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809. Expressive Words

Description

Sometimes people repeat letters to represent extra feeling. For example:

• "hello" -> "heeellooo"
• "hi" -> "hiiii"

In these strings like "heeellooo", we have groups of adjacent letters that are all the same: "h", "eee", "ll", "ooo".

You are given a string s and an array of query strings words. A query word is stretchy if it can be made to be equal to s by any number of applications of the following extension operation: choose a group consisting of characters c, and add some number of characters c to the group so that the size of the group is three or more.

• For example, starting with "hello", we could do an extension on the group "o" to get "hellooo", but we cannot get "helloo" since the group "oo" has a size less than three. Also, we could do another extension like "ll" -> "lllll" to get "helllllooo". If s = "helllllooo", then the query word "hello" would be stretchy because of these two extension operations: query = "hello" -> "hellooo" -> "helllllooo" = s.

Return the number of query strings that are stretchy.

 

Example 1:

Input: s = "heeellooo", words = ["hello", "hi", "helo"]
Output: 1
Explanation: 
We can extend "e" and "o" in the word "hello" to get "heeellooo".
We can't extend "helo" to get "heeellooo" because the group "ll" is not size 3 or more.

Example 2:

Input: s = "zzzzzyyyyy", words = ["zzyy","zy","zyy"]
Output: 3

 

Constraints:

• 1 <= s.length, words.length <= 100
• 1 <= words[i].length <= 100
• s and words[i] consist of lowercase letters.

Solutions

• Java
• C++
• Python
• Go

• class Solution {
      public int expressiveWords(String s, String[] words) {
          int ans = 0;
          for (String t : words) {
              if (check(s, t)) {
                  ++ans;
              }
          }
          return ans;
      }
  
      private boolean check(String s, String t) {
          int m = s.length(), n = t.length();
          if (n > m) {
              return false;
          }
          int i = 0, j = 0;
          while (i < m && j < n) {
              if (s.charAt(i) != t.charAt(j)) {
                  return false;
              }
              int k = i;
              while (k < m && s.charAt(k) == s.charAt(i)) {
                  ++k;
              }
              int c1 = k - i;
              i = k;
              k = j;
              while (k < n && t.charAt(k) == t.charAt(j)) {
                  ++k;
              }
              int c2 = k - j;
              j = k;
              if (c1 < c2 || (c1 < 3 && c1 != c2)) {
                  return false;
              }
          }
          return i == m && j == n;
      }
  }
  

• class Solution {
  public:
      int expressiveWords(string s, vector<string>& words) {
          auto check = [](string& s, string& t) -> int {
              int m = s.size(), n = t.size();
              if (n > m) return 0;
              int i = 0, j = 0;
              while (i < m && j < n) {
                  if (s[i] != t[j]) return 0;
                  int k = i;
                  while (k < m && s[k] == s[i]) ++k;
                  int c1 = k - i;
                  i = k, k = j;
                  while (k < n && t[k] == t[j]) ++k;
                  int c2 = k - j;
                  j = k;
                  if (c1 < c2 || (c1 < 3 && c1 != c2)) return 0;
              }
              return i == m && j == n;
          };
  
          int ans = 0;
          for (string& t : words) ans += check(s, t);
          return ans;
      }
  };
  

• class Solution:
      def expressiveWords(self, s: str, words: List[str]) -> int:
          def check(s, t):
              m, n = len(s), len(t)
              if n > m:
                  return False
              i = j = 0
              while i < m and j < n:
                  if s[i] != t[j]:
                      return False
                  k = i
                  while k < m and s[k] == s[i]:
                      k += 1
                  c1 = k - i
                  i, k = k, j
                  while k < n and t[k] == t[j]:
                      k += 1
                  c2 = k - j
                  j = k
                  if c1 < c2 or (c1 < 3 and c1 != c2):
                      return False
              return i == m and j == n
  
          return sum(check(s, t) for t in words)
  
  

• func expressiveWords(s string, words []string) (ans int) {
  	check := func(s, t string) bool {
  		m, n := len(s), len(t)
  		if n > m {
  			return false
  		}
  		i, j := 0, 0
  		for i < m && j < n {
  			if s[i] != t[j] {
  				return false
  			}
  			k := i
  			for k < m && s[k] == s[i] {
  				k++
  			}
  			c1 := k - i
  			i, k = k, j
  			for k < n && t[k] == t[j] {
  				k++
  			}
  			c2 := k - j
  			j = k
  			if c1 < c2 || (c1 != c2 && c1 < 3) {
  				return false
  			}
  		}
  		return i == m && j == n
  	}
  	for _, t := range words {
  		if check(s, t) {
  			ans++
  		}
  	}
  	return ans
  }
  

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