Count Non-Decreasing Subarrays After K Operations

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ArraysSliding Windows

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

For each subarray of nums, you can apply up to k operations on it. In each operation, you increment any element of the subarray by 1.

Note that each subarray is considered independently, meaning changes made to one subarray do not persist to another.

Return the number of subarrays that you can make non-decreasing after performing at most k operations.

An array is said to be non-decreasing if each element is greater than or equal to its previous element, if it exists.

Example 1:

Input: nums = [6,3,1,2,4,4], k = 7

Output: 17

Explanation:

Out of all 21 possible subarrays of nums, only the subarrays [6, 3, 1], [6, 3, 1, 2], [6, 3, 1, 2, 4] and [6, 3, 1, 2, 4, 4] cannot be made non-decreasing after applying up to k = 7 operations. Thus, the number of non-decreasing subarrays is 21 - 4 = 17.

Example 2:

Input: nums = [6,3,1,3,6], k = 4

Output: 12

Explanation:

The subarray [3, 1, 3, 6] along with all subarrays of nums with three or fewer elements, except [6, 3, 1], can be made non-decreasing after k operations. There are 5 subarrays of a single element, 4 subarrays of two elements, and 2 subarrays of three elements except [6, 3, 1], so there are 1 + 5 + 4 + 2 = 12 subarrays that can be made non-decreasing.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
1 <= k <= 10⁹

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 size of the input array, and what is the range of values within the array and for K?
If no non-decreasing subarrays exist after applying K operations, what should the function return?
Can I modify the original input array, or do I need to preserve it?
What constitutes an 'operation'? Can I increment an element beyond the initial range specified?
By 'non-decreasing', do you mean strictly increasing or is it okay to have consecutive elements with the same value?

Brute Force Solution

Approach

The brute force method involves checking every possible subarray to see if, after making at most K changes to its elements, it becomes non-decreasing. We check each subarray individually, making sure we count only the valid ones.

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

Consider every possible subarray within the given sequence of numbers. Think of it as selecting every possible slice of the sequence.
For each of these subarrays, determine the minimum number of changes needed to make it non-decreasing (where each number is greater than or equal to the number before it).
To find the minimum number of changes, consider every number that could be the value that the elements in our subarray could be set to. Set all the elements of our subarray to that specific value and determine how many operations it would take to make all elements of our subarray equal to it. The number of changes is the number of elements in our subarray that do not match the value we are considering setting the elements to.
If the number of changes required is less than or equal to K (the maximum allowed changes), then the subarray is considered a valid non-decreasing subarray.
Keep a running count of all the valid non-decreasing subarrays we find.
After checking all possible subarrays, the final count represents the total number of non-decreasing subarrays achievable with at most K operations.

Code Implementation

def count_non_decreasing_subarrays(numbers, max_operations):
    subarray_count = 0

    for start_index in range(len(numbers)):
        for end_index in range(start_index, len(numbers)):
            subarray = numbers[start_index : end_index + 1]
            changes_needed = float('inf')

            # Find the minimum operations needed for current subarray
            for potential_value in subarray:
                current_changes = 0
                for number in subarray:
                    if number != potential_value:
                        current_changes += 1

                changes_needed = min(changes_needed, current_changes)

            # Count if the operations needed are within the limit
            if changes_needed <= max_operations:
                is_non_decreasing = True
                for index in range(1, len(subarray)):
                    if subarray[index] < subarray[index - 1]:
                        is_non_decreasing = False
                        break

                if is_non_decreasing:
                    subarray_count += 1

    return subarray_count

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm iterates through all possible subarrays, which takes O(n^2) time. For each subarray, it iterates through each element in the subarray to determine the minimum number of changes required to make it non-decreasing. This inner loop has a cost of O(n) since we are considering setting all elements to each possible value within the subarray, which in the worst case could be n different values. Therefore, the overall time complexity is O(n^2) * O(n), which simplifies to O(n^3).

Space Complexity

O(1) – The provided solution iterates through subarrays and calculates the number of changes needed. It does not use any auxiliary data structures that scale with the input size N. The space used is limited to a few variables for indexing and counting, which remain constant regardless of the size of the input array. Therefore, the space complexity is O(1).

Optimal Solution

Approach

The goal is to find how many groups of numbers can become non-decreasing with a limited number of changes. Instead of checking every possible group, we use a sliding window that efficiently checks only the valid groups, avoiding redundant calculations. This sliding window expands or contracts based on the changes needed to ensure the group follows the rules.

