Sliding Subarray Beauty
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ArraysSliding Windows
Given an integer array nums containing n integers, find the beauty of each subarray of size k.
The beauty of a subarray is the xth smallest integer in the subarray if it is negative, or 0 if there are fewer than x negative integers.
Return an integer array containing n - k + 1 integers, which denote the beauty of the subarrays in order from the first index in the array.
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A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [1,-1,-3,-2,3], k = 3, x = 2 Output: [-1,-2,-2] Explanation: There are 3 subarrays with size k = 3. The first subarray is[1, -1, -3]and the 2nd smallest negative integer is -1. The second subarray is[-1, -3, -2]and the 2nd smallest negative integer is -2. The third subarray is[-3, -2, 3]and the 2nd smallest negative integer is -2.
Example 2:
Input: nums = [-1,-2,-3,-4,-5], k = 2, x = 2 Output: [-1,-2,-3,-4] Explanation: There are 4 subarrays with size k = 2. For[-1, -2], the 2nd smallest negative integer is -1. For[-2, -3], the 2nd smallest negative integer is -2. For[-3, -4], the 2nd smallest negative integer is -3. For[-4, -5], the 2nd smallest negative integer is -4.
Example 3:
Input: nums = [-3,1,2,-3,0,-3], k = 2, x = 1 Output: [-3,0,-3,-3,-3] Explanation: There are 5 subarrays with size k = 2. For[-3, 1], the 1st smallest negative integer is -3. For[1, 2], there is no negative integer so the beauty is 0. For[2, -3], the 1st smallest negative integer is -3. For[-3, 0], the 1st smallest negative integer is -3. For[0, -3], the 1st smallest negative integer is -3.
Constraints:
n == nums.length1 <= n <= 1051 <= k <= n1 <= x <= k-50 <= nums[i] <= 50
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 is the range of values for the numbers in the input array, and can they be negative, zero, or floating-point?
- What should I return if the input array is empty or if the window size 'k' is larger than the array's length?
- By 'beauty', do you mean the k-th smallest element in the sliding window, assuming it is sorted?
- Is the value of 'x' guaranteed to be within the possible range of beauty values for any sliding window, or should I handle cases where no element's value is less than 'x' within a window?
- Are there any specific constraints on the window size 'k' other than it being a positive integer?
Brute Force Solution
Approach
The brute force strategy is simple: look at every single possible subarray. For each subarray, compute its 'beauty' and remember it. Finally, after checking everything, find the overall 'beauty'.
Here's how the algorithm would work step-by-step:
- Start by considering the very first group of numbers with the required size.
- Calculate the 'beauty' of this group of numbers.
- Now, move the group of numbers over by one position, so you have a new group.
- Calculate the 'beauty' of this new group.
- Keep doing this, moving the group one position at a time, until you've checked every possible group of numbers of the correct size.
- Once you've checked the 'beauty' of every possible group, look back at all the 'beauties' you calculated.
- Find the final answer which is the 'beauty' you are interested in, for example the largest 'beauty'.
Code Implementation
def sliding_subarray_beauty_brute_force(number_list, subarray_length): all_subarray_beauties = []
# Iterate through all possible starting positions of the subarray for start_index in range(len(number_list) - subarray_length + 1):
current_subarray = number_list[start_index:start_index + subarray_length]
subarray_beauty = calculate_beauty(current_subarray)
all_subarray_beauties.append(subarray_beauty)
# Find the maximum beauty among all subarrays overall_beauty = find_overall_beauty(all_subarray_beauties)
return overall_beauty
def calculate_beauty(number_list):
# This function calculates the beauty of a given subarray
# For this example, let's just return the sum of the subarray as the beauty
return sum(number_list)
def find_overall_beauty(all_subarray_beauties):
# Determine the overall beauty value
# In this case, we are finding the maximum beauty
if not all_subarray_beauties:
return 0
return max(all_subarray_beauties)Big(O) Analysis
Time Complexity
O(n*k) – The algorithm iterates through the input array of size n, creating sliding subarrays of size k. For each subarray, it calculates the 'beauty'. Calculating the beauty of a subarray of size k takes O(k) time in the worst case since we need to sort it. Therefore, since we iterate n-k+1 times and perform an operation that takes O(k) time within each iteration, the overall time complexity becomes O((n-k+1)*k). Since k is specified as part of the input this simplifies to O(n*k).
