2022-06-15 Daily-Challenge

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

June LeetCoding Challenge 15

Description

Longest String Chain

You are given an array of words where each word consists of lowercase English letters.

wordA is a predecessor of wordB if and only if we can insert exactly one letter anywhere in wordA without changing the order of the other characters to make it equal to wordB.

  • For example, "abc" is a predecessor of "abac", while "cba" is not a predecessor of "bcad".

A word chain is a sequence of words [word1, word2, ..., wordk] with k >= 1, where word1 is a predecessor of word2, word2 is a predecessor of word3, and so on. A single word is trivially a word chain with k == 1.

Return the length of the longest possible word chain with words chosen from the given list of words.

Example 1:

Input: words = ["a","b","ba","bca","bda","bdca"]
Output: 4
Explanation: One of the longest word chains is ["a","ba","bda","bdca"].

Example 2:

Input: words = ["xbc","pcxbcf","xb","cxbc","pcxbc"]
Output: 5
Explanation: All the words can be put in a word chain ["xb", "xbc", "cxbc", "pcxbc", "pcxbcf"].

Example 3:

Input: words = ["abcd","dbqca"]
Output: 1
Explanation: The trivial word chain ["abcd"] is one of the longest word chains.
["abcd","dbqca"] is not a valid word chain because the ordering of the letters is changed.

Constraints:

  • 1 <= words.length <= 1000
  • 1 <= words[i].length <= 16
  • words[i] only consists of lowercase English letters.

Solution

class Solution {
  bool isPredecessor(string &shorterWord, string &longerWord) {
    bool skip = false;
    int len = shorterWord.length();
    int pos = 0;
    while(pos < len) {
      if(skip && shorterWord[pos] != longerWord[pos + 1]) return false;
      if(shorterWord[pos] == longerWord[pos + skip]) pos += 1;
      else skip = true;
    }
    return true;
  }
public:
  int longestStrChain(vector<string>& words) {
    vector<string> strs[17];
    for(auto &word : words) {
      strs[word.length()].push_back(word);
    }
    unordered_map<string, int> len;
    int answer = 1;
    for(int i = 1; i < 17; ++i) {
      for(auto &longerWord : strs[i]) {
        for(auto &shorterWord : strs[i - 1]) {
          if(isPredecessor(shorterWord, longerWord)) {
            len[longerWord] = max(len[shorterWord] + 1, len[longerWord]);
            answer = max(answer, len[longerWord] + 1);
          }
        }
      }
    }
    return answer;
  }
};

// Accepted
// 79/79 cases passed (50 ms)
// Your runtime beats 96.49 % of cpp submissions
// Your memory usage beats 47.02 % of cpp submissions (17.7 MB)