leetcode - 2981. Find Longest Special Substring That Occurs Thrice I

KpLn_HJL2024-12-11 16:43

Description

You are given a string s that consists of lowercase English letters.

A string is called special if it is made up of only a single character. For example, the string "abc" is not special, whereas the strings "ddd", "zz", and "f" are special.

Return the length of the longest special substring of s which occurs at least thrice, or -1 if no special substring occurs at least thrice.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

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Input: s = "aaaa"
Output: 2
Explanation: The longest special substring which occurs thrice is "aa": substrings "aaaa", "aaaa", and "aaaa".
It can be shown that the maximum length achievable is 2.

Example 2:

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Input: s = "abcdef"
Output: -1
Explanation: There exists no special substring which occurs at least thrice. Hence return -1.

Example 3:

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Input: s = "abcaba"
Output: 1
Explanation: The longest special substring which occurs thrice is "a": substrings "abcaba", "abcaba", and "abcaba".
It can be shown that the maximum length achievable is 1.

Constraints:

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3 <= s.length <= 50
s consists of only lowercase English letters.

Solution

Brute Force

Because of the low constrains, a very simple brute force way would be: we get all the special subarrays, and then calculate each one's frequency, then get the largest length of the subarrays that meet the requirements.

Time complexity: o ( n 3 ) o(n^3) o(n3)

Space complexity: o ( 1 ) o(1) o(1)

With a more strict constrain, see this 2982. Find Longest Special Substring That Occurs Thrice II

Code

Brute Force

python3 复制代码

class Solution:
    def maximumLength(self, s: str) -> int:
        res = -1
        # s[i:j+1]
        for i in range(len(s)):
            for j in range(i, len(s)):
                if s[j] != s[i]:
                    break
                sub_string = s[i: j + 1]
                sub_string_fre = 0
                for k in range(0, len(s)):
                    if k + j + 1 - i > len(s):
                        break
                    if sub_string == s[k: k + j + 1 - i]:
                        sub_string_fre += 1
                if sub_string_fre >= 3:
                    res = max(res, j + 1 - i)
        return res