Find the Power of K-Size Subarrays I

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ArraysSliding Windows

You are given an array of integers nums of length n and a positive integer k.

The power of an array is defined as:

Its maximum element if all of its elements are consecutive and sorted in ascending order.
-1 otherwise.

You need to find the power of all subarrays of nums of size k.

Return an integer array results of size n - k + 1, where results[i] is the power of nums[i..(i + k - 1)].

Example 1:

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

Output: [3,4,-1,-1,-1]

Explanation:

There are 5 subarrays of nums of size 3:

[1, 2, 3] with the maximum element 3.
[2, 3, 4] with the maximum element 4.
[3, 4, 3] whose elements are not consecutive.
[4, 3, 2] whose elements are not sorted.
[3, 2, 5] whose elements are not consecutive.

Example 2:

Input: nums = [2,2,2,2,2], k = 4

Output: [-1,-1]

Example 3:

Input: nums = [3,2,3,2,3,2], k = 2

Output: [-1,3,-1,3,-1]

Constraints:

1 <= n == nums.length <= 500
1 <= nums[i] <= 10⁵
1 <= k <= n

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 possible ranges for the integer values within the input array?
Can the input array contain negative numbers, zeros, or floating-point numbers?
Is the input array guaranteed to have at least k elements, or could it be smaller?
What should the function return if no subarray of size k exists that meets the problem criteria?
Could you please elaborate on the specific criteria used to define the 'power' of a k-size subarray?

Brute Force Solution

Approach

Imagine you have a list of numbers, and you want to find groups of a specific size. The brute force method looks at every possible group of that size within the list. It checks all of them individually to see if they have the property you're looking for.

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

Consider the first possible group of the specified size at the beginning of the list.
Calculate the value we're interested in for that group (the 'power' in this case).
Now, shift the group by one position to the right and consider the next possible group.
Calculate the value for this new group.
Keep sliding the group to the right, one position at a time, and calculating the value for each group.
Repeat this process until you've considered every possible group of the specified size within the list.
Finally, go through all the calculated values and find the one that meets the specific condition you are looking for (e.g., the maximum value).

Code Implementation

def find_power_of_k_size_subarrays_brute_force(numbers, subarray_size):
    all_subarray_powers = []
    number_of_elements = len(numbers)

    # Iterate through all possible starting positions of the subarray.
    for start_index in range(number_of_elements - subarray_size + 1):

        current_subarray_power = 1
        # Calculate power of current subarray.
        for index_in_subarray in range(subarray_size):
            current_subarray_power *= numbers[start_index + index_in_subarray]

        # Store power of the current subarray.
        all_subarray_powers.append(current_subarray_power)

    # Initialize maximum_power with negative infinity for safe comparison
    maximum_power = float('-inf')

    # Find the maximum power among all subarrays
    for subarray_power in all_subarray_powers:
        if subarray_power > maximum_power:

            # Update maximum_power if current power is larger
            maximum_power = subarray_power

    # Returns the largest calculated subarray power.
    return maximum_power

Big(O) Analysis

Time Complexity

O(n*k) – The provided brute force approach iterates through the array to consider each possible subarray of size k. For an array of size n, there are approximately n-k+1 such subarrays. For each subarray of size k, we perform calculations to determine its 'power'. Therefore, we are iterating through each of the n-k+1 subarrays of size k, resulting in approximately (n-k+1)*k operations. Since k is a fixed size this simplifies to O(n*k).

Space Complexity

O(1) – The brute force approach, as described, iterates through subarrays of size K and calculates a value for each. It doesn't explicitly state any auxiliary data structures are created to store intermediate subarrays or any other data related to the array itself. The maximum value calculated can be stored in a single variable. Therefore, the space complexity is constant and does not depend on the input size N (the length of the list).

Optimal Solution

Approach

The optimal approach calculates a value for each group of numbers of a specific size without recalculating the sum repeatedly. Instead of doing a full recalculation, we update our running calculation based on the previous group.

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

Calculate the value of the first group of numbers.
Then, to get the value of the next group, subtract the number that is leaving the group and add the number that is joining the group.
Repeat this process, sliding the group along the number list, always updating the running total instead of recomputing it from scratch.
Each time the group is moved, perform whatever function/calculation you need to do with the new total.
This clever approach saves a lot of work by reusing previous calculations.

Code Implementation

def find_power_of_k_size_subarrays(number_list, subarray_size):    if not number_list or subarray_size <= 0 or subarray_size > len(number_list):
        return []

    result = []
    current_sum = 0

    # Calculate the sum of the first subarray.
    for i in range(subarray_size):
        current_sum += number_list[i]

    result.append(current_sum**2)

    # Slide the window and update the sum.
    for i in range(len(number_list) - subarray_size):
        current_sum -= number_list[i]

        # Add the next element in the array.
        current_sum += number_list[i + subarray_size]

        result.append(current_sum**2)

    return result

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n once. Inside the loop, only constant-time operations (addition, subtraction) are performed to update the running calculation for each subarray. Thus, the time complexity is directly proportional to the number of elements in the array, resulting in O(n).

Space Complexity

O(1) – The provided explanation focuses on a sliding window approach where we update a running value. This method does not create any additional data structures whose size scales with the input array's size (N). It utilizes a fixed number of variables to store intermediate calculations or indices for the current window, independent of N. Therefore, the auxiliary space remains constant, leading to a space complexity of O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty list or appropriate error code if the input array is null or empty as no subarrays can be formed.
k is zero or negative	Return an empty list or raise an IllegalArgumentException if k is zero or negative because a subarray size must be positive.
k is greater than the array length	Return an empty list because you can't form a subarray bigger than the array itself.
Array contains only one element and k is 1	Return a list containing the single element if k is 1 and the array has one element.
Array contains all identical values	The algorithm should still correctly generate the power for each k-sized window, correctly handling edge cases based on formula.
Input array with large integers leading to potential integer overflow in power calculation	Use a data type that can accommodate larger values like 'long' in Java or handle overflow explicitly.
Large input array to test for time complexity and memory usage	Ensure the algorithm's time complexity meets requirements (e.g., sliding window approach for O(n) time) and memory usage is optimized.
Array contains negative numbers and zero	The power calculation must correctly handle negative numbers and zero values, as their powers behave differently compared to positive integers.

Null or empty input array

How to Handle:

Return an empty list or appropriate error code if the input array is null or empty as no subarrays can be formed.

k is zero or negative

How to Handle:

Return an empty list or raise an IllegalArgumentException if k is zero or negative because a subarray size must be positive.

k is greater than the array length

How to Handle:

Return an empty list because you can't form a subarray bigger than the array itself.

Array contains only one element and k is 1

How to Handle:

Return a list containing the single element if k is 1 and the array has one element.

Array contains all identical values

How to Handle:

The algorithm should still correctly generate the power for each k-sized window, correctly handling edge cases based on formula.

Input array with large integers leading to potential integer overflow in power calculation

How to Handle:

Use a data type that can accommodate larger values like 'long' in Java or handle overflow explicitly.

Large input array to test for time complexity and memory usage

How to Handle:

Ensure the algorithm's time complexity meets requirements (e.g., sliding window approach for O(n) time) and memory usage is optimized.

Array contains negative numbers and zero

How to Handle:

The power calculation must correctly handle negative numbers and zero values, as their powers behave differently compared to positive integers.