Maximize Subarrays After Removing One Conflicting Pair

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Arrays

You are given an integer n which represents an array nums containing the numbers from 1 to n in order. Additionally, you are given a 2D array conflictingPairs, where conflictingPairs[i] = [a, b] indicates that a and b form a conflicting pair.

Remove exactly one element from conflictingPairs. Afterward, count the number of non-empty subarrays of nums which do not contain both a and b for any remaining conflicting pair [a, b].

Return the maximum number of subarrays possible after removing exactly one conflicting pair.

Example 1:

Input: n = 4, conflictingPairs = [[2,3],[1,4]]

Output: 9

Explanation:

Remove [2, 3] from conflictingPairs. Now, conflictingPairs = [[1, 4]].
There are 9 subarrays in nums where [1, 4] do not appear together. They are [1], [2], [3], [4], [1, 2], [2, 3], [3, 4], [1, 2, 3] and [2, 3, 4].
The maximum number of subarrays we can achieve after removing one element from conflictingPairs is 9.

Example 2:

Input: n = 5, conflictingPairs = [[1,2],[2,5],[3,5]]

Output: 12

Explanation:

Remove [1, 2] from conflictingPairs. Now, conflictingPairs = [[2, 5], [3, 5]].
There are 12 subarrays in nums where [2, 5] and [3, 5] do not appear together.
The maximum number of subarrays we can achieve after removing one element from conflictingPairs is 12.

Constraints:

2 <= n <= 10⁵
1 <= conflictingPairs.length <= 2 * n
conflictingPairs[i].length == 2
1 <= conflictingPairs[i][j] <= n
conflictingPairs[i][0] != conflictingPairs[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 size of the `nums` array and the range of values within `nums` and `k`? Are negative numbers allowed?
If it's not possible to achieve more subarrays summing to `k` after removing any pair, what should the function return? Should it return the number of subarrays we can achieve without removing any elements?
If multiple pairs result in the same maximum number of subarrays, is any one of those pairs acceptable?
Can the array `nums` be empty, or can `k` be zero?
Are duplicate values allowed in the `nums` array, and if so, how should they be handled when considering pairs to remove?

Brute Force Solution

Approach

The brute force way to solve this problem is to try every single possibility. We will examine all possible removals of one conflicting pair, and for each removal, we will count how many nice groups we have left.

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

First, make a list of all the pairs that cause problems.
Next, one at a time, remove each of those problematic pairs from the list.
After removing a pair, figure out how many groups are now 'nice' or non-conflicting.
Write down this 'niceness' number for the removal you just tried.
Do this for every problematic pair from the first list.
Finally, look at all the 'niceness' numbers you wrote down and pick the biggest one. That tells you which removal results in the most 'nice' groups.

Code Implementation

def maximize_subarrays_brute_force(array_of_numbers):
    conflicting_pairs = []
    for first_index in range(len(array_of_numbers) - 1):
        if array_of_numbers[first_index] >= array_of_numbers[first_index + 1]:
            conflicting_pairs.append((first_index, first_index + 1))

    max_nice_subarrays = 0
    for conflicting_pair in conflicting_pairs:
        # Create a copy to not modify the original
        temp_array = array_of_numbers[:]

        # Remove conflicting pair indices from the copy
        index_one, index_two = conflicting_pair
        del temp_array[max(index_one, index_two)]
        del temp_array[min(index_one, index_two)]

        nice_subarrays_count = 1
        if not temp_array:
            max_nice_subarrays = max(max_nice_subarrays, nice_subarrays_count)
            continue

        # Count subarrays after removal
        for index in range(len(temp_array) - 1):
            if temp_array[index] < temp_array[index + 1]:
                nice_subarrays_count += 1

        # Record maximum
        max_nice_subarrays = max(max_nice_subarrays, nice_subarrays_count)

    # If no conflicting pairs exist, return original count
    if not conflicting_pairs:
        nice_subarrays_count = 1
        for index in range(len(array_of_numbers) - 1):
            if array_of_numbers[index] < array_of_numbers[index + 1]:
                nice_subarrays_count += 1
        return nice_subarrays_count

    return max_nice_subarrays

Big(O) Analysis

Time Complexity

O(n^3) – First, finding all conflicting pairs involves iterating through the array and comparing each element with every other, resulting in O(n^2) operations. Then, for each conflicting pair (in the worst case, proportional to n^2), we remove it and re-evaluate the number of nice subarrays. Checking the number of nice subarrays after each removal takes O(n) time because we need to iterate through the array once. Since the re-evaluation step is nested inside the loop that considers each conflicting pair, we have an O(n^2) operation within which an O(n) operation is performed; therefore, it would be O(n^2) * O(n) = O(n^3).

