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