LeetCode in Kotlin

2405. Optimal Partition of String

Medium

Given a string s, partition the string into one or more substrings such that the characters in each substring are unique. That is, no letter appears in a single substring more than once.

Return the minimum number of substrings in such a partition.

Note that each character should belong to exactly one substring in a partition.

Example 1:

Input: s = “abacaba”

Output: 4

Explanation:

Two possible partitions are (“a”,”ba”,”cab”,”a”) and (“ab”,”a”,”ca”,”ba”).

It can be shown that 4 is the minimum number of substrings needed.

Example 2:

Input: s = “ssssss”

Output: 6

Explanation:

The only valid partition is (“s”,”s”,”s”,”s”,”s”,”s”).

Constraints:

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

Solution

class Solution {
    fun partitionString(s: String): Int {
        var count = 1
        var arr = BooleanArray(26)
        for (c in s.toCharArray()) {
            if (arr[c.code - 'a'.code]) {
                count++
                arr = BooleanArray(26)
            }
            arr[c.code - 'a'.code] = true
        }
        return count
    }
}

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