Space Complexity

O(N^2) – The solution first identifies all conflicting pairs, which in the worst case, could involve every pair of elements in the input. This results in storing a list of conflicting pairs with a maximum size proportional to N^2, where N is the number of elements in the input. The algorithm then iterates through this list, removing each pair one at a time. The space used to store the list of conflicting pairs dominates the space complexity. Thus, the auxiliary space complexity is O(N^2).

Optimal Solution

Approach

The core idea is to find the pair of conflicting activities that cause the fewest number of subarrays when removed. We don't actually remove anything; instead, we count the subarrays if we *pretend* we've removed a pair, and keep track of the best outcome.

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

First, count the number of subarrays there are in the original schedule without removing any activities.
Now, consider each possible conflicting pair of activities, one at a time.
For each conflicting pair, temporarily ignore those two activities and count how many subarrays would exist in the modified schedule.
Keep track of the largest number of subarrays we find after 'removing' each pair.
After going through all possible conflicting pairs, the largest number of subarrays we found is the answer.

Code Implementation

def maximize_subarrays(activities):
    number_of_activities = len(activities)
    maximum_subarrays = 0

    # Calculate initial number of subarrays
    initial_subarrays = count_subarrays(activities)
    maximum_subarrays = initial_subarrays

    for i in range(number_of_activities):
        for j in range(i + 1, number_of_activities):
            # Identify potential conflicting pairs
            if activities[i][1] > activities[j][0] and activities[j][1] > activities[i][0]:

                # Create a temporary schedule excluding the pair
                temp_activities = []
                for k in range(number_of_activities):
                    if k != i and k != j:
                        temp_activities.append(activities[k])

                # Count subarrays without the conflicting pair
                number_of_subarrays = count_subarrays(temp_activities)
                maximum_subarrays = max(maximum_subarrays, number_of_subarrays)

    return maximum_subarrays

def count_subarrays(activities):
    if not activities:
        return 0

    activities.sort(key=lambda x: x[0])
    number_of_subarrays = 1
    last_end_time = activities[0][1]

    # Count subarrays based on non-overlapping intervals
    for i in range(1, len(activities)):
        if activities[i][0] >= last_end_time:
            # If start time is greater than last end time, increment
            number_of_subarrays += 1
            last_end_time = activities[i][1]
        else:
            last_end_time = min(last_end_time, activities[i][1])

    return number_of_subarrays

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm first counts the initial number of subarrays, which takes O(n) time. Then, it iterates through all possible pairs of conflicting activities, which involves a nested loop. For each pair, it iterates through the schedule again to count the number of subarrays assuming the pair is removed. Thus, checking each conflicting pair takes O(n) time, and since there are O(n^2) conflicting pairs to check, the overall time complexity becomes O(n^2) * O(n) = O(n^3).

Space Complexity

O(1) – The algorithm iterates through conflicting pairs but doesn't create any additional data structures that scale with the input size (N, representing the number of activities). It only stores a few variables such as the maximum number of subarrays found so far and potentially indices of the best conflicting pair. Therefore, the auxiliary space used remains constant regardless of the input size.

Edge Cases

Case	How to Handle
Empty input array	Return 0 as no subarrays can be formed.
Array with one element	Return 0 as no pair can be removed.
Array where no subarrays with sum k can be formed initially or after removing any pair.	The algorithm should return 0 in this case.
Integer overflow when calculating subarray sums.	Use long data type to store sums to prevent overflow.
Array with all elements equal to k, but length not a multiple of k.	Removing any pair should result in more k-sum subarrays.
Array with negative numbers	The algorithm should correctly handle negative numbers when calculating subarray sums.
Large array size leading to potential memory issues with storing sums or indices	Optimize memory usage by only storing necessary information and avoiding unnecessary data structures.
Cases where removing different pairs leads to the same maximum number of subarrays	The algorithm is designed to find a maximum, not all maximums, and can simply return one.

Empty input array

How to Handle:

Return 0 as no subarrays can be formed.

Array with one element

How to Handle:

Return 0 as no pair can be removed.

Array where no subarrays with sum k can be formed initially or after removing any pair.

How to Handle:

The algorithm should return 0 in this case.

Integer overflow when calculating subarray sums.

How to Handle:

Use long data type to store sums to prevent overflow.

Array with all elements equal to k, but length not a multiple of k.

How to Handle:

Removing any pair should result in more k-sum subarrays.

Array with negative numbers

How to Handle:

The algorithm should correctly handle negative numbers when calculating subarray sums.

Large array size leading to potential memory issues with storing sums or indices

How to Handle:

Optimize memory usage by only storing necessary information and avoiding unnecessary data structures.

Cases where removing different pairs leads to the same maximum number of subarrays

How to Handle:

The algorithm is designed to find *a* maximum, not *all* maximums, and can simply return one.