Overview

This problem involves simulating a greedy traversal on a graph where the “rooms” are length-\(N\) binary strings (without three consecutive identical characters) arranged in lexicographic order, and rooms differing by exactly one character are connected by corridors. The goal is to determine the number of steps to reach a target room by always greedily moving to the unvisited adjacent room with the smallest room number.

Analysis

Number of Valid Strings

Let \(f(N)\) denote the number of length-\(N\) strings with no three or more consecutive identical characters. Then \(f(N) = f(N-1) + f(N-2)\) (a Fibonacci-type recurrence) holds. With \(f(1) = 2, f(2) = 4\), we get \(f(26) \approx 392{,}836\) for \(N = 26\). This is small enough to enumerate all strings.

Determining Room Numbers

When valid strings are arranged in lexicographic order, their index directly becomes their room number. By generating them in lexicographic order, the array index corresponds to the room number.

Adjacency Relations

For each position in a string, if flipping that character produces a valid string (no three consecutive identical characters), those two rooms are connected by a corridor. The validity after flipping can be checked in \(O(1)\) by examining only the 2 characters before and after the flipped position.

Traversal Simulation

The rule “move to the unvisited adjacent room with the smallest room number” is a simple greedy search without backtracking. If the adjacency list for each room is pre-sorted in ascending order of room numbers, we simply scan from the beginning and choose the first unvisited room.

Algorithm

1. Enumerate all valid strings: Recursively generate all valid strings in lexicographic order and store them in an array valid.
2. String-to-index mapping: Use a hash map to look up the array index (= room number − 1) from each string.
3. Build adjacency lists: For each string, flip all \(N\) positions, and if the result is valid, add the corresponding index to the adjacency list. Sort the lists in ascending order.
4. Greedy simulation: Start from the starting room, and at each step scan the adjacency list from the beginning to move to the first unvisited room. Output the step count when the target room is reached, or output \(-1\) if stuck in a dead end.

Complexity

• Time complexity: \(O(M \cdot N \log N)\) (dominated by sorting the adjacency lists. The simulation itself is \(O(M \cdot N)\). Here \(M \approx 392{,}836\), \(N \leq 26\))
• Space complexity: \(O(M \cdot N)\) (for storing the adjacency lists)

Implementation Notes

• Lexicographic generation: By appending '0' → '1' in order during recursion, strings are naturally generated in lexicographic order. Only check whether the character matches the previous two characters.

• Validity check after flipping: When position \(j\) is flipped, check whether three consecutive identical characters occur in the range \(j-2, j-1, j, j+1, j+2\) (3 patterns to check).

• Using a hash map: First locally determine whether the flipped string is valid, and only look up the index in the hash map when it is valid, improving efficiency.

• Visited array: Manage the visit state of each room with vector<bool>. Since the simulation terminates in at most \(M\) steps, it runs sufficiently fast.

  Source Code

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

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int N;
    string s, t;
    cin >> N >> s >> t;
    
    // Generate all valid strings in lexicographic order
    vector<string> valid;
    valid.reserve(300000);
    function<void(string&)> generate = [&](string& cur) {
        if ((int)cur.size() == N) {
            valid.push_back(cur);
            return;
        }
        for (char c : {'0', '1'}) {
            int len = cur.size();
            if (len >= 2 && cur[len-1] == c && cur[len-2] == c) continue;
            cur.push_back(c);
            generate(cur);
            cur.pop_back();
        }
    };
    string empty = "";
    generate(empty);
    
    int M = valid.size();
    
    // Map string to index using hash map
    unordered_map<string, int> str_to_idx;
    str_to_idx.reserve(M * 2);
    for (int i = 0; i < M; i++) {
        str_to_idx[valid[i]] = i;
    }
    
    int start = str_to_idx[s];
    int target = str_to_idx[t];
    
    if (start == target) {
        cout << 0 << endl;
        return 0;
    }
    
    // For each room, compute sorted list of neighbor indices
    vector<vector<int>> neighbors(M);
    for (int i = 0; i < M; i++) {
        string& str = valid[i];
        for (int j = 0; j < N; j++) {
            char orig = str[j];
            char flipped = (orig == '0') ? '1' : '0';
            // Check validity of flipping position j
            bool ok = true;
            if (j >= 2 && flipped == str[j-1] && flipped == str[j-2]) ok = false;
            if (ok && j >= 1 && j < N-1 && flipped == str[j-1] && flipped == str[j+1]) ok = false;
            if (ok && j < N-2 && flipped == str[j+1] && flipped == str[j+2]) ok = false;
            if (ok) {
                str[j] = flipped;
                auto it = str_to_idx.find(str);
                if (it != str_to_idx.end()) {
                    neighbors[i].push_back(it->second);
                }
                str[j] = orig;
            }
        }
        sort(neighbors[i].begin(), neighbors[i].end());
    }
    
    // Simulate the greedy walk
    vector<bool> visited(M, false);
    int current = start;
    visited[current] = true;
    int steps = 0;
    
    while (true) {
        int next = -1;
        for (int nb : neighbors[current]) {
            if (!visited[nb]) {
                next = nb;
                break;
            }
        }
        if (next == -1) {
            cout << -1 << endl;
            return 0;
        }
        steps++;
        visited[next] = true;
        current = next;
        if (current == target) {
            cout << steps << endl;
            return 0;
        }
    }
}

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