Today I have done leetcode's September LeetCoding Challenge with cpp.

September LeetCoding Challenge 25

Description

Sum of Prefix Scores of Strings

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" is 2, 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

struct TrieNode {
  int count = 0;
  TrieNode *child[26] = {};
};
void insert(TrieNode *root, const string &word) {
  TrieNode *cur = root;
  for(auto c : word) {
    if(!cur->child[c - 'a']) cur->child[c - 'a'] = new TrieNode();
    cur = cur->child[c - 'a'];
    cur->count += 1;
  }
}
int count(TrieNode *root, const string &word) {
  TrieNode *cur = root;
  int result = 0;
  for(auto c : word) {
    if(!cur->child[c - 'a']) break;
    cur = cur->child[c - 'a'];
    result += cur->count;
  }
  return result;
}
class Solution {
public:
  vector<int> sumPrefixScores(vector<string>& words) {
    TrieNode *root = new TrieNode();
    for(const auto &word : words) {
      insert(root, word);
    }
    vector<int> answer;
    answer.reserve(words.size());
    for(const auto &word : words) {
      answer.push_back(count(root, word));
    }
    return answer;
  }
};