# 【LeetCode】 2405. Optimal Partition of String

## Description
> 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.

> Constraints:
`1 <= s.length <= 105`
`s` consists of only English lowercase letters.

> 給一字串 `s`，將該字串切割為一段或是多段的子字串，且每段子字串中的字母都得是唯一的。也就是說，任意一個子字串中，不會有字母出現超過一次。
> 回傳分割後子字串數量的**最小值**。
> 注意，在一個合理的分割後每個字元都應該剛好屬於一個子字串

> 限制：
`1 <= s.length <= 105`
`s` 只會包含小寫的英文字母

## Example:

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").

## Solution

* 用貪婪演算法就可以解了
* 從字串頭開始往後面遍歷，檢查該字元是否出現在子字串中
* 如果沒有，將該字元加入子字串
* 如果有，將子字串清空，答案數量加一(開始下一個子字串的概念)
* 實作上為了加速，子字串的部分可用 hash map，加速搜尋字元是否出現過的部分

### Code

class Solution {
public:
    int partitionString(string s) {
        set<char> unique_char;
        int count = 0;
        for(int i = 0; i < s.length(); i++)
        {
            if(unique_char.count(s[i]) != 0)
            {
                count++;
                unique_char.clear();
            }
            unique_char.insert(s[i]);
        }
        return count + 1;
    }
};

###### tags: `LeetCode` `C++`