LeetCode in Kotlin

3597. Partition String

Medium

Given a string s, partition it into unique segments according to the following procedure:

Return an array of strings segments, where segments[i] is the ith segment created.

Example 1:

Input: s = “abbccccd”

Output: [“a”,”b”,”bc”,”c”,”cc”,”d”]

Explanation:

Here is your table, converted from HTML to Markdown:

Index Segment After Adding Seen Segments Current Segment Seen Before? New Segment Updated Seen Segments
0 “a” [] No ”” [“a”]
1 “b” [“a”] No ”” [“a”, “b”]
2 “b” [“a”, “b”] Yes “b” [“a”, “b”]
3 “bc” [“a”, “b”] No ”” [“a”, “b”, “bc”]
4 “c” [“a”, “b”, “bc”] No ”” [“a”, “b”, “bc”, “c”]
5 “c” [“a”, “b”, “bc”, “c”] Yes “c” [“a”, “b”, “bc”, “c”]
6 “cc” [“a”, “b”, “bc”, “c”] No ”” [“a”, “b”, “bc”, “c”, “cc”]
7 “d” [“a”, “b”, “bc”, “c”, “cc”] No ”” [“a”, “b”, “bc”, “c”, “cc”, “d”]

Hence, the final output is ["a", "b", "bc", "c", "cc", "d"].

Example 2:

Input: s = “aaaa”

Output: [“a”,”aa”]

Explanation:

Here is your table converted to Markdown:

Index Segment After Adding Seen Segments Current Segment Seen Before? New Segment Updated Seen Segments
0 “a” [] No ”” [“a”]
1 “a” [“a”] Yes “a” [“a”]
2 “aa” [“a”] No ”” [“a”, “aa”]
3 “a” [“a”, “aa”] Yes “a” [“a”, “aa”]

Hence, the final output is ["a", "aa"].

Constraints:

Solution

class Solution {
    private class Trie {
        var tries: Array<Trie?> = arrayOfNulls<Trie>(26)
    }

    fun partitionString(s: String): List<String> {
        val trie = Trie()
        val res: MutableList<String> = ArrayList()
        var node: Trie = trie
        var i = 0
        var j = 0
        while (i < s.length && j < s.length) {
            val idx = s[j].code - 'a'.code
            if (node.tries[idx] == null) {
                res.add(s.substring(i, j + 1))
                node.tries[idx] = Trie()
                i = j + 1
                j = i
                node = trie
            } else {
                node = node.tries[idx]!!
                j++
            }
        }
        return res
    }
}