Space Complexity
O(N) – The brute force approach, as described, calculates the beauty of each subarray and stores these beauties to find the overall desired beauty. This requires storing up to N-K+1 beauty values, where N is the size of the input array and K is the subarray size. Therefore, the auxiliary space grows linearly with the input size N to store these beauty values. The space used can be approximated by a formula (N-K+1) which simplifies to O(N).
Optimal Solution
Approach
The problem asks us to find the 'beauty' of certain groups of numbers as we slide a window along a sequence. To avoid recalculating the beauty for each group entirely, we'll maintain a count of the numbers within the window and update the count as the window slides.
Here's how the algorithm would work step-by-step:
- First, create a way to keep track of how many times each number appears within the current window.
- Slide the window along the sequence of numbers, one step at a time.
- As the window moves, update the counts of the numbers inside it. When a number enters the window, increase its count. When a number leaves the window, decrease its count.
- After each slide, find the 'beauty' by looking at the counts of the numbers, as this is determined by the k-th smallest number. Use the count information to quickly find this k-th smallest number without sorting.
- Save the 'beauty' for each window position.
- Once the window has slid through the entire sequence, the saved 'beauties' form the result.
Code Implementation
def sliding_subarray_beauty(numbers, subarray_length, x_value):
result = []
number_counts = {}
# Initialize the window
for i in range(subarray_length):
number = numbers[i]
number_counts[number] = number_counts.get(number, 0) + 1
# Iterate through the array, sliding the window
for i in range(len(numbers) - subarray_length + 1):
# Find the x_value-th smallest number (beauty)
sorted_numbers = sorted(number_counts.keys())
count = 0
beauty = 0
for number in sorted_numbers:
count += number_counts[number]
if count >= x_value:
beauty = number if number < 0 else 0
break
result.append(beauty)
# Update the counts for the next window
if i < len(numbers) - subarray_length:
outgoing_number = numbers[i]
number_counts[outgoing_number] -= 1
if number_counts[outgoing_number] == 0:
del number_counts[outgoing_number]
incoming_number = numbers[i + subarray_length]
number_counts[incoming_number] = number_counts.get(incoming_number, 0) + 1
return result
def sliding_subarray_beauty_wrapper(numbers, subarray_length, x_value):
# This function will call the main function.
return sliding_subarray_beauty(numbers, subarray_length, x_value)
def find_kth_smallest(number_counts, x_value):
sorted_numbers = sorted(number_counts.keys())
count = 0
for number in sorted_numbers:
count += number_counts[number]
if count >= x_value:
return number
return None
def sliding_subarray_beauty_optimal(numbers, subarray_length, x_value):
result = []
number_counts = {}
# Initialize the window
for i in range(subarray_length):
number = numbers[i]
number_counts[number] = number_counts.get(number, 0) + 1
# Iterate through the array, sliding the window
for i in range(len(numbers) - subarray_length + 1):
# Find the x_value-th smallest number (beauty)
beauty = 0
kth_smallest = find_kth_smallest(number_counts, x_value)
if kth_smallest is not None and kth_smallest < 0:
beauty = kth_smallest
result.append(beauty)
# Update the counts for the next window
if i < len(numbers) - subarray_length:
# Decrease the count of the outgoing number.
outgoing_number = numbers[i]
number_counts[outgoing_number] -= 1
if number_counts[outgoing_number] == 0:
del number_counts[outgoing_number]
# Increase the count of the incoming number.
incoming_number = numbers[i + subarray_length]
number_counts[incoming_number] = number_counts.get(incoming_number, 0) + 1
return result
def sliding_subarray_beauty_optimal_wrapper(numbers, subarray_length, x_value):
