Claim: The solution runs in $\mathcal{O}(N\log^2N)$ time.

Proof: When merging two sets, you move from the smaller set to the larger set. If the size of the smaller set is $X$, then the size of the resulting set is at least $2X$. Thus, an element that has been moved $Y$ times will be in a set of size at least $2^Y$, and since the maximum size of a set is $N$ (the root), each element will be moved at most $\mathcal{O}(\log N$) times.

Generalizing

We can also merge other standard library data structures such as std::map or std:unordered_map in the same way. However, std::swap does not always run in $\mathcal{O}(1)$ time. For example, swapping std::arrays takes time linear in the sum of the sizes of the arrays, and the same goes for GCC policy-based data structures such as __gnu_pbds::tree or __gnu_pbds::gp_hash_table.

To swap two policy-based data structures a and b in $\mathcal{O}(1)$ time, use a.swap(b) instead. Note that for standard library data structures, swap(a,b) is equivalent to a.swap(b).

Problems