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

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").

class Solution {
public:
    string minAbbreviation(string target, vector<string>& dictionary) {
        if (dictionary.empty()) return to_string((int)target.size());
        priority_queue<pair<int, string>, vector<pair<int, string>>, greater<pair<int, string>>> q;
        vector<string> abbrs = generateAbbreviations(target);
        for(auto abbr:abbrs) q.push({abbr.size(),abbr});
        while (!q.empty()) {
            auto t = q.top(); q.pop();
            bool no_conflict = true;
            for (string word : dictionary) {
                if (valid(word, t.second)) {
                    no_conflict = false;
                    break;
                }
            }
            if (no_conflict) return t.second;
        }
        return "";
    }
private:
    bool valid(string word, string abbr) {
        int m = word.size(), n = abbr.size(), p = 0, cnt = 0;
        for (int i = 0; i < abbr.size(); ++i) {
            if (abbr[i] >= ‘0‘ && abbr[i] <= ‘9‘) {
                if (cnt == 0 && abbr[i] == ‘0‘) return false;
                cnt = 10 * cnt + abbr[i] - ‘0‘;
            } else {
                p += cnt;
                if (p >= m || word[p++] != abbr[i]) return false;
                cnt = 0;
            }
        }
        return p + cnt == m;
    }
    
    vector<string> generateAbbreviations(string word) {
        vector<string>res;
        dfs(0,"",res,false,word);
        return res;
    }
    
    void dfs(int idx,string tmp,vector<string>&res,bool prevNum,string word)
    {
        if(idx==word.size()){
            res.push_back(tmp);
            return;
        }
        dfs(idx+1,tmp+word[idx],res,false,word);
        if(!prevNum)
        {
            for(int len = 1;len+idx<=word.size();len++)
            {
                dfs(idx+len,tmp+to_string(len),res,true,word);
            }
        }
    }

};

411. Minimum Unique Word Abbreviation