leetcode - 3223. Minimum Length of String After Operations

Description

You are given a string s.

You can perform the following process on s any number of times:

Choose an index i in the string such that there is at least one character to the left of index i that is equal to s $i$ , and at least one character to the right that is also equal to s $i$ .
Delete the closest character to the left of index i that is equal to s $i$ .
Delete the closest character to the right of index i that is equal to s $i$ .

Return the minimum length of the final string s that you can achieve.

Example 1:

复制代码

Input: s = "abaacbcbb"

Output: 5

Explanation:
We do the following operations:

Choose index 2, then remove the characters at indices 0 and 3. The resulting string is s = "bacbcbb".
Choose index 3, then remove the characters at indices 0 and 5. The resulting string is s = "acbcb".
Example 2:

Input: s = "aa"

Output: 2

复制代码

Explanation:
We cannot perform any operations, so we return the length of the original string.

Constraints:

复制代码

1 <= s.length <= 2 * 10^5
s consists only of lowercase English letters.

Solution

Because we only need to get the minimum length of the string, so we don't actually need to decide which character we need to delete. Just count the frequency of all characters, and then for anything above 2, if it's an odd number, we have 1 (keep deleting 2s until it's below 3), for even numbers, we have 2 (keep deleting 2s until it's below 3). Add those numbers up and we have our result.

Update: for frequencies below 3, it's the same: for 1 we add 1, for 2 we add 2.

Time complexity: o ( n ) o(n) o(n)

Space complexity: o ( n ) ≈ o ( 1 ) o(n) \approx o(1) o(n)≈o(1), because we have at most 26 lowercase characters.

Code

python3 复制代码

class Solution:
    def minimumLength(self, s: str) -> int:
        ch_cnt = collections.Counter(s)
        res = 0
        for ch, cnt in ch_cnt.items():
            res += (1 if cnt & 1 == 1 else 2)
        return res