- Difficulty: Medium
- Tags: LeetCode, Medium, Array, Two Pointers, String, leetcode-809, O(n + s), O(l + s)
Problem
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"
. Ifs = "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
andwords[i]
consist of lowercase letters.