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

January LeetCoding Challenge 8

Description

Count Prefix and Suffix Pairs I

You are given a 0-indexed string array words.

Let's define a boolean function isPrefixAndSuffix that takes two strings, str1 and str2:

• isPrefixAndSuffix(str1, str2) returns true if str1 is both a prefix and a suffix of str2, and false otherwise.

For example, isPrefixAndSuffix("aba", "ababa") is true because "aba" is a prefix of "ababa" and also a suffix, but isPrefixAndSuffix("abc", "abcd") is false.

Return an integer denoting the number of index pairs (i, j) such that i < j, and isPrefixAndSuffix(words[i], words[j]) is true.

Example 1:

Input: words = ["a","aba","ababa","aa"]
Output: 4
Explanation: In this example, the counted index pairs are:
i = 0 and j = 1 because isPrefixAndSuffix("a", "aba") is true.
i = 0 and j = 2 because isPrefixAndSuffix("a", "ababa") is true.
i = 0 and j = 3 because isPrefixAndSuffix("a", "aa") is true.
i = 1 and j = 2 because isPrefixAndSuffix("aba", "ababa") is true.
Therefore, the answer is 4.

Example 2:

Input: words = ["pa","papa","ma","mama"]
Output: 2
Explanation: In this example, the counted index pairs are:
i = 0 and j = 1 because isPrefixAndSuffix("pa", "papa") is true.
i = 2 and j = 3 because isPrefixAndSuffix("ma", "mama") is true.
Therefore, the answer is 2.  

Example 3:

Input: words = ["abab","ab"]
Output: 0
Explanation: In this example, the only valid index pair is i = 0 and j = 1, and isPrefixAndSuffix("abab", "ab") is false.
Therefore, the answer is 0.

Constraints:

• 1 <= words.length <= 50
• 1 <= words[i].length <= 10
• words[i] consists only of lowercase English letters.

Solution

struct TrieNode {
  int count = 0;
  TrieNode *child[26 * 26] = {};
};
class Solution {
public:
  int countPrefixSuffixPairs(vector<string>& words) {
    TrieNode *root = new TrieNode;
    long long answer = 0;
    for(const auto &word : words) {
      TrieNode *current = root;
      for(int i = 0, n = word.length(); i < n; ++i) {
        if(!current->child[(word[i] - 'a') * 26 + word[n - 1 - i] - 'a']) {
          current->child[(word[i] - 'a') * 26 + word[n - 1 - i] - 'a'] = new TrieNode();
        }
        current = current->child[(word[i] - 'a') * 26 + word[n - 1 - i] - 'a'];
        answer += current->count;
      }
      current->count += 1;
    }  
    return answer;
  }
};

// Accepted
// 594/594 cases passed (32 ms)
// Your runtime beats 7.63 % of cpp submissions
// Your memory usage beats 5.08 % of cpp submissions (127.5 MB)