How do you solve Maximum Score of Non-overlapping Intervals on LeetCode?

Maximum Score of Non-overlapping Intervals (LeetCode #3414) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting intervals by their end times helps in efficiently finding non-overlapping intervals.

What is the time complexity of Maximum Score of Non-overlapping Intervals?

The optimal solution for Maximum Score of Non-overlapping Intervals runs in O(n log n) time and uses O(n) space. The complexity is O(n log n) due to the sorting step, and O(n) for the dynamic programming array and previous index tracking.

What is the brute force solution for Maximum Score of Non-overlapping Intervals?

The brute force approach for Maximum Score of Non-overlapping Intervals is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible combinations of intervals to find the maximum score. This involves checking every possible selection of up to 4 intervals to see which combination yields the highest weight sum. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Score of Non-overlapping Intervals?

Maximum Score of Non-overlapping Intervals uses the following concepts: Array, Binary Search, Dynamic Programming, Sorting. The recommended patterns to study are: Dynamic Programming, Sorting.

What are the interview tips for Maximum Score of Non-overlapping Intervals?

Always clarify the problem constraints and requirements before diving into coding. Think about edge cases, such as intervals that just touch each other. Discuss your thought process clearly while coding, as it helps interviewers understand your approach.

What are common mistakes when solving Maximum Score of Non-overlapping Intervals?

Failing to check for overlaps correctly, leading to incorrect combinations. Not sorting the intervals before processing, which can lead to missed optimal solutions.

#3414

Maximum Score of Non-overlapping Intervals

Hard

ArrayBinary SearchDynamic ProgrammingSortingDynamic ProgrammingSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

By sorting the intervals by their end times and using dynamic programming, we can efficiently find the maximum score of non-overlapping intervals. This approach allows us to build solutions incrementally and avoid unnecessary checks.

⚙️

Algorithm

4 steps

1Step 1: Sort the intervals based on their end times.
2Step 2: Use a dynamic programming array to store the maximum score achievable up to each interval.
3Step 3: For each interval, check previous intervals to find the last non-overlapping interval and update the DP array accordingly.
4Step 4: Track the indices of the intervals chosen to reconstruct the solution.

solution.py19 lines

1def maxScoreIntervals(intervals):
2    intervals.sort(key=lambda x: x[1])
3    n = len(intervals)
4    dp = [0] * n
5    prev = [-1] * n
6    for i in range(n):
7        dp[i] = intervals[i][2]
8        for j in range(i):
9            if intervals[j][1] <= intervals[i][0]:
10                if dp[j] + intervals[i][2] > dp[i]:
11                    dp[i] = dp[j] + intervals[i][2]
12                    prev[i] = j
13    max_score = max(dp)
14    index = dp.index(max_score)
15    result = []
16    while index != -1:
17        result.append(intervals[index][0])
18        index = prev[index]
19    return sorted(result)

ℹ

Complexity note: The complexity is O(n log n) due to the sorting step, and O(n) for the dynamic programming array and previous index tracking.

1Sorting intervals by their end times helps in efficiently finding non-overlapping intervals.
2Dynamic programming allows us to build solutions incrementally, avoiding redundant checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.