#3036

Hard

LeetCode Problem Workspace

Number of Subarrays That Match a Pattern II

Count subarrays matching a pattern of relative values using array transformation and rolling hash techniques.

array rolling hash string matching hash function

Problem Statement

You are given an array nums and a pattern consisting of -1, 0, and 1. You need to determine the number of subarrays in nums that match this pattern. A subarray matches the pattern if the relative comparisons between consecutive elements in nums follow the conditions described by the elements of the pattern.

For instance, if the pattern is [1, 0, -1], the first comparison in the subarray should be less than the second element, the second element should be equal to the third element, and the third element should be greater than the fourth. The task is to count how many subarrays of size m + 1 in nums follow the exact pattern.

Examples

Example 1

Input: nums = [1,2,3,4,5,6], pattern = [1,1]

Output: 4

The pattern [1,1] indicates that we are looking for strictly increasing subarrays of size 3. In the array nums, the subarrays [1,2,3], [2,3,4], [3,4,5], and [4,5,6] match this pattern. Hence, there are 4 subarrays in nums that match the pattern.

Example 2

Input: nums = [1,4,4,1,3,5,5,3], pattern = [1,0,-1]

Output: 2

Here, the pattern [1,0,-1] indicates that we are looking for a sequence where the first number is smaller than the second, the second is equal to the third, and the third is greater than the fourth. In the array nums, the subarrays [1,4,4,1], and [3,5,5,3] match this pattern. Hence, there are 2 subarrays in nums that match the pattern.

Constraints

2 <= n == nums.length <= 106
1 <= nums[i] <= 109
1 <= m == pattern.length < n
-1 <= pattern[i] <= 1

Solution Approach

Array Transformation

Transform the array nums into a new array nums2 where each element represents the relative comparison between consecutive elements in nums. Specifically, 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]. This transformation makes it easier to match subarrays to the pattern.

Rolling Hash Approach

Using the transformed array nums2, apply a rolling hash technique to efficiently compare subarrays. Instead of checking each subarray individually, use a hash value to represent subarrays and quickly compare them to the pattern. This method reduces time complexity for large arrays.

Pattern Matching

With the transformed array and rolling hash, check each subarray of length m + 1 to see if it matches the given pattern. For a match, the relative values in nums2 should correspond exactly to the elements in the pattern.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the specific approach used. If using the array transformation and rolling hash techniques, the complexity is O(n), where n is the length of nums. This is much more efficient than a brute-force solution. The space complexity is O(n) due to the transformation array and hash storage.

What Interviewers Usually Probe

Tests the candidate's ability to transform arrays efficiently.
Assesses understanding of rolling hash techniques for optimization.
Evaluates pattern matching skills and their application to subarrays.

Common Pitfalls or Variants

Common pitfalls

Incorrectly transforming the array nums into nums2 based on relative comparisons.
Failure to optimize the solution with rolling hash, resulting in time complexity that is too high.
Not accounting for edge cases, such as when nums has the smallest or largest possible length.

Follow-up variants

Using sliding window techniques instead of rolling hash for comparison.
Allowing for additional constraints in the pattern, such as different values for matching conditions.
Changing the input array structure, for example, making nums a list of strings instead of integers.

FAQ

What is the best approach to solve 'Number of Subarrays That Match a Pattern II'?

The best approach is to transform the array into a new one that represents relative comparisons between consecutive elements and then use a rolling hash for efficient pattern matching.

How can rolling hash help in solving this problem?

Rolling hash helps by reducing the need to compare each subarray element by element. Instead, it computes a hash for each subarray, allowing for quick comparisons.

What is the complexity of the solution?

The time complexity is O(n) and space complexity is O(n), where n is the length of the input array nums.

Can this problem be solved without using rolling hash?

Yes, but using a rolling hash significantly optimizes the solution, especially for large arrays, making it much more efficient than a brute force approach.

What should I do if I encounter edge cases in 'Number of Subarrays That Match a Pattern II'?

Ensure that the edge cases, such as small array sizes and various pattern values, are handled correctly in both the array transformation and pattern matching steps.

terminal

Solution

Solution 1

#### Python3

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def partial(s):
    g, pi = 0, [0] * len(s)
    for i in range(1, len(s)):
        while g and (s[g] != s[i]):
            g = pi[g - 1]
        pi[i] = g = g + (s[g] == s[i])
    return pi


def match(s, pat):
    pi = partial(pat)
    g, idx = 0, []
    for i in range(len(s)):
        while g and pat[g] != s[i]:
            g = pi[g - 1]
        g += pat[g] == s[i]
        if g == len(pi):
            idx.append(i + 1 - g)
            g = pi[g - 1]
    return idx


def string_find(s, pat):
    pi = partial(pat)
    g = 0
    for i in range(len(s)):
        while g and pat[g] != s[i]:
            g = pi[g - 1]
        g += pat[g] == s[i]
        if g == len(pi):
            return True
    return False


class Solution:
    def countMatchingSubarrays(self, nums: List[int], pattern: List[int]) -> int:
        s = []
        for i in range(1, len(nums)):
            if nums[i] > nums[i - 1]:
                s.append(1)
            elif nums[i] == nums[i - 1]:
                s.append(0)
            else:
                s.append(-1)
        return len(match(s, pattern))

Continue Practicing

#3034 number of subarrays that match a pattern i #3042 count prefix and suffix pairs i #3045 count prefix and suffix pairs ii