# 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").
``````

``````class Solution {
public:
string minAbbreviation(string target, vector<string>& dictionary) {
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;
}
};
``````

Valid Word Abbreviation

Generalized Abbreviation

Unique Word Abbreviation

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

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

LeetCode All in One 题目讲解汇总(持续更新中…)

 微信打赏 Venmo 打赏
（欢迎加入博主的知识星球，博主将及时答疑解惑，并分享刷题经验与总结，试运营期间前五十位可享受半价优惠～）

×

Help us with donation