How do you solve Maximum Number of Events That Can Be Attended II on LeetCode?

Maximum Number of Events That Can Be Attended II (LeetCode #1751) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k * log(n)) time complexity and O(n * k) space complexity. Key insight: Sorting events by their end day allows for efficient overlap checking.

What is the brute force solution for Maximum Number of Events That Can Be Attended II?

The brute force approach for Maximum Number of Events That Can Be Attended II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of events to find the maximum value we can achieve by attending up to k events. This is straightforward but inefficient, especially as the number of events increases. The optimal Optimal Solution approach improves this to O(n * k * log(n)).

What algorithm or data structure is used to solve Maximum Number of Events That Can Be Attended II?

Maximum Number of Events That Can Be Attended II uses the following concepts: Array, Binary Search, Dynamic Programming, Sorting. The recommended patterns to study are: Dynamic Programming, Binary Search, Greedy Algorithms.

What are the interview tips for Maximum Number of Events That Can Be Attended II?

Always clarify the constraints and edge cases before diving into coding. Explain your thought process and approach clearly as you code. Practice writing clean and efficient code, as readability is key in interviews.

What are common mistakes when solving Maximum Number of Events That Can Be Attended II?

Failing to check for overlapping events correctly. Not considering the case where fewer than k events can be attended.

#1751

Maximum Number of Events That Can Be Attended II

Hard

ArrayBinary SearchDynamic ProgrammingSortingDynamic ProgrammingBinary SearchGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n * k * log(n))Space O(n * k)

The optimal approach uses dynamic programming combined with binary search to efficiently calculate the maximum value of attending events. By sorting the events and using a DP table, we can avoid redundant calculations and quickly find the best options.

⚙️

Algorithm

4 steps

1Step 1: Sort the events by their end day.
2Step 2: Initialize a DP table where dp[i][j] represents the maximum value obtainable by attending up to j events from the first i events.
3Step 3: For each event, decide whether to attend it or not. If attending, find the last non-overlapping event using binary search.
4Step 4: Update the DP table based on the choice made.

solution.py24 lines

1def maxValue(events, k):
2    events.sort(key=lambda x: x[1])
3    n = len(events)
4    dp = [[0] * (k + 1) for _ in range(n + 1)]
5
6    for i in range(1, n + 1):
7        for j in range(1, k + 1):
8            # Option 1: Don't attend the current event
9            dp[i][j] = dp[i - 1][j]
10            # Option 2: Attend the current event
11            last_non_overlap = binary_search(events, i - 1, events[i - 1][0])
12            dp[i][j] = max(dp[i][j], dp[last_non_overlap][j - 1] + events[i - 1][2])
13
14    return dp[n][k]
15
16def binary_search(events, right, start):
17    left = 0
18    while left <= right:
19        mid = (left + right) // 2
20        if events[mid][1] < start:
21            left = mid + 1
22        else:
23            right = mid - 1
24    return right

ℹ

Complexity note: The time complexity is O(n * k * log(n)) because we iterate through n events and for each event, we perform a binary search, which takes log(n) time. The space complexity is O(n * k) due to the DP table.

1Sorting events by their end day allows for efficient overlap checking.
2Using dynamic programming helps to build solutions incrementally and avoid redundant calculations.

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