You are given an array words
of size n
consisting of non-empty strings.
We define the score of a string word
as the number of strings words[i]
such that word
is a prefix of words[i]
.
- For example, if
words = ["a", "ab", "abc", "cab"]
, then the score of"ab"
is2
, since"ab"
is a prefix of both"ab"
and"abc"
.
Return an array answer
of size n
where answer[i]
is the sum of scores of every non-empty prefix of words[i]
.
Note that a string is considered as a prefix of itself.
Example 1:
Input: words = ["abc","ab","bc","b"] Output: [5,4,3,2] Explanation: The answer for each string is the following: - "abc" has 3 prefixes: "a", "ab", and "abc". - There are 2 strings with the prefix "a", 2 strings with the prefix "ab", and 1 string with the prefix "abc". The total is answer[0] = 2 + 2 + 1 = 5. - "ab" has 2 prefixes: "a" and "ab". - There are 2 strings with the prefix "a", and 2 strings with the prefix "ab". The total is answer[1] = 2 + 2 = 4. - "bc" has 2 prefixes: "b" and "bc". - There are 2 strings with the prefix "b", and 1 string with the prefix "bc". The total is answer[2] = 2 + 1 = 3. - "b" has 1 prefix: "b". - There are 2 strings with the prefix "b". The total is answer[3] = 2.
Example 2:
Input: words = ["abcd"] Output: [4] Explanation: "abcd" has 4 prefixes: "a", "ab", "abc", and "abcd". Each prefix has a score of one, so the total is answer[0] = 1 + 1 + 1 + 1 = 4.
Constraints:
1 <= words.length <= 1000
1 <= words[i].length <= 1000
words[i]
consists of lowercase English letters.
Solution: Trie
Insert all the words into a tire whose node val is the number of substrings that have the current prefix.
During query time, sum up the values along the prefix path.
Time complexity: O(sum(len(word))
Space complexity: O(sum(len(word))
C++
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// Author: Huahua struct Trie { Trie* ch[26] = {}; int cnt = 0; void insert(string_view s) { Trie* cur = this; for (char c : s) { if (!cur->ch[c - 'a']) cur->ch[c - 'a'] = new Trie(); cur = cur->ch[c - 'a']; ++cur->cnt; } } int query(string_view s) { Trie* cur = this; int ans = 0; for (char c : s) { cur = cur->ch[c - 'a']; ans += cur->cnt; } return ans; } }; class Solution { public: vector<int> sumPrefixScores(vector<string>& words) { Trie* root = new Trie(); for (const string& w : words) root->insert(w); vector<int> ans; for (const string& w : words) ans.push_back(root->query(w)); return ans; } }; |