为何检查含重复词的列表匹配方法时间复杂度为2*O(n log n)+O(m log m)

First, let's ground this in the problem at hand: we need to confirm every word in searchWords (including its duplicate count) has at least as many occurrences in newsPaperWords. Your initial intuition and your interviewer's feedback both make sense—they just refer to two common, distinct implementations of this check.

Your Linear-Time Intuition (O(n + m)) #

Your estimate of 2*O(n)+O(m) likely comes from a hash map (dictionary) frequency counting approach, which is a standard linear-time solution:

• Step 1: Traverse newsPaperWords and count how many times each word appears. This takes O(m) time, where m is the length of newsPaperWords.
• Step 2: Traverse searchWords and count its word frequencies. This takes O(n) time, where n is the length of searchWords.
• Step 3: Iterate through the frequency map of searchWords, verifying each word's count in newsPaperWords meets or exceeds the required number. This takes O(k) time (worst case O(n)), where k is the number of unique words in searchWords.

Adding these up gives a total time complexity of O(m + n)—which aligns with your initial thought (you probably grouped the two linear traversals as 2*O(n) when simplifying variables).

Your Interviewer's Sorting-Based Approach (O(n log n + m log m)) #

The 2*O(n log n)+O(m log m) complexity refers to a sort + two-pointer implementation, another valid way to solve the problem:

• Step 1: Sort newsPaperWords. Sorting a list of length m takes O(m log m) time.
• Step 2: Sort searchWords. Sorting a list of length n takes O(n log n) time.
• Step 3: Use two pointers to traverse both sorted lists, matching words and confirming each entry in searchWords has enough corresponding occurrences in newsPaperWords. This final traversal takes O(m + n) time, which is negligible compared to the sorting steps.

The dominant terms here are the two sorting operations, leading to the total time complexity your interviewer cited: O(n log n + m log m) (the "2*" likely references the two separate sorting passes on the input lists).

Key Tradeoffs Between the Two Methods #

• Hash map method: Faster linear time, but requires extra space to store frequency counts (space complexity O(k + l), where k and l are the number of unique words in each list).
• Sort + two-pointer method: No extra space needed (if using in-place sorting), but has higher time complexity due to the sorting overhead.

Both are valid solutions—you were thinking of the hash map approach, while your interviewer was referencing the sorting-based alternative.

内容的提问来源于stack exchange，提问作者antonro