LeetCode in Kotlin

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.

Solution

@Suppress("NAME_SHADOWING")
class Solution {
    fun expressiveWords(s: String, words: Array<String>): Int {
        var ans = 0
        for (w in words) {
            if (check(s, w)) {
                ans++
            }
        }
        return ans
    }

    private fun check(s: String, w: String): Boolean {
        var i = 0
        var j = 0
        while (i < s.length && j < w.length) {
            val ch1 = s[i]
            val ch2 = w[j]
            val len1 = getLen(s, i)
            val len2 = getLen(w, j)
            if (ch1 == ch2) {
                if (len1 == len2 || len1 >= 3 && len2 < len1) {
                    i += len1
                    j += len2
                } else {
                    return false
                }
            } else {
                return false
            }
        }
        return i == s.length && j == w.length
    }

    private fun getLen(value: String, i: Int): Int {
        var i = i
        i += 1
        var count = 1
        for (j in i until value.length) {
            if (value[j] == value[i - 1]) {
                count++
            } else {
                break
            }
        }
        return count
    }
}

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