LeetCode in Kotlin

3597. Partition String

Medium

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

• Start building a segment beginning at index 0.
• Continue extending the current segment character by character until the current segment has not been seen before.
• Once the segment is unique, add it to your list of segments, mark it as seen, and begin a new segment from the next index.
• Repeat until you reach the end of s.

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:

• 1 <= s.length <= 105
• s contains only lowercase English letters.

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
    }
}

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