How do you solve Number of Subarrays That Match a Pattern II on LeetCode?

Number of Subarrays That Match a Pattern II (LeetCode #3036) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Transforming the problem into a simpler form can significantly reduce complexity.

What is the time complexity of Number of Subarrays That Match a Pattern II?

The optimal solution for Number of Subarrays That Match a Pattern II runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse the nums array a couple of times, making it efficient for large inputs.

What is the brute force solution for Number of Subarrays That Match a Pattern II?

The brute force approach for Number of Subarrays That Match a Pattern II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible subarray of size m + 1 in nums to see if it matches the pattern. This is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Subarrays That Match a Pattern II?

Number of Subarrays That Match a Pattern II uses the following concepts: Array, Rolling Hash, String Matching, Hash Function. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Subarrays That Match a Pattern II?

Always start with a brute-force solution to understand the problem better. Look for patterns in the problem that can simplify your approach. Practice transforming problems into different forms to find efficient solutions.

What are common mistakes when solving Number of Subarrays That Match a Pattern II?

Not correctly handling edge cases where the pattern length is close to the nums length. Failing to optimize the approach after identifying a brute-force solution.

#3036

Number of Subarrays That Match a Pattern II

Hard

ArrayRolling HashString MatchingHash FunctionHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution transforms the nums array into a new array that represents the relationships between consecutive elements. This allows us to directly compare subarrays to the pattern using a sliding window approach, making it much faster.

⚙️

Algorithm

3 steps

1Step 1: Create a new array nums2 where nums2[i] = 1 if nums[i + 1] > nums[i], nums2[i] = 0 if nums[i + 1] == nums[i], and nums2[i] = -1 if nums[i + 1] < nums[i].
2Step 2: Use a sliding window of size m to count how many times the pattern appears in nums2.
3Step 3: Compare each window of size m in nums2 to the pattern and count matches.

solution.py21 lines

1# Full working Python code
2
3def count_matching_subarrays(nums, pattern):
4    n = len(nums)
5    m = len(pattern)
6    nums2 = [0] * (n - 1)
7    for i in range(n - 1):
8        if nums[i + 1] > nums[i]:
9            nums2[i] = 1
10        elif nums[i + 1] == nums[i]:
11            nums2[i] = 0
12        else:
13            nums2[i] = -1
14    count = 0
15    for i in range(n - m - 1):
16        if nums2[i:i + m] == pattern:
17            count += 1
18    return count
19
20# Example usage
21print(count_matching_subarrays([1, 2, 3, 4, 5, 6], [1, 1]))  # Output: 4

ℹ

Complexity note: The time complexity is O(n) because we only traverse the nums array a couple of times, making it efficient for large inputs.

1Transforming the problem into a simpler form can significantly reduce complexity.
2Using a sliding window approach allows for efficient comparisons.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.