2981. Find Longest Special Substring That Occurs Thrice I

Problem

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.

https://leetcode.com/problems/find-longest-special-substring-that-occurs-thrice-i/

Example 1:

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:

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

Example 3:

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:

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

Test Cases

python
class Solution:
    def maximumLength(self, s: str) -> int:

solution_test.py下载完整源码
py
import pytest

from solution import Solution


@pytest.mark.parametrize('s, expected', [
    ("aaaa", 2),
    ("abcdef", -1),
    ("abcaba", 1),

    ("biaei", -1),
    ("eccdnmcnkl", 1),
])
class Test:
    def test_solution(self, s, expected):
        sol = Solution()
        assert sol.maximumLength(s) == expected

Thoughts

首先对于某个字符的一个连续片段（如 "aaaa"），若长度为 n，则其中包含「出现至少三次的特殊子字符串」的最长长度为 n - 2（如 Example 1 所示）。

如果某字符有两个片段（如 "aa****aaa"），设长度分别为 m 和 n（不妨设 m ≤ n）。如果 m < n，则长为 n 的片段中至少可以出现两次长度为 m 的特殊子字符串，加上长为 m 的片段就可以满足「出现至少三次」。如果 m = n，则两个片段中各出现两次长度为 n - 1 的特殊子字符串，加起来可以满足「出现至少三次」。

如果某字符有三个片段（如 "aa***aa***aaa"），设长度分别为 k、m 和 n（不妨设 k ≤ m ≤ n）。如果 k = m = n，则显然长为 n 的特殊子字符串满足「出现至少三次」。如果 k < m ≤ n，那就对 m 和 n 按上边两个片段的逻辑处理即可。

如果有更多的片段，只需要看最长的两个和三个即可。

总的来看，设某个字符最长的三个片段的长度分别为 k、m 和 n（不妨设 0 ≤ k ≤ m ≤ n），那么该字符的「出现至少三次的特殊子字符串」的最长长度为：

\begin{cases} n & \text{if }k=m=n \\ n-1 & \text{if }k<m=n \\ max(m,n-2) & \text{if }k\le m<n \end{cases}

遍历 s，对每个字符用大小为 3 的最小堆记录最长的三个片段的长度，根据上边式子计算该字符「出现至少三次的特殊子字符串」的最长长度，最后对所有字符取最大即可。

时间复杂度 O(n)（这里 n 表示 s 的长度），空间复杂度 O(1)（实际上是 s 中不同字符的数量，题目限定了仅小写英文字符，则可以认为是常数；否则也可以认为是 O(n)）。

Code

solution.py下载完整源码
py
from collections import defaultdict
from heapq import heapreplace


class Solution:
    def maximumLength(self, s: str) -> int:
        n = len(s)
        top3s: dict[str, list[int]] = defaultdict(lambda: [0, 0, 0]) # char: its three longest continuous pieces.
        prev = s[0]
        cnt = 1
        for i in range(1, n + 1):
            if i < n and (c := s[i]) == prev:
                cnt += 1
                continue

            if cnt > top3s[prev][0]:
                heapreplace(top3s[prev], cnt)

            if i < n:
                prev = c
                cnt = 1

        max_len = 0
        for x, y, z in top3s.values():
            if y == z:
                cnt = z if z == x else z - 1
            else:
                if y > z:
                    y, z = z, y
                cnt = max(y, z - 2)

            if cnt > max_len:
                max_len = cnt

        return max_len or -1

#Problem

#Test Cases

#Thoughts

#Code

Problem

Test Cases

Thoughts

Code