Number of Subarrays That Match a Pattern I

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You are given a 0-indexed integer array nums of size n, and a 0-indexed integer array pattern of size m consisting of integers -1, 0, and 1.

A subarray nums[i..j] of size m + 1 is said to match the pattern if the following conditions hold for each element pattern[k]:

nums[i + k + 1] > nums[i + k] if pattern[k] == 1.
nums[i + k + 1] == nums[i + k] if pattern[k] == 0.
nums[i + k + 1] < nums[i + k] if pattern[k] == -1.

Return the count of subarrays in nums that match the pattern.

Example 1:

Input: nums = [1,2,3,4,5,6], pattern = [1,1]
Output: 4
Explanation: 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
Explanation: 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 <= 100
1 <= nums[i] <= 10⁹
1 <= m == pattern.length < n
-1 <= pattern[i] <= 1

Try coding it in LeetCode

Number of Subarrays That Match a Pattern I

Medium

Asked by:

29 views

Topics:

Arrays

You are given a 0-indexed integer array nums of size n, and a 0-indexed integer array pattern of size m consisting of integers -1, 0, and 1.

A subarray nums[i..j] of size m + 1 is said to match the pattern if the following conditions hold for each element pattern[k]:

nums[i + k + 1] > nums[i + k] if pattern[k] == 1.
nums[i + k + 1] == nums[i + k] if pattern[k] == 0.
nums[i + k + 1] < nums[i + k] if pattern[k] == -1.

Return the count of subarrays in nums that match the pattern.

Example 1:

Input: nums = [1,2,3,4,5,6], pattern = [1,1]
Output: 4
Explanation: 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
Explanation: 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 <= 100
1 <= nums[i] <= 10⁹
1 <= m == pattern.length < n
-1 <= pattern[i] <= 1

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the lengths of `nums` and `pattern`? What is the maximum size they can be?
What is the range of integer values within the `nums` array? Can I expect negative numbers, zeros, or just positive integers?
Is the `pattern` array guaranteed to only contain the values -1, 0, and 1?
If no subarrays match the pattern, what value should I return?
Could you provide a few more examples of the input and corresponding output to ensure I fully understand the problem requirements?

Brute Force Solution

Approach

We want to find how many times a specific pattern appears in a larger sequence of numbers. The most straightforward approach is to check every possible place where the pattern might start in the number sequence. We'll slide the pattern along the sequence and see if it matches at each position.

Here's how the algorithm would work step-by-step:

Imagine you have a pattern to match and a longer sequence of numbers.
Start with the very beginning of the sequence and see if the pattern matches starting there.
Then, shift the pattern over by one number and check if it matches at that new position.
Continue shifting the pattern, one number at a time, and checking for a match each time.
Keep doing this until the pattern has reached the end of the sequence, making sure it can completely fit at each position.
Count how many times you found a perfect match between the pattern and the sequence.
The final count represents the total number of places where the pattern appears in the sequence.

Code Implementation

def number_of_subarrays_that_match_a_pattern_one(numbers, pattern):
    number_of_matches = 0
    pattern_length = len(pattern)
    numbers_length = len(numbers)

    # Iterate through all possible starting positions of the pattern in the numbers array
    for start_index in range(numbers_length - pattern_length):

        match = True

        # Compare the pattern to the corresponding subarray of numbers
        for pattern_index in range(pattern_length):
            # Need to check all values in pattern
            if numbers[start_index + pattern_index] != pattern[pattern_index]:
                match = False
                break

        # If the entire pattern matches at this starting position, increment the count
        if match:
            number_of_matches += 1

    # Return the total number of matches found
    return number_of_matches

Big(O) Analysis

Time Complexity

O(n*m) – Let n be the length of the input array 'nums' and m be the length of the pattern array. The algorithm iterates through the 'nums' array, starting from index 0 up to n-m (inclusive), effectively sliding the pattern across the array. For each starting position, it compares the pattern with a subarray of length m in 'nums'. Therefore, the time complexity is proportional to (n-m+1) * m. In the worst-case scenario where m is significantly smaller than n, this can be approximated as n*m. Alternatively, if m and n are proportional, the complexity becomes O(n^2).

Space Complexity

O(1) – The provided plain English explanation describes an iterative process that slides the pattern along the sequence and checks for matches. This approach does not require any auxiliary data structures to store intermediate results or visited locations. Only a few variables are likely needed to keep track of the current position being checked, which takes up a constant amount of space. Therefore, the space complexity is independent of the input size and is considered constant.

