Maximum Score of Non-overlapping Intervals

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Dynamic ProgrammingArrays

You are given a 2D integer array intervals, where intervals[i] = [l_i, r_i, weight_i]. Interval i starts at position l_i and ends at r_i, and has a weight of weight_i. You can choose up to 4 non-overlapping intervals. The score of the chosen intervals is defined as the total sum of their weights.

Return the lexicographically smallest array of at most 4 indices from intervals with maximum score, representing your choice of non-overlapping intervals.

Two intervals are said to be non-overlapping if they do not share any points. In particular, intervals sharing a left or right boundary are considered overlapping.

Example 1:

Input: intervals = [[1,3,2],[4,5,2],[1,5,5],[6,9,3],[6,7,1],[8,9,1]]

Output: [2,3]

Explanation:

You can choose the intervals with indices 2, and 3 with respective weights of 5, and 3.

Example 2:

Input: intervals = [[5,8,1],[6,7,7],[4,7,3],[9,10,6],[7,8,2],[11,14,3],[3,5,5]]

Output: [1,3,5,6]

Explanation:

You can choose the intervals with indices 1, 3, 5, and 6 with respective weights of 7, 6, 3, and 5.

Constraints:

1 <= intevals.length <= 5 * 10⁴
intervals[i].length == 3
intervals[i] = [l_i, r_i, weight_i]
1 <= l_i <= r_i <= 10⁹
1 <= weight_i <= 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 is the format of the intervals? Is it a list of tuples, each representing the start and end time (e.g., `[(start1, end1), (start2, end2), ...]`)? What data types are the start and end times (integers, floats, etc.)?
Are the intervals guaranteed to be sorted by start time, end time, or neither? If not sorted, can I sort them?
Are overlapping intervals allowed in the input? If so, how should I handle them (e.g., should I merge them or treat them as distinct intervals with potential for scoring)?
What are the constraints on the start and end times of the intervals? Can they be negative, zero, or very large? Also, what is the range of possible scores for each interval?
If there are no non-overlapping intervals, or if the input list is empty, what should I return? Should I return 0, or null, or throw an exception?

Brute Force Solution

Approach

The brute force approach to finding the maximum score involves checking every possible combination of intervals. We explore all possible subsets of intervals, checking for overlaps and calculating the score for each valid combination. Eventually we'll check every possible set of intervals.

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

First, consider choosing no intervals at all. Calculate the score, which would be zero in this case.
Next, consider choosing only one interval. Do this for each interval individually and calculate the score.
Now, consider choosing two intervals. Go through every possible pair of intervals. For each pair, check if they overlap. If they don't overlap, calculate the score of this combination.
Repeat this process for choosing three intervals, four intervals, and so on, up to choosing all the intervals. In each step, check for overlaps and calculate the score when intervals are non-overlapping.
As you explore each possible combination of intervals and calculate their scores, remember to only consider non-overlapping intervals. If any combination includes overlaps, you should discard it.
Finally, after evaluating every possible combination of non-overlapping intervals, compare all the calculated scores. The highest score is the answer.

Code Implementation

def maximum_score_of_non_overlapping_intervals(intervals): 
    maximum_score = 0
    number_of_intervals = len(intervals)

    for i in range(1 << number_of_intervals):
        current_intervals = []
        current_score = 0

        for j in range(number_of_intervals):
            if (i >> j) & 1:
                current_intervals.append(intervals[j])

        # Check for overlaps. It's crucial to avoid intervals clashing
        is_valid = True
        for first_interval_index in range(len(current_intervals)): 
            for second_interval_index in range(first_interval_index + 1, len(current_intervals)): 
                start_one, end_one, score_one = current_intervals[first_interval_index]
                start_two, end_two, score_two = current_intervals[second_interval_index]
                if not (end_one <= start_two or end_two <= start_one): 
                    is_valid = False
                    break
            if not is_valid:
                break

        # Only calculate score if set of intervals are non-overlapping
        if is_valid: 
            for interval_start, interval_end, interval_score in current_intervals:
                current_score += interval_score

            maximum_score = max(maximum_score, current_score)

    return maximum_score

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach involves examining every possible subset of the input intervals. Given n intervals, there are 2^n possible subsets (each interval can either be included or excluded). For each subset, we need to check for overlaps among the selected intervals, which takes O(n) time in the worst case. Consequently, the overall time complexity is O(n * 2^n) which is simplified to O(2^n) because the exponential term dominates.

