Overview

This problem is a shortest path problem where strings (bead sequences) are treated as vertices and each operation as an edge.
Since the cost of each operation is 1, performing a breadth-first search (BFS) from the initial state \(S\) allows us to find the minimum number of operations to reach the target \(T\).

Key Observations

There are two important points:

1. Minimizing the number of operations → This is “shortest distance in an unweighted graph,” so BFS is appropriate
   
2. Since \(N \le 25\), states (strings of length \(N\) consisting of A/B) can be represented as bit strings for speedup

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Each state is a binary string of length \(N\), so the theoretical number of states is at most \(2^N\) (approximately 33 million for \(N=25\)).
Rather than enumerating all states from the beginning, we explore only the states reachable from \(S\) using BFS.

Why a Naive Implementation is Too Slow

If we perform operations directly on strings, string copying occurs frequently at each transition, making the overall search slow.
Therefore, the code encodes strings as uint32_t bitmasks.
- A is mapped to 0, B to 1 - The character at position \(i\) is managed by bit \(i\)

This allows rotations and pattern duplication to be implemented primarily with bit operations, reducing the constant factor.

Handling Operations

• Left rotation / Right rotation: Computed in one step using bit shifts
• Pattern duplication: Choose \(d\) such that \(d \mid N,\, d<N\), and repeat the first \(d\) bits as a period across the entire length

By only enqueuing unvisited states during BFS, the first time we reach \(T\) gives us the minimum number of operations.
If \(T\) is never reached, the answer is -1.

Algorithm

1. Read input. If \(S=T\), output 0 and terminate.
2. Convert \(S, T\) to bitmasks.
3. Enumerate the proper divisors \(d\) of \(N\) (\(1 \le d < N,\, N \bmod d=0\)).
4. Execute BFS:
   • From the current state cur, generate:
     • Left rotation state
     • Right rotation state
     • Pattern duplication state for each \(d\)
   • If unvisited, register with distance +1 and enqueue
   • If a generated state matches the goal goal, immediately output and terminate
5. If not found by the end of BFS, output -1.

Complexity

Let \(M\) be the number of states reached from \(S\), and let \(\tau(N)-1\) be the number of proper divisors.

• The number of transitions per state is \(2 + (\tau(N)-1)\)
• However, since generating one pattern duplication takes \(O(N)\) in this implementation, the overall complexity is roughly
  Time complexity: \(O\!\left(M \cdot (\tau(N)\cdot N)\right)\)
  (Sufficiently practical since \(N \le 25\))
• Space complexity: \(O(M)\) (for dist and the queue)

Implementation Notes

• Since N<=25, all bits can be safely stored in a uint32_t.

• Mapping bit \(i\) to “position \(i\) of the string” makes the duplication operation src = i % d easy to write.

• BFS visit management uses unordered_map<uint32_t,int> (or unordered_set + distance array), enabling fast unvisited checks.

• Checking the goal state “at the moment of generation” rather than “when dequeued” is slightly faster.

  Source Code

#include <bits/stdc++.h>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N;
    string S, T;
    cin >> N >> S >> T;

    if (S == T) {
        cout << 0 << '\n';
        return 0;
    }

    // Enumerate all binary strings of length N as states (N<=25).
    // BFS from S with operations:
    // 1) left rotation
    // 2) right rotation
    // 3) pattern copy for any divisor d of N, d < N

    // Encode string as uint32_t bitmask, bit i = 1 if s[i]=='B'
    auto encode = [&](const string& s) -> uint32_t {
        uint32_t x = 0;
        for (int i = 0; i < N; i++) {
            if (s[i] == 'B') x |= (1u << i);
        }
        return x;
    };

    auto left_rot = [&](uint32_t x) -> uint32_t {
        uint32_t first = x & 1u;
        x >>= 1;
        if (first) x |= (1u << (N - 1));
        return x;
    };

    auto right_rot = [&](uint32_t x) -> uint32_t {
        uint32_t last = (x >> (N - 1)) & 1u;
        x = ((x << 1) & ((N == 32) ? 0xFFFFFFFFu : ((1u << N) - 1u)));
        if (last) x |= 1u;
        return x;
    };

    vector<int> divisors;
    for (int d = 1; d < N; d++) {
        if (N % d == 0) divisors.push_back(d);
    }

    auto replicate = [&](uint32_t x, int d) -> uint32_t {
        // Take prefix length d and repeat N/d times
        uint32_t res = 0;
        for (int i = 0; i < N; i++) {
            int src = i % d;
            uint32_t bit = (x >> src) & 1u;
            res |= (bit << i);
        }
        return res;
    };

    uint32_t start = encode(S);
    uint32_t goal = encode(T);

    unordered_map<uint32_t, int> dist;
    dist.reserve(1 << 20);
    queue<uint32_t> q;

    dist[start] = 0;
    q.push(start);

    while (!q.empty()) {
        uint32_t cur = q.front(); q.pop();
        int cd = dist[cur];

        uint32_t n1 = left_rot(cur);
        if (!dist.count(n1)) {
            dist[n1] = cd + 1;
            if (n1 == goal) {
                cout << cd + 1 << '\n';
                return 0;
            }
            q.push(n1);
        }

        uint32_t n2 = right_rot(cur);
        if (!dist.count(n2)) {
            dist[n2] = cd + 1;
            if (n2 == goal) {
                cout << cd + 1 << '\n';
                return 0;
            }
            q.push(n2);
        }

        for (int d : divisors) {
            uint32_t n3 = replicate(cur, d);
            if (!dist.count(n3)) {
                dist[n3] = cd + 1;
                if (n3 == goal) {
                    cout << cd + 1 << '\n';
                    return 0;
                }
                q.push(n3);
            }
        }
    }

    cout << -1 << '\n';
    return 0;
}

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This editorial was generated by gpt-5.3-codex.