Sometimes people repeat letters to represent extra feeling, such as "hello" -> "heeellooo", "hi" ->
"hiiii". In these strings like "heeellooo", we have groups of adjacent letters that are all the same:
"h", "eee", "ll", "ooo".
For some given string S, 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 3 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 size less than 3. 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.
Given a list of query words, return the number of words that are stretchy.
Example:
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.
Notes:
0 <= len(S) <= 100.
0 <= len(words) <= 100.
0 <= len(words[i]) <= 100.
S and all words in words consist only of lowercase letters
public int expressiveWords(String S, String[] words) {
int result = 0;
for(String w : words) {
if(isStretchy(S, w)) {
result++;
}
}
return result;
}
public boolean isStretchy(String t, String s) {
if(s.length() > t.length() || t.length() < 3) return false;
int i = 0, j = 0;
int cnt = 0;
while(i < t.length() || j < s.length()) {
if(i == t.length() && j < s.length()) return false; // t: eeeeeeee
// s: ed
if(j < s.length() && t.charAt(i) == s.charAt(j)) {
i++;
j++;
continue;
}
if(i == 0 && j == 0) return false;
if(t.charAt(i) == t.charAt(i-1)) { // if this is a valid expansion position
char c = t.charAt(i);
// check if there are three consecutive c, if yes, consume them all
if(i-2 >= 0 && t.charAt(i-2) == c || i+1 < t.length() && t.charAt(i+1) == c) {
while(i < t.length() && t.charAt(i) == c) i++;
if(j == s.length()) return i == t.length();
continue;
}
}
return false;
}
return true;
}