固定长度最频繁子串：最快算法描述、正确性证明及复杂度计算

Hey there! Let's break down the fastest approach to solve this problem—no full implementation required, just clear, efficient logic that leverages the limited character set (only 'a' and 'b') for maximum speed.

Algorithm Description #

We'll use a binary mapping + sliding window counting approach, optimized specifically for the two-character constraint:

• Step 1: Map substrings to unique integers
  Treat 'a' as 0 and 'b' as 1. Every k-length substring can be converted to a unique integer between 0 and 2^k - 1 (since each of the k positions has exactly 2 possible values).
• Step 2: Initialize a count tracker
  Use an array (faster than a hash table here, since we know the exact range of possible values) of size 2^k, initialized to 0. If k is very large (where 2^k exceeds n), switch to a hash table to avoid wasted space.
• Step 3: Process the first substring
  Convert the first k characters to their binary integer equivalent, then increment the corresponding index in the count array.
• Step 4: Slide through the rest of the string
  For each subsequent position:
  1. Update the current integer by removing the leftmost character's contribution: shift the integer left by 1 bit, then apply a mask of (1 << k) - 1 to keep only the last k bits (prevents overflow).
  2. Add the binary value of the new rightmost character (0 for 'a', 1 for 'b').
  3. Increment the count array at this new integer's index.
• Step 5: Extract the result
  Iterate through the count array to find the index with the highest value. Convert this index back to a k-length string by mapping each bit to 'a' (0) or 'b' (1) (note: the highest bit corresponds to the first character of the substring). If multiple indices have the same max count, return any corresponding string.

Correctness Proof #

Let's verify this approach works as intended:

1. Unique 1:1 mapping: Every k-length 'a'/'b' substring maps to exactly one integer between 0 and 2^k -1, and vice versa. There's no overlap or ambiguity—different substrings get different integers, so our count array tracks each substring's occurrences accurately.
2. Sliding window accuracy: When we slide the window, we're effectively removing the leftmost character and adding the new rightmost one. The bit-shift and mask operation perfectly simulates this: shifting left drops the leftmost bit, and the mask ensures we only keep k bits. This update takes O(1) time per step, so we don't miss any substrings or miscount.
3. Count validity: We process all n - k + 1 possible k-length substrings, each exactly once. The count array's values are the true occurrence counts of each substring, so picking the index with the highest count gives us the most frequent substring.

Complexity Analysis #

• Time Complexity: O(n)
  • Processing the first substring takes O(k) time.
  • Sliding through the remaining n - k positions takes O(n - k) time (each step is O(1)).
  • Finding the max count takes O(2^k) time, but since 2^k is at most n (when k ≤ log2(n)), this is dominated by the O(n) traversal. For larger k, we'd use a hash table, and the max lookup is O(n) in the worst case, still keeping the total time O(n).
• Space Complexity: O(min(2^k, n))
  • For small k (where 2^k ≤ n), we use an array of size 2^k.
  • For large k (where 2^k > n), we use a hash table that stores at most n - k + 1 entries (since there can't be more unique substrings than the number of total substrings).

Example Walkthrough #

Take the input string ababaaaabb and k=3:

1. First substring aba → binary 010 → integer 2 → count[2] = 1.
2. Next substring bab → 101 → integer 5 → count[5] =1.
3. Next substring aba → 010 → count[2] =2.
4. Continuing through the rest of the string, aba ends up with the highest count (2), so we convert 2 back to aba as the result.

内容的提问来源于stack exchange，提问作者qalis