Leetcode•Oct 07, 2026

Expressive Words

Hazrat Ali

Leetcode

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

Solution
var expressiveWords = function(s, words) {
    function isStretchy(word) {
        let i = 0;
        let j = 0;

        while (i < s.length && j < word.length) {
            if (s[i] !== word[j]) return false;

            let startS = i;
            let startW = j;
            while (i < s.length && s[i] === s[startS]) {
                i++;
            }
            while (j < word.length && word[j] === word[startW]) {
                j++;
            }

            const countS = i - startS;
            const countW = j - startW;

            if (countW > countS) return false;

            if (countS !== countW && countS < 3) {
                return false;
            }
        }

        return i === s.length && j === word.length;
    }

    let answer = 0;

    for (const word of words) {
        if (isStretchy(word)) {
            answer++;
        }
    }

    return answer;
};
 

Comments