题解 | 宝石手串

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <climits>
using namespace std;

int main() {
    int T;
    cin >> T;
    while (T--) {
        int n;
        string s;
        cin >> n >> s;

        bool found_adjacent = false;
        for (int i = 0; i < n; i++) {
            if (s[i] == s[(i + 1) % n]) {
                found_adjacent = true;
                break;
            }
        }
        if (found_adjacent) {
            cout << 0 << endl;
            continue;
        }

        vector<int> char_count(26, 0);
        for (char c : s) {
            char_count[c - 'a']++;
        }
        bool all_unique = true;
        for (int count : char_count) {
            if (count > 1) {
                all_unique = false;
                break;
            }
        }
        if (all_unique) {
            cout << -1 << endl;
            continue;
        }

        vector<vector<int>> positions(26);
        for (int i = 0; i < n; i++) {
            positions[s[i] - 'a'].push_back(i);
        }

        int min_gap = INT_MAX;
        for (int i = 0; i < 26; i++) {
            int k = positions[i].size();
            if (k <= 1) continue;
            sort(positions[i].begin(), positions[i].end());
            for (int j = 0; j < k; j++) {
                int gap;
                if (j < k - 1) {
                    gap = positions[i][j + 1] - positions[i][j] - 1;
                } else {
                    gap = positions[i][0] + n - positions[i][j] - 1;
                }
                if (gap < min_gap) {
                    min_gap = gap;
                }
            }
        }
        cout << min_gap << endl;
    }
    return 0;
}