1750. Minimum Length of String After Deleting Similar Ends
Description
Given a string s
consisting only of characters 'a'
, 'b'
, and 'c'
. You are asked to apply the following algorithm on the string any number of times:
- Pick a non-empty prefix from the string
s
where all the characters in the prefix are equal. - Pick a non-empty suffix from the string
s
where all the characters in this suffix are equal. - The prefix and the suffix should not intersect at any index.
- The characters from the prefix and suffix must be the same.
- Delete both the prefix and the suffix.
Return the minimum length of s
after performing the above operation any number of times (possibly zero times).
Example 1:
Input: s = "ca" Output: 2 Explanation: You can't remove any characters, so the string stays as is.
Example 2:
Input: s = "cabaabac" Output: 0 Explanation: An optimal sequence of operations is: - Take prefix = "c" and suffix = "c" and remove them, s = "abaaba". - Take prefix = "a" and suffix = "a" and remove them, s = "baab". - Take prefix = "b" and suffix = "b" and remove them, s = "aa". - Take prefix = "a" and suffix = "a" and remove them, s = "".
Example 3:
Input: s = "aabccabba" Output: 3 Explanation: An optimal sequence of operations is: - Take prefix = "aa" and suffix = "a" and remove them, s = "bccabb". - Take prefix = "b" and suffix = "bb" and remove them, s = "cca".
Constraints:
1 <= s.length <= 105
s
only consists of characters'a'
,'b'
, and'c'
.
Solutions
Solution 1: Two pointers
We define two pointers $i$ and $j$ to point to the head and tail of the string $s$ respectively, then move them to the middle until the characters pointed to by $i$ and $j$ are not equal, then $\max(0, j - i + 1)$ is the answer.
The time complexity is $O(n)$ and the space complexity is $O(1)$. Where $n$ is the length of the string $s$.
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