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Kotlin-based LeetCode algorithm problem solutions, regularly updated
809\. Expressive Words
Medium
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.