Detailed mathematical approach

Problem Summary

Start from the source phrase "thereisasyetinsufficientdataforameaningfulanswer". Its letters form a multiset, and we consider every distinct word over the alphabet from a to z whose length is at most \(15\) and whose letter multiplicities do not exceed the available counts.

These words are ordered lexicographically, with the usual dictionary convention that a word comes before any longer word having it as a prefix. The task is to determine the ranks of five specified words, combine those ranks arithmetically, and then reconstruct the unique word at the resulting rank.

A brute-force list of all admissible words would be far too large. The successful approach is to count the size of every lexicographic subtree, then use those counts both for ranking and for unranking.

Mathematical Approach

Let \(s_\ell\) be the multiplicity of letter \(\ell\) in the source phrase, and let \(L=15\) be the maximum allowed length. For any current prefix, write \(\mathbf c=(c_a,\dots,c_z)\) for the remaining letter counts and \(r\) for the number of positions still available.

Step 1: Count the suffixes below a prefix

Define \(F(\mathbf c,r)\) as the number of valid suffixes that can still be appended when the remaining stock is \(\mathbf c\) and at most \(r\) more letters may be used. The empty suffix is allowed, so it already contributes one possibility.

Therefore

$F(\mathbf c,0)=1,$

$F(\mathbf c,r)=1+\sum_{\ell:\,c_\ell>0}F(\mathbf c-\mathbf e_\ell,r-1).$

The leading \(1\) means “stop here”. Every term in the sum means “choose one available letter first, then count everything below that child”. This is exactly the size of the lexicographic subtree rooted at the current prefix.

Step 2: Canonicalize states by sorted multiplicities

The exact letter labels do not matter for subtree size; only the multiset of remaining multiplicities matters. If two states differ only by renaming letters, they generate isomorphic subtrees and must have the same count.

So the implementations replace \(\mathbf c\) by a canonical representative \(\lambda(\mathbf c)\): take the positive entries of \(\mathbf c\), sort them in nonincreasing order, and discard trailing zeros. The dynamic program is memoized on the pair \((\lambda(\mathbf c),r)\).

This is a major compression step. For example, the states \((2,1,1,0,\dots)\) and \((1,2,1,0,\dots)\) become the same canonical key \((2,1,1)\). If equal parts appear, they are still summed separately in the recurrence because they correspond to different actual letters with the same remaining multiplicity.

Step 3: Compute the rank of a word

Let \(P_{\mathbf c,r}(w)\) denote the 1-based position of word \(w\) inside the subtree determined by \((\mathbf c,r)\). The empty word has position \(1\):

$P_{\mathbf c,r}(\varepsilon)=1.$

If \(w=w_1w_2\dots w_t\) is nonempty, then every legal first letter smaller than \(w_1\) contributes an entire subtree that lies before \(w\). Hence

$P_{\mathbf c,r}(w)=1+\sum_{\ell\lt w_1,\ c_\ell>0}F(\lambda(\mathbf c-\mathbf e_\ell),r-1)+P_{\mathbf c-\mathbf e_{w_1},\,r-1}(w_2\dots w_t).$

The ranking arithmetic in the implementations is performed with the 0-based value \(R(w)=P(w)-1\). Subtracting \(1\) removes the root empty word and makes addition and subtraction of ranks cleaner.

Step 4: Invert the process by unranking

To recover a word from its 0-based rank \(R\), first switch to the 1-based position \(Q=R+1\). If \(Q=1\), the correct continuation is empty. Otherwise discard that empty option by replacing \(Q\leftarrow Q-1\).

Now scan letters in increasing order. For each candidate letter \(\ell\) with \(c_\ell>0\), compute

$T_\ell=F(\lambda(\mathbf c-\mathbf e_\ell),r-1).$

If \(Q>T_\ell\), the entire \(\ell\)-subtree lies before the target, so set \(Q\leftarrow Q-T_\ell\) and continue. Otherwise \(\ell\) is the next letter of the answer, and the same procedure recurses one level deeper. Because the subtree counts are exact, this reconstruction is unique.

Step 5: Apply the problem-specific linear combination

The five words supplied by the problem are legionary, calorimeters, annihilate, orchestrated, and fluttering. If their 0-based ranks are \(R_1,R_2,R_3,R_4,R_5\), then the target rank is

$T=R_1+R_2-R_3+R_4-R_5.$

The final answer is the word obtained by unranking \(T\) in the full initial state defined by the source phrase and the depth limit \(15\).

Worked Example

Take a toy multiset with two a's, one b, and maximum length \(2\). The canonical state is \((2,1)\). Its subtree size satisfies

$F((2,1),2)=1+F((1,1),1)+F((2),1).$

Now \(F((1,1),1)=3\) because the admissible suffixes are \(\varepsilon\), a, and b, while \(F((2),1)=2\) because the admissible suffixes are \(\varepsilon\) and a. Therefore

$F((2,1),2)=1+3+2=6.$

The six words in order are

$\varepsilon,\ a,\ aa,\ ab,\ b,\ ba.$

So \(P(ab)=4\) and \(R(ab)=3\). Running the inverse procedure on rank \(3\) returns ab again. This miniature example mirrors the full solution: subtree counting first, then ranking or unranking by skipping whole branches at a time.

How the Code Works

The C++, Python, and Java implementations begin by counting the source letters once and fixing the maximum depth at \(15\). They then memoize subtree sizes using the canonical sorted multiplicities together with the remaining depth.

Each ranking query processes its word from left to right. At every position it adds the sizes of all lexicographically earlier sibling branches, consumes the actual next letter, and continues on the smaller remaining state. This produces a 1-based tree position, which is converted to a 0-based rank for the final arithmetic.

The unranking phase performs the complementary search. It tests candidate letters in alphabetical order, subtracts whole subtree sizes when those branches lie entirely before the target, and descends into the first branch that contains the desired position.

All five rank queries and the final unranking call share the same memoized subtree counts, so the expensive combinatorial work is done once and then reused.

Complexity Analysis

Let \(\mathcal S\) be the set of reachable canonical states \((\lambda,r)\). Each such state is evaluated once, and each evaluation inspects at most \(26\) letter slots, so the dynamic programming phase costs \(O(26|\mathcal S|)\) time and \(O(|\mathcal S|)\) memory. Since the sorting step is over at most \(26\) entries, it only affects the constant factor.

If a queried word has length \(m\le 15\), then one ranking or unranking operation examines at most \(26\) candidates at each of the \(m\) levels, so its cost is \(O(26m)\) after memoization. In practice, runtime is dominated by building the cache of subtree sizes, not by the final six queries.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=480
2. Lexicographic order: Wikipedia — Lexicographic order
3. Dynamic programming: Wikipedia — Dynamic programming
4. Multiset: Wikipedia — Multiset
5. Ranking and unranking: Wikipedia — Ranking