Space Complexity

O(1) – The brute force approach described explores all subsets of intervals, but the plain English explanation doesn't specify any explicit data structures to store these subsets. Although it mentions calculating and comparing scores, it implies these calculations and comparisons happen on the fly. Therefore, we can assume that the space usage is dominated by a few variables to keep track of the current combination's score and the maximum score encountered so far, both of which take constant space, regardless of the number of intervals (N). Hence, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The best strategy for this problem focuses on making choices one step at a time. We'll process the intervals in a specific order and decide at each step whether to include the current interval in our solution, always aiming for the maximum possible score. This avoids having to explore all the different combinations of intervals.

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

First, organize the intervals based on when they finish, from earliest to latest.
Start by considering the first interval in your sorted list.
Decide: Should you include this interval in your solution? If including it improves your score (compared to not including it), then definitely include it.
Move to the next interval. Before deciding to include it, make sure it doesn't overlap with any interval you've already included. If it does overlap, skip it.
Keep repeating this process: If an interval doesn't overlap with any earlier selected intervals, include it if it improves the score.
By carefully choosing the best intervals one by one, you'll eventually find the arrangement that gives you the highest possible score without having to try every combination.

Code Implementation

def maximum_score_of_non_overlapping_intervals(intervals):
    # Sort intervals by end time in ascending order.
    intervals.sort(key=lambda interval: interval[1])
    
    selected_intervals = []
    total_score = 0

    for interval_start, interval_end, interval_score in intervals:
        # Check for overlap with already selected intervals.
        overlap = False
        for selected_start, selected_end, _ in selected_intervals:
            if interval_start < selected_end and interval_end > selected_start:
                overlap = True
                break

        if not overlap:
            # Include the interval if no overlap.
            selected_intervals.append((interval_start, interval_end, interval_score))
            total_score += interval_score

    return total_score

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm first sorts the intervals by their finish times, which takes O(n log n) time. Then, it iterates through the sorted list of n intervals. For each interval, it checks for overlaps with previously selected intervals. In the worst case, checking for overlaps can take O(n) time if we need to compare the current interval with all previously selected intervals. However, because we're maintaining a sorted list implicitly through the selection process (intervals selected will always have earlier finish times), we could use a binary search to find the latest non-overlapping interval in O(log n) time. Since the overlap checking happens inside the loop that iterates through n intervals, the overlap checking contributes O(n log n). Because O(n log n) (sorting) + O(n log n) (overlap checking) simplifies to O(n log n), that is the overall complexity.

Space Complexity

O(1) – The provided approach sorts the intervals in place and iterates through them. It primarily involves comparing intervals and updating a score based on non-overlapping conditions. There are no auxiliary data structures like lists, maps, or recursion impacting space complexity beyond a constant number of variables used for tracking the current interval and the best score seen so far. Thus, the space used remains constant, irrespective of the number of intervals (N).

Edge Cases

Case	How to Handle
Empty input list of intervals	Return 0 as there are no intervals to select a score from.
Intervals with identical start and end times (duration 0)	Treat these intervals as valid and potentially part of an optimal solution.
Intervals sorted by start time are extremely skewed, with one very large interval.	The dynamic programming approach considers all possibilities regardless of input skew.
Intervals with negative start or end times.	The algorithm should correctly handle negative values if the problem allows for them, or error if they are invalid input
Large number of intervals leading to potential stack overflow in recursive approaches	Prefer dynamic programming (iterative) over recursive solutions to avoid stack overflow issues.
Intervals with overlapping or touching endpoints.	Clearly define whether intervals that share endpoints are considered overlapping or not, and handle accordingly in the overlap check.
Integer overflow when calculating scores with large interval values	Use a data type that can accommodate the potential maximum score value, such as long.
All intervals have negative scores.	Return 0, as it's better to select no intervals than any that would decrease the overall score.

Empty input list of intervals

How to Handle:

Return 0 as there are no intervals to select a score from.

Intervals with identical start and end times (duration 0)

How to Handle:

Treat these intervals as valid and potentially part of an optimal solution.

Intervals sorted by start time are extremely skewed, with one very large interval.

How to Handle:

The dynamic programming approach considers all possibilities regardless of input skew.

Intervals with negative start or end times.

How to Handle:

The algorithm should correctly handle negative values if the problem allows for them, or error if they are invalid input

Large number of intervals leading to potential stack overflow in recursive approaches

How to Handle:

Prefer dynamic programming (iterative) over recursive solutions to avoid stack overflow issues.

Intervals with overlapping or touching endpoints.

How to Handle:

Clearly define whether intervals that share endpoints are considered overlapping or not, and handle accordingly in the overlap check.

Integer overflow when calculating scores with large interval values

How to Handle:

Use a data type that can accommodate the potential maximum score value, such as long.

All intervals have negative scores.

How to Handle:

Return 0, as it's better to select no intervals than any that would decrease the overall score.