Overview

We consider 01 strings of length \(N\) that satisfy the conditions as rooms, and continuously move to “the unvisited adjacent room with the smallest room number” according to the specified rules.
Since this traversal is a single non-branching path, we can simply simulate the order of visits.

Analysis

Each room is a 01 string of length \(N\).
It is convenient to treat this as an \(N\)-bit integer.

For example, when \(N=3\):

• 001 \(\rightarrow 1\)
• 010 \(\rightarrow 2\)
• 011 \(\rightarrow 3\)
• 100 \(\rightarrow 4\)

This is the correspondence we use.

The key insight here is that for 01 strings of the same length, lexicographic order and numerical order coincide.
In other words, even without computing the actual room number, selecting the candidate with the smallest integer value is equivalent to selecting the room with the “smallest room number.”

However, not all \(2^N\) strings exist as rooms.
Strings like 000 or 111, where the same character appears \(3\) or more times consecutively, are invalid.

Therefore, we manage whether each bit string is a valid room using a valid array in advance.

Also, this problem is not about checking reachability with standard BFS or DFS.
Takahashi does not backtrack to continue searching; he simply moves to exactly one room according to the fixed rule each time.
Therefore, we need to directly simulate the actual movement.

Algorithm

First, convert strings \(s, t\) to integer bitmasks respectively.

• start: the starting room
• target: the goal room

If start == target, then it is already visited at departure, so the answer is \(0\).

Next, enumerate all valid rooms.

We use DFS to determine characters from left to right.

The state consists of:

• pos: how many characters have been determined so far
• last: the previous character
• run: how many times the previous character has appeared consecutively
• mask: the bit string constructed so far

If the next character b is the same as last and run == 2 already, then placing it would create \(3\) consecutive identical characters, so it cannot be placed.
We recurse only in other cases.

When we have constructed a string of length \(N\), we mark that mask as a valid room with valid[mask] = true.

After that, we simulate the traversal.

Let the current room be cur.
Rooms that differ from cur by exactly 1 bit are potentially adjacent rooms.

For each bit position bit:

nxt = cur ^ (1 << bit)

This gives the room where only the bit-th position of cur is flipped.

For this nxt, if:

• It is a valid room
• It has not been visited yet

then it is a movement candidate.

Among these, we select the one with the smallest integer value.
This corresponds to selecting the room with the smallest lexicographic order, i.e., the smallest room number.

If no destination exists, the traversal ends.
We increment the move count with each move, and if we reach target, we output that count.

If target is never reached, we output \(-1\).

Complexity

Let \(V\) be the number of valid rooms.
The number of 01 strings without \(3\) or more consecutive identical characters is on the order of Fibonacci numbers, so \(V = O(\varphi^N)\).

• Enumeration of valid rooms: \(O(V)\)
• Traversal simulation: checking at most \(N\) adjacent candidates per room, so \(O(NV)\)
• Array initialization: \(O(2^N)\)

Therefore:

• Time complexity: \(O(2^N + NV)\)
• Space complexity: \(O(2^N)\)

Since \(N \leq 26\), preparing an array of size \(2^N\) is feasible.

Implementation Notes

• Strings are converted to integers by packing bits from left to right.

• Since the length is fixed, lexicographic order of 01 strings coincides with numerical order of integers.

• There is no need to compute the actual room number values.

• Adjacent rooms can be found with cur ^ (1 << bit).

• Among unvisited valid adjacent rooms, we select the one with the smallest integer value.

• Since the traversal never backtracks, we simulate a single path according to the rules, not BFS or DFS.

  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;

    auto toMask = [&](const string& x) {
        int m = 0;
        for (char c : x) m = (m << 1) | (c - '0');
        return m;
    };

    int start = toMask(s);
    int target = toMask(t);

    if (start == target) {
        cout << 0 << '\n';
        return 0;
    }

    int M = 1 << N;
    vector<unsigned char> valid(M, 0), visited(M, 0);

    auto dfs = [&](auto&& self, int pos, int last, int run, int mask) -> void {
        if (pos == N) {
            valid[mask] = 1;
            return;
        }
        for (int b = 0; b <= 1; b++) {
            if (b == last && run == 2) continue;
            self(self, pos + 1, b, (b == last ? run + 1 : 1), (mask << 1) | b);
        }
    };

    dfs(dfs, 0, -1, 0, 0);

    int cur = start;
    visited[cur] = 1;
    int moves = 0;

    while (true) {
        int best = -1;

        for (int bit = 0; bit < N; bit++) {
            int nxt = cur ^ (1 << bit);
            if (valid[nxt] && !visited[nxt]) {
                if (best == -1 || nxt < best) best = nxt;
            }
        }

        if (best == -1) break;

        cur = best;
        visited[cur] = 1;
        moves++;

        if (cur == target) {
            cout << moves << '\n';
            return 0;
        }
    }

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

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This editorial was generated by gpt-5.5-xhigh.