1408. String Matching in an Array

Problem

Given an array of string words, return all strings in words that is a substring of another word. You can return the answer in any order.

A substring is a contiguous sequence of characters within a string.

https://leetcode.com/problems/string-matching-in-an-array/

Example 1:

Input: words = ["mass","as","hero","superhero"]
Output: ["as","hero"]
Explanation: "as" is substring of "mass" and "hero" is substring of "superhero".
["hero","as"] is also a valid answer.

Example 2:

Input: words = ["leetcode","et","code"]
Output: ["et","code"]
Explanation: "et", "code" are substring of "leetcode".

Example 3:

Input: words = ["blue","green","bu"]
Output: []
Explanation: No string of words is substring of another string.

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 30
• words[i] contains only lowercase English letters.
• All the strings of words are unique.

Test Cases

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class Solution:
    def stringMatching(self, words: List[str]) -> List[str]:

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import pytest

from solution import Solution


@pytest.mark.parametrize('words, expected', [
    (["mass","as","hero","superhero"], ["as","hero"]),
    (["leetcode","et","code"], ["et","code"]),
    (["blue","green","bu"], []),

    (["leetcoder","leetcode","od","hamlet","am"], ["leetcode","od","am"])
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, words, expected):
    assert sorted(sol.stringMatching(words.copy())) == sorted(expected)

Thoughts

先对 words 按单词长度排序，可以减少后续的循环次数，并能很好地避免重复输出。

用两重循环比较任意两个字符串，看较长的是否包含较短的。

时间复杂度 O(n²)，额外的空间复杂度 O(1)。

Code

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class Solution:
    def stringMatching(self, words: list[str]) -> list[str]:
        words.sort(key=lambda w: len(w))
        res = []
        n = len(words)
        for i in range(n - 1):
            a = words[i]
            for j in range(i + 1, n):
                b = words[j]
                if a in b:
                    res.append(a)
                    break

        return res

参考资料

• 反向引用

  • 3042. Count Prefix and Suffix Pairs I

许可协议

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。

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