LeetCode #1353 — MEDIUM

Maximum Number of Events That Can Be Attended

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array of events where events[i] = [startDay_i, endDay_i]. Every event i starts at startDay_iand ends at endDay_i.

You can attend an event i at any day d where startDay_i <= d <= endDay_i. You can only attend one event at any time d.

Return the maximum number of events you can attend.

Example 1:

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.

Example 2:

Input: events= [[1,2],[2,3],[3,4],[1,2]]
Output: 4

Constraints:

1 <= events.length <= 10⁵
events[i].length == 2
1 <= startDay_i <= endDay_i <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an array of events where events[i] = [startDayi, endDayi]. Every event i starts at startDayi and ends at endDayi. You can attend an event i at any day d where startDayi <= d <= endDayi. You can only attend one event at any time d. Return the maximum number of events you can attend.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[[1,2],[2,3],[3,4]]

Example 2

[[1,2],[2,3],[3,4],[1,2]]

Core Insight

What unlocks the optimal approach

Sort the events by the start time and in case of tie by the end time in ascending order.
Loop over the sorted events. Attend as much as you can and keep the last day occupied. When you try to attend new event keep in mind the first day you can attend a new event in.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1353: Maximum Number of Events That Can Be Attended
class Solution {
    public int maxEvents(int[][] events) {
        Map<Integer, List<Integer>> g = new HashMap<>();
        int l = Integer.MAX_VALUE, r = 0;
        for (int[] event : events) {
            int s = event[0], e = event[1];
            g.computeIfAbsent(s, k -> new ArrayList<>()).add(e);
            l = Math.min(l, s);
            r = Math.max(r, e);
        }
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        int ans = 0;
        for (int s = l; s <= r; s++) {
            while (!pq.isEmpty() && pq.peek() < s) {
                pq.poll();
            }
            for (int e : g.getOrDefault(s, List.of())) {
                pq.offer(e);
            }
            if (!pq.isEmpty()) {
                pq.poll();
                ans++;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1353: Maximum Number of Events That Can Be Attended
func maxEvents(events [][]int) (ans int) {
	g := map[int][]int{}
	l, r := math.MaxInt32, 0
	for _, event := range events {
		s, e := event[0], event[1]
		g[s] = append(g[s], e)
		l = min(l, s)
		r = max(r, e)
	}

	pq := &hp{}
	heap.Init(pq)
	for s := l; s <= r; s++ {
		for pq.Len() > 0 && pq