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Problem solution · Python

Number of Subarrays That Match a Pattern II

Number of Subarrays That Match a Pattern II: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
49 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Number of Subarrays That Match a Pattern II, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 49 lines of Python from the credited upstream file 3036.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeNumber of Subarrays That Match a Pattern II · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  # Same as 3034. Number of Subarrays That Match a Pattern I  def countMatchingSubarrays(self, nums: list[int], pattern: list[int]) -> int:    def getNum(a: int, b: int) -> int:      if a < b:        return 1      if a > b:        return -1      return 0     numsPattern = [getNum(a, b) for a, b in itertools.pairwise(nums)]    return self._kmp(numsPattern, pattern)   def _kmp(self, nums: list[int], pattern: list[int]) -> int:    """Returns the number of occurrences of the pattern in `nums`."""     def getLPS(nums: list[int]) -> list[int]:      """      Returns the lps array, where lps[i] is the length of the longest prefix of      nums[0..i] which is also a suffix of this substring.      """      lps = [0] * len(nums)      j = 0      for i in range(1, len(nums)):        while j > 0 and nums[j] != nums[i]:          j = lps[j - 1]        if nums[i] == nums[j]:          lps[i] = j + 1          j += 1      return lps     lps = getLPS(pattern)    res = 0    i = 0  # s' index    j = 0  # pattern's index    while i < len(nums):      if nums[i] == pattern[j]:        i += 1        j += 1        if j == len(pattern):          res += 1          j = lps[j - 1]      elif j != 0:  # Mismatch after j matches.        # Don't match lps[0..lps[j - 1]] since they will match anyway.        j = lps[j - 1]      else:        i += 1    return res

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