Leetcode #1353: Maximum Number of Events That Can Be Attended

In this guide, we solve Leetcode #1353 Maximum Number of Events That Can Be Attended in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You are given an array of events where events[i] = [startDayi, endDayi]. Every event i starts at startDayi and ends at endDayi.

Quick Facts

Difficulty: Medium
Premium: No
Tags: Greedy, Array, Sorting, Heap (Priority Queue)

Intuition

A locally optimal choice leads to a globally optimal result for this structure.

That means we can commit to decisions as we scan without backtracking.

Approach

Sort or preprocess if needed, then repeatedly take the best available local choice.

Maintain the minimal state necessary to validate the greedy decision.

Steps:

Sort or preprocess as needed.
Iterate and pick the best local option.
Track the current solution.

Example

Input: events = [[1,2],[2,3],[3,4]]
Output: 3
Explanation: You can attend all the three events.
One way to attend them all is as shown.
Attend the first event on day 1.
Attend the second event on day 2.
Attend the third event on day 3.

Python Solution

class Solution:
    def maxEvents(self, events: List[List[int]]) -> int:
        g = defaultdict(list)
        l, r = inf, 0
        for s, e in events:
            g[s].append(e)
            l = min(l, s)
            r = max(r, e)
        pq = []
        ans = 0
        for s in range(l, r + 1):
            while pq and pq[0] < s:
                heappop(pq)
            for e in g[s]:
                heappush(pq, e)
            if pq:
                heappop(pq)
                ans += 1
        return ans

Complexity

The time complexity is $O(M \times \log n)$ , and the space complexity is $O(n)$ , where $M$ is the maximum end time and $n$ is the number of events. The space complexity is $O(n)$ , where $M$ is the maximum end time and $n$ is the number of events.

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Python Solution

class Solution:
    def maxEvents(self, events: List[List[int]]) -> int:
        g = defaultdict(list)
        l, r = inf, 0
        for s, e in events:
            g[s].append(e)
            l = min(l, s)
            r = max(r, e)
        pq = []
        ans = 0
        for s in range(l, r + 1):
            while pq and pq[0] < s:
                heappop(pq)
            for e in g[s]:
                heappush(pq, e)
            if pq:
                heappop(pq)
                ans += 1
        return ans

Leetcode #1353: Maximum Number of Events That Can Be Attended

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #1353: Maximum Number of Events That Can Be Attended

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview