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

Number of Subarrays That Match a Pattern I (LeetCode #3034) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding the pattern conditions is crucial for matching subarrays.

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

The optimal solution for Number of Subarrays That Match a Pattern I runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only traverse the nums array once, checking conditions based on the pattern without nested loops.

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

The brute force approach for Number of Subarrays That Match a Pattern I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray of size m + 1 in the nums array to see if it matches the pattern. This is straightforward but can be slow for larger arrays. 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 I?

Number of Subarrays That Match a Pattern I 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 I?

Clearly explain your thought process while coding. Consider edge cases and test your solution with them. Be prepared to discuss the time and space complexity of your solution.

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

Not resetting the valid length when a condition fails. Overlooking edge cases where the pattern might not match at all.

#3034

Number of Subarrays That Match a Pattern I

Medium

ArrayRolling HashString MatchingHash FunctionHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution uses a single pass through the nums array while maintaining a count of valid subarrays based on the pattern. This reduces the need for nested loops and improves efficiency.

⚙️

Algorithm

5 steps

1Step 1: Initialize a count to 0 and a variable to track the length of valid subarrays.
2Step 2: Iterate through nums while checking conditions based on the pattern.
3Step 3: If a condition is satisfied, increment the length of the valid subarray.
4Step 4: If a condition fails, reset the length and check if it matches the pattern.
5Step 5: Return the total count of valid subarrays.

solution.py18 lines

1# Full working Python code
2
3def count_matching_subarrays(nums, pattern):
4    count = 0
5    m = len(pattern)
6    n = len(nums)
7    valid_length = 0
8    for i in range(n - 1):
9        if i < n - 1 and (pattern[valid_length] == 1 and nums[i + 1] > nums[i] or 
10                           pattern[valid_length] == 0 and nums[i + 1] == nums[i] or 
11                           pattern[valid_length] == -1 and nums[i + 1] < nums[i]):
12            valid_length += 1
13            if valid_length == m:
14                count += 1
15        else:
16            valid_length = 0
17    return count
18

ℹ

Complexity note: The time complexity is O(n) because we only traverse the nums array once, checking conditions based on the pattern without nested loops.

1Understanding the pattern conditions is crucial for matching subarrays.
2The optimal solution leverages a single pass to reduce time complexity.

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