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

Start with a window covering the beginning of the list of numbers.
Expand the window to include more numbers until the numbers inside need more changes than we are allowed to become non-decreasing.
If the window needs too many changes, shrink it from the beginning until the changes needed are within our limit.
When the window's changes are within the limit, record the size of the window since this contributes to the number of valid groups.
Slide the window forward by one position and repeat the expansion and contraction steps.
Continue this process until you've gone through all the numbers.
The total number of recorded window sizes represents the total number of non-decreasing groups you can form with the changes allowed.

Code Implementation

def count_non_decreasing_subarrays_after_k_operations(numbers, allowed_changes):
    array_length = len(numbers)
    start_index = 0
    end_index = 0
    changes_needed = 0
    subarray_count = 0

    for end_index in range(array_length):
        # Expand the window until too many changes are needed.
        changes_needed = calculate_changes_needed(numbers, start_index, end_index)

        while changes_needed > allowed_changes:
            # Contract the window from the start to reduce changes.
            start_index += 1
            changes_needed = calculate_changes_needed(numbers, start_index, end_index)

        # Accumulate the size of valid windows.
        subarray_count += (end_index - start_index + 1)

    return subarray_count

def calculate_changes_needed(numbers, start_index, end_index):
    if start_index > end_index:
        return 0

    subarray = numbers[start_index:end_index + 1]
    changes = 0
    if not subarray:
        return 0

    # Find the most frequent element in the subarray.
    most_frequent_element = find_most_frequent_element(subarray)
    
    # The number of elements that are not the most frequent need changing.
    for element in subarray:
        if element != most_frequent_element:
            changes += 1

    return changes

def find_most_frequent_element(subarray):
    element_counts = {}
    for element in subarray:
        element_counts[element] = element_counts.get(element, 0) + 1

    most_frequent_element = None
    max_count = 0

    # Locate most frequent element by its count.
    for element, count in element_counts.items():
        if count > max_count:
            most_frequent_element = element
            max_count = count

    return most_frequent_element

Big(O) Analysis

Time Complexity

O(n) – The algorithm uses a sliding window approach to iterate through the array. The outer loop implicitly advances the window's starting position, while the inner operations of expanding and contracting the window (calculating required changes) ensure that each element is visited at most a constant number of times. Therefore, the overall time complexity is proportional to the number of elements in the array, resulting in O(n) time complexity.

Space Complexity

O(1) – The sliding window approach outlined in the plain English explanation primarily uses a constant number of variables to track the window's start and end positions, as well as potentially the number of changes needed. No auxiliary data structures that scale with the input size N (where N is the number of elements in the list) are mentioned or implied. Therefore, the space complexity remains constant, independent of the input size.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0, as there are no subarrays in an empty array or null array.
Array with only one element	Return 1, since a single element is considered a non-decreasing subarray of length 1.
K is zero (no operations allowed)	The problem simplifies to counting the number of non-decreasing subarrays in the original array.
K is very large (larger than any possible adjustment needed)	The maximum possible non-decreasing subarrays will eventually be selected after sufficient allowed operations.
Input array is already sorted in non-decreasing order	The algorithm should correctly count the existing non-decreasing subarrays without requiring any operations.
All elements in the array are the same	The entire array is a non-decreasing subarray, and any value of K will not change the total possible subarrays; the answer will be n*(n+1)/2.
Integer overflow when calculating subarray counts with large n	Use appropriate data types (e.g., long) to store the counts to prevent overflow.
Array contains negative numbers, zeros, and positive numbers	The comparison logic should correctly handle all number types, including negative and zero values.

Null or empty input array

How to Handle:

Return 0, as there are no subarrays in an empty array or null array.

Array with only one element

How to Handle:

Return 1, since a single element is considered a non-decreasing subarray of length 1.

K is zero (no operations allowed)

How to Handle:

The problem simplifies to counting the number of non-decreasing subarrays in the original array.

K is very large (larger than any possible adjustment needed)

How to Handle:

The maximum possible non-decreasing subarrays will eventually be selected after sufficient allowed operations.

Input array is already sorted in non-decreasing order

How to Handle:

The algorithm should correctly count the existing non-decreasing subarrays without requiring any operations.

All elements in the array are the same

How to Handle:

The entire array is a non-decreasing subarray, and any value of K will not change the total possible subarrays; the answer will be n*(n+1)/2.

Integer overflow when calculating subarray counts with large n

How to Handle:

Use appropriate data types (e.g., long) to store the counts to prevent overflow.

Array contains negative numbers, zeros, and positive numbers

How to Handle:

The comparison logic should correctly handle all number types, including negative and zero values.