2405. Optimal Partition of String

#### QUESTION:

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 {
func partitionString(_ s: String) -> Int {
var result:[String] = []
var index:Int = 0
var arr:[Character] = Array(s)
while index < arr.count {
var tmp:[Character] = []
while index < arr.count && !tmp.contains(arr[index]) {
tmp.append(arr[index])
index += 1
}
result.append(String(tmp))
}
return result.count
}
}
``````