Today I have done Maximum Number of Occurrences of a Substring and leetcode's April LeetCoding Challenge with cpp.

Maximum Number of Occurrences of a Substring

Description

Given a string s, return the maximum number of ocurrences of any substring under the following rules:

• The number of unique characters in the substring must be less than or equal to maxLetters.
• The substring size must be between minSize and maxSize inclusive.

Example 1:

Input: s = "aababcaab", maxLetters = 2, minSize = 3, maxSize = 4
Output: 2
Explanation: Substring "aab" has 2 ocurrences in the original string.
It satisfies the conditions, 2 unique letters and size 3 (between minSize and maxSize).

Example 2:

Input: s = "aaaa", maxLetters = 1, minSize = 3, maxSize = 3
Output: 2
Explanation: Substring "aaa" occur 2 times in the string. It can overlap.

Example 3:

Input: s = "aabcabcab", maxLetters = 2, minSize = 2, maxSize = 3
Output: 3

Example 4:

Input: s = "abcde", maxLetters = 2, minSize = 3, maxSize = 3
Output: 0

Constraints:

• 1 <= s.length <= 10^5
• 1 <= maxLetters <= 26
• 1 <= minSize <= maxSize <= min(26, s.length)
• s only contains lowercase English letters.

Solution

why this problem is medium?

class Solution {
public:
  int maxFreq(string s, int maxLetters, int minSize, int maxSize) {
    int len = s.length();
    for(int i = minSize; i <= maxSize; ++i) {
      vector<int> cnt(26);
      unordered_map<string, int> sCnt;
      int letters = 0;
      for(int j = 0; j < i; ++j) {
        if(!cnt[s[j] - 'a']) letters += 1;
        cnt[s[j] - 'a'] += 1;
      }
      if(letters <= maxLetters) sCnt[s.substr(0, i)] += 1;
      for(int j = i; j < len; ++j) {
        if(!cnt[s[j] - 'a']) letters += 1;
        cnt[s[j] - 'a'] += 1;
        cnt[s[j - i] - 'a'] -= 1;
        if(!cnt[s[j - i] - 'a']) letters -= 1;
        if(letters <= maxLetters) sCnt[s.substr(j - i + 1, i)] += 1;
      }
      if(sCnt.size()) {
        int answer = 0;
        for(auto &[s, c] : sCnt) answer = max(answer, c);
        return answer;
      }
    }
    return 0;
  }
};

April LeetCoding challenge 9

Description

Verifying an Alien Dictionary

In an alien language, surprisingly they also use english lowercase letters, but possibly in a different order. The order of the alphabet is some permutation of lowercase letters.

Given a sequence of words written in the alien language, and the order of the alphabet, return true if and only if the given words are sorted lexicographicaly in this alien language.

Example 1:

Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
Output: true
Explanation: As 'h' comes before 'l' in this language, then the sequence is sorted.

Example 2:

Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
Output: false
Explanation: As 'd' comes after 'l' in this language, then words[0] > words[1], hence the sequence is unsorted.

Example 3:

Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
Output: false
Explanation: The first three characters "app" match, and the second string is shorter (in size.) According to lexicographical rules "apple" > "app", because 'l' > '∅', where '∅' is defined as the blank character which is less than any other character (More info).

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 20
• order.length == 26
• All characters in words[i] and order are English lowercase letters.

Solution

class Solution {
public:
  bool isAlienSorted(vector<string>& words, string order) {
    int orderIndex[128] = {};
    int pos = 0;
    for(auto c : order) orderIndex[c] = pos++;
    return is_sorted(words.begin(), words.end(), [&](const string &a, const string &b) {
      for(int i = 0; i < min(a.length(), b.length()); ++i) {
        if(a[i] != b[i]) return orderIndex[a[i]] < orderIndex[b[i]];
      }
      return a.length() < b.length();
    });
  }
};