Optimal Solution

Approach

Instead of comparing the pattern with every possible subarray, the optimal strategy focuses on sliding the pattern along the input. By computing a numerical representation of the input array's changes and then comparing that to our pattern, we only need to check each starting position once.

Here's how the algorithm would work step-by-step:

First, transform the input array into a new array that represents the relative changes between consecutive numbers (either 'increase' or 'decrease'). This new array will have one fewer element than the original.
Now, slide the pattern along this transformed array. At each position, check if the pattern matches the corresponding section of the transformed array.
If a match is found, increment the count. Move to the next starting position and repeat the process until the end of the transformed array is reached.
The final count represents the number of subarrays that match the specified pattern.

Code Implementation

def number_of_subarrays_that_match_a_pattern(nums, pattern):
    pattern_matches_count = 0
    modified_nums = []

    # Build modified array representing relative changes.
    for i in range(len(nums) - 1):
        if nums[i] < nums[i + 1]:
            modified_nums.append(1)
        elif nums[i] > nums[i + 1]:
            modified_nums.append(-1)
        else:
            modified_nums.append(0)

    # Iterate through the modified array with a sliding window
    for i in range(len(modified_nums) - len(pattern) + 1):
        subarray_matches = True

        # Verify if current sub-array matches pattern
        for j in range(len(pattern)):
            if modified_nums[i + j] != pattern[j]:
                subarray_matches = False
                break

        # Count the pattern if it matches the current sub-array.
        if subarray_matches:
            pattern_matches_count += 1

    return pattern_matches_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm first transforms the input array of size n into a new array of size n-1. This transformation takes O(n) time. Then, it iterates through the transformed array of size n-1, sliding the pattern along it. For each starting position, it compares the pattern (of fixed length m, where m is pattern.length) with the corresponding subarray of the transformed array. This comparison takes O(m) time. Since we slide the pattern at most n-m times, the total time complexity is O(n) + O((n-m) * m). Since m is constant (pattern size), this simplifies to O(n).

Space Complexity

O(N) – The algorithm transforms the input array of size N into a new array representing relative changes, which will have a size of N-1. This transformed array is stored in memory and constitutes the primary auxiliary space usage. Since the size of the transformed array scales linearly with the input size N, the space complexity is O(N).

Edge Cases

Case	How to Handle
nums is null or empty	Return 0 since no subarrays are possible.
pattern is null or empty	If nums has at least one subarray with length 1 return nums.length -1, otherwise return 0.
nums.length is less than pattern.length + 1	Return 0 since a subarray of the required length cannot exist.
pattern contains values other than -1, 0, or 1	The problem statement should define how to handle invalid pattern values; otherwise, assume only -1, 0, and 1 are allowed, and an error should be thrown or the invalid pattern element skipped during comparison.
nums contains very large or very small numbers (potential integer overflow during comparison)	Consider using a wider data type (e.g., long) or normalizing the nums array if comparisons result in overflow.
nums contains only identical values and pattern contains only 0s	The number of matching subarrays should be correctly calculated based on the length of nums and pattern.
pattern is very long and nums is also very long (potential performance issues)	Ensure the solution has reasonable time complexity, avoiding unnecessary nested loops by using efficient comparison methods.
nums or pattern contains negative, positive and zero values	The solution must correctly compare the elements and their relations with each other according to pattern even if the arrays contain all kinds of integer types.

nums is null or empty

How to Handle:

Return 0 since no subarrays are possible.

pattern is null or empty

How to Handle:

If nums has at least one subarray with length 1 return nums.length -1, otherwise return 0.

nums.length is less than pattern.length + 1

How to Handle:

Return 0 since a subarray of the required length cannot exist.

pattern contains values other than -1, 0, or 1

How to Handle:

The problem statement should define how to handle invalid pattern values; otherwise, assume only -1, 0, and 1 are allowed, and an error should be thrown or the invalid pattern element skipped during comparison.

nums contains very large or very small numbers (potential integer overflow during comparison)

How to Handle:

Consider using a wider data type (e.g., long) or normalizing the nums array if comparisons result in overflow.

nums contains only identical values and pattern contains only 0s

How to Handle:

The number of matching subarrays should be correctly calculated based on the length of nums and pattern.

pattern is very long and nums is also very long (potential performance issues)

How to Handle:

Ensure the solution has reasonable time complexity, avoiding unnecessary nested loops by using efficient comparison methods.

nums or pattern contains negative, positive and zero values

How to Handle:

The solution must correctly compare the elements and their relations with each other according to pattern even if the arrays contain all kinds of integer types.