3042. Count Prefix and Suffix Pairs I

Problem

You are given a 0-indexed string array words.

Let’s define a boolean function isPrefixAndSuffix that takes two strings, str1 and str2:

isPrefixAndSuffix(str1, str2) returns true if str1 is both a prefix and a suffix of str2, and false otherwise.

A prefix of a string is a substring that starts from the beginning of the string and extends to any point within it.

A suffix of a string is a substring that begins at any point in the string and extends to its end.

For example, isPrefixAndSuffix("aba", "ababa") is true because "aba" is a prefix of "ababa" and also a suffix, but isPrefixAndSuffix("abc", "abcd") is false.

Return an integer denoting the number of index pairs (i, j) such that i < j, and isPrefixAndSuffix(words[i], words[j]) is true.

https://leetcode.com/problems/count-prefix-and-suffix-pairs-i/

Example 1:

Input: words = ["a","aba","ababa","aa"]
Output: 4
Explanation: In this example, the counted index pairs are:
i = 0 and j = 1 because isPrefixAndSuffix("a", "aba") is true.
i = 0 and j = 2 because isPrefixAndSuffix("a", "ababa") is true.
i = 0 and j = 3 because isPrefixAndSuffix("a", "aa") is true.
i = 1 and j = 2 because isPrefixAndSuffix("aba", "ababa") is true.
Therefore, the answer is 4.

Example 2:

Input: words = ["pa","papa","ma","mama"]
Output: 2
Explanation: In this example, the counted index pairs are:
i = 0 and j = 1 because isPrefixAndSuffix("pa", "papa") is true.
i = 2 and j = 3 because isPrefixAndSuffix("ma", "mama") is true.
Therefore, the answer is 2.

Example 3:

Input: words = ["abab","ab"]
Output: 0
Explanation: In this example, the only valid index pair is i = 0 and j = 1, and isPrefixAndSuffix("abab", "ab") is false.
Therefore, the answer is 0.

Constraints:

1 <= words.length <= 50
1 <= words[i].length <= 10
words[i] consists only of lowercase English letters.

Test Cases

1 2	class Solution: def countPrefixSuffixPairs(self, words: List[str]) -> int:

solution_test.py

import pytest

from solution import Solution


@pytest.mark.parametrize('words, expected', [
    (["a","aba","ababa","aa"], 4),
    (["pa","papa","ma","mama"], 2),
    (["abab","ab"], 0),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, words, expected):
    assert sol.countPrefixSuffixPairs(words) == expected

Thoughts

直接类似 1408. String Matching in an Array 那样，遍历所有的单词对。

时间复杂度 O(n²)，空间复杂度 O(1)。

本题单词长度可以认为是常数，否则如果单词长度为 k，则时间复杂度 O(k * n²)。

不过本题的 k 和 n 都很小，优化了时间复杂度可能反倒系数很大导致运行更慢。

Code

solution.py

from itertools import combinations


class Solution:
    def countPrefixSuffixPairs(self, words: list[str]) -> int:
        return sum(1 for s1, s2 in combinations(words, 2) if s2.startswith(s1) and s2.endswith(s1))