411. Minimum Unique Word Abbreviation

A string such as "word" contains the following abbreviations:

["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]

Given a target string and a set of strings in a dictionary, find an abbreviation of this target string with the smallest possible length such that it does not conflict with abbreviations of the strings in the dictionary.

Each number or letter in the abbreviation is considered length = 1. For example, the abbreviation "a32bc" has length = 4.

Note:

• In the case of multiple answers as shown in the second example below, you may return any one of them.

• Assume length of target string = m, and dictionary size = n. You may assume that m ≤ 21, n ≤ 1000, and log2(n) + m ≤ 20.

Examples:

"apple", ["blade"] -> "a4" (because "5" or "4e" conflicts with "blade")

"apple", ["plain", "amber", "blade"] -> "1p3" (other valid answers include "ap3", "a3e", "2p2", "3le", "3l1").

priority_queue<pair<int, string>, vector<pair<int, string> >, greater<pair<int, string> > > generate(const string& target) {
    priority_queue<pair<int, string>, vector<pair<int, string> >, greater<pair<int, string> > > res;
    for (int i = 0; i < pow(2, target.length()); ++i) {
        string cur;
        int cnt = 0, len = 0;
        for (int j = 0; j < target.length(); ++j) {
            if ((i >> j) & 1) ++cnt;
            else {
                if (cnt) {
                    cur += to_string(cnt);
                    cnt = 0;
                    ++len;
                }
                cur += target[j];
                ++len;
            }
        }
        if (cnt) {
            cur += to_string(cnt);
            ++len;
        }
        res.push(make_pair(len, cur));
    }
    return res;
}
bool valid(const string& word, const string& abbr) {
    int m = word.length(), n = abbr.length(), pos = 0, cnt = 0;
    for (int i = 0; i < n; ++i) {
        if (abbr[i] >= '0' && abbr[i] <= '9') {
            if (cnt == 0 && abbr[i] == '0') return false;
            cnt = cnt * 10 + abbr[i] - '0';
        } else {
            pos += cnt;
            if (pos >= m || word[pos++] != abbr[i]) return false;
            cnt = 0;
        }
    }
    return pos + cnt == m;
}
string minAbbreviation(string target, vector<string>& dictionary) {
    if (dictionary.empty()) return to_string(static_cast<int>(target.length()));
    priority_queue<pair<int, string>, vector<pair<int, string> >, greater<pair<int, string> > > pq;
    pq = generate(target);
    while (!pq.empty()) {
        auto t = pq.top(); pq.pop();
        bool no_conflict = true;
        for (const string& word : dictionary) {
            if (valid(word, t.second)) {
                no_conflict = false;
                break;
            }
        }
        if (no_conflict) return t.second;
    }
    return "";
}