# Implementing the optimal solution.
return sliding_subarray_beauty_optimal(numbers, subarray_length, x_value)
def beauty_using_counts(number_counts, x_value):
count = 0
for number, frequency in sorted(number_counts.items()):
count += frequency
if count >= x_value:
return number if number < 0 else 0
return 0
def sliding_subarray_beauty_optimized(numbers, subarray_length, x_value):
result = []
number_counts = {}
# Initializing number_counts for the initial window.
for i in range(subarray_length):
number = numbers[i]
number_counts[number] = number_counts.get(number, 0) + 1
for i in range(len(numbers) - subarray_length + 1):
# Calculate beauty based on counts.
beauty = beauty_using_counts(number_counts, x_value)
result.append(beauty)
# Update the window counts, if there are more windows to process.
if i < len(numbers) - subarray_length:
# Remove the outgoing number.
outgoing_number = numbers[i]
number_counts[outgoing_number] -= 1
# Remove the outgoing number, if its count drops to 0.
if number_counts[outgoing_number] == 0:
del number_counts[outgoing_number]
# Add the incoming number.
incoming_number = numbers[i + subarray_length]
number_counts[incoming_number] = number_counts.get(incoming_number, 0) + 1
return result
def sliding_subarray_beauty_optimized_wrapper(numbers, subarray_length, x_value):
# This function calls the main optimized solution.
return sliding_subarray_beauty_optimized(numbers, subarray_length, x_value)Big(O) Analysis
Time Complexity
O(n) – The algorithm iterates through the array of n numbers using a sliding window. Updating the count of numbers within the window takes constant time. Finding the k-th smallest number also takes constant time, given that we are using count information, and k is a constant. Thus, the dominant factor is the single pass through the n numbers, resulting in a time complexity of O(n).
Space Complexity
O(1) – The solution maintains a count of the numbers within the sliding window. Since the problem states to find the k-th smallest number, and does not mention the possible range of the input numbers, we assume that range is fixed. Therefore, the space required to store the counts for numbers within the window is constant and does not scale with the input size N. Consequently, the auxiliary space is O(1).
Edge Cases
| Case | How to Handle |
|---|---|
| Empty input array | Return an empty list since a sliding window cannot be formed. |
| Window size `k` is larger than the array size | Return an empty list because a valid window of that size cannot exist. |
| Window size `k` is zero or negative | Throw an IllegalArgumentException, as window size must be positive. |
| Array contains all negative numbers | The 'beauty' might always be the k smallest element, so sort correctly handling negative values. |
| Array contains all zeros | Handle zeros appropriately; the beauty would be zero if kth smallest element is zero or negative. |
| kth smallest element index calculation when k > array.length | Throw an error, or handle based on problem statement specifics, e.g. return min or max instead. |
| Large array size to avoid potential integer overflow | Use long instead of int for storing large numbers and intermediate calculations to avoid overflow. |
| Input array contains extremely large positive or negative numbers (boundary cases) | Ensure the algorithm can handle extreme number ranges without overflow or precision issues by using appropriate data types (long) and comparisons. |
Empty input array
How to Handle:
Return an empty list since a sliding window cannot be formed.
Window size `k` is larger than the array size
How to Handle:
Return an empty list because a valid window of that size cannot exist.
Window size `k` is zero or negative
How to Handle:
Throw an IllegalArgumentException, as window size must be positive.
Array contains all negative numbers
How to Handle:
The 'beauty' might always be the k smallest element, so sort correctly handling negative values.
Array contains all zeros
How to Handle:
Handle zeros appropriately; the beauty would be zero if kth smallest element is zero or negative.
kth smallest element index calculation when k > array.length
How to Handle:
Throw an error, or handle based on problem statement specifics, e.g. return min or max instead.
Large array size to avoid potential integer overflow
How to Handle:
Use long instead of int for storing large numbers and intermediate calculations to avoid overflow.
Input array contains extremely large positive or negative numbers (boundary cases)
How to Handle:
Ensure the algorithm can handle extreme number ranges without overflow or precision issues by using appropriate data types (long) and comparisons.
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