# A Deeper Look at the Small-to-Large Heuristic

The small-to-large heuristic is well known among the competitive programming community, but most problems that require it are straightforward applications of set merging. Here, we introduce a different way to think about the heuristic so that it may be applicable to a wider variety of problems. Consider the problem Non-boring Sequences. In short, the problem asks to determine whether all consecutive subsequences of a sequence \(A\) contain a unique element (such sequences are termed “non-boring”). Surely, after a quick read, we can start to brainstorm some segment tree or other data structure based solution, but what if we try some brute force approaches as well?

Consider a recursive `check`

function taking parameters \(l\) and \(r\) for whether the subsequence \(A[l, r]\) is non-boring. For some \(i\) in \([l, r]\), if \(A_i\) is unique (the previous and next appearance of \(A_i\) in \(A\) occurs outside of \([l, r]\)), then all subsequences of \(A[l, r]\) “crossing” \(i\) are non-boring. Thus, it suffices to check that both \(A[l, i-1]\) and \(A[i+1, r]\) are non-boring with a recursive call to `check`

(note that we only need to recurse for one such \(i\) if it exists, think about why this is).

Naively, the algorithm above runs in \(\mathcal{O}(N^2)\), far too slow for the given constraint of \(N \leq 200\;000\). However, what if we try a different order of looping \(i\) in the `check`

function? Instead of looping \(i\) in the order \([l, l+1, l+2, .., r]\), we will loop \(i\) “outside-in”, in the order \([l, r, l+1, r-1, l+2, r-2, ...]\). As it turns out, this provides us with an \(\mathcal{O}(N \log N)\) algorithm, which is a dramatic improvement from what we thought would be \(\mathcal{O}(N^2)\)!

To prove this, consider the tree formed by our decisions for \(i\) for each \(i\) recursively chosen by `check`

. This will be a binary tree, where the number of children in the left child is the size of the left side of our split \([l, i-1]\), and the number of children in the right child is the size of \([i+1, r]\). At each step of the algorithm, we are essentially “unmerging” a set of objects into the left and right children, giving each child the corresponding number of objects to its size. Note that this unmerging happens in a time complexity proportional to the size of the smaller child, by nature of us looping outside-in. However, considering the reverse process, this is exactly the process of small-to-large set merging, which is \(\mathcal{O}(N \log N)\)! Thus, we have obtained the correct complexity of our algorithm, and this problem is solved with barely any pain or book-code. Below shows a C++ implementation of `check`

, where `lst`

and `nxt`

store the index of the previous and next appearance of \(A_i\) respectively:

In conclusion, it may be worth the time to consider seemingly brute force solutions to some problems, as long as there is a merging or unmerging process that can happen proportional to the size of the smaller set, capitalizing on the small-to-large heuristic when it seems like the last thing one could do.