[LeetCode] 411. Minimum Unique Word Abbreviation 最短的獨一無二的單詞縮寫

 

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 thesmallest 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").

 

這道題實際上是之前那兩道Valid Word Abbreviation和Generalized Abbreviation的合體，我們的思路其實很簡單，首先找出target的所有的單詞縮寫的形式，然后按照長度來排序，小的排前面，我們用優先隊列來自動排序，里面存一個pair，保存單詞縮寫及其長度，然后我們從最短的單詞縮寫開始，跟dictionary中所有的單詞一一進行驗證，利用Valid Word Abbreviation中的方法，看其是否是合法的單詞的縮寫，如果是，說明有沖突，直接break，進行下一個單詞縮寫的驗證，參見代碼如下：

 

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;
        q = generate(target);
        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 "";
    }
    priority_queue<pair<int, string>, vector<pair<int, string>>, greater<pair<int, string>>> generate(string target) {
        priority_queue<pair<int, string>, vector<pair<int, string>>, greater<pair<int, string>>> res;
        for (int i = 0; i < pow(2, target.size()); ++i) {
            string out = "";
            int cnt = 0, size = 0;
            for (int j = 0; j < target.size(); ++j) {
                if ((i >> j) & 1) ++cnt;
                else {
                    if (cnt != 0) {
                        out += to_string(cnt);
                        cnt = 0;
                        ++size;
                    }
                    out += target[j];
                    ++size;
                }
            }
            if (cnt > 0) {
                out += to_string(cnt);
                ++size;
            }
            res.push({size, out});
        }
        return res;
    }
    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;
    }
};

 

類似題目：

Valid Word Abbreviation

Generalized Abbreviation

Unique Word Abbreviation

 

參考資料：

https://leetcode.com/problems/minimum-unique-word-abbreviation/

https://discuss.leetcode.com/topic/61457/c-bit-manipulation-dfs-solution

 

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