LeetCode in Kotlin

3163. String Compression III

Medium

Given a string word, compress it using the following algorithm:

• Begin with an empty string comp. While word is not empty, use the following operation:
  • Remove a maximum length prefix of word made of a single character c repeating at most 9 times.
  • Append the length of the prefix followed by c to comp.

Return the string comp.

Example 1:

Input: word = “abcde”

Output: “1a1b1c1d1e”

Explanation:

Initially, comp = "". Apply the operation 5 times, choosing "a", "b", "c", "d", and "e" as the prefix in each operation.

For each prefix, append "1" followed by the character to comp.

Example 2:

Input: word = “aaaaaaaaaaaaaabb”

Output: “9a5a2b”

Explanation:

Initially, comp = "". Apply the operation 3 times, choosing "aaaaaaaaa", "aaaaa", and "bb" as the prefix in each operation.

• For prefix "aaaaaaaaa", append "9" followed by "a" to comp.
• For prefix "aaaaa", append "5" followed by "a" to comp.
• For prefix "bb", append "2" followed by "b" to comp.

Constraints:

• 1 <= word.length <= 2 * 105
• word consists only of lowercase English letters.

Solution

class Solution {
    fun compressedString(word: String): String {
        val builder = StringBuilder()
        var last = word[0]
        var count = 1
        var i = 1
        val l = word.length
        while (i < l) {
            if (word[i] == last) {
                count++
                if (count == 10) {
                    builder.append(9).append(last)
                    count = 1
                }
            } else {
                builder.append(count).append(last)
                last = word[i]
                count = 1
            }
            i++
        }
        builder.append(count).append(last)
        return builder.toString()
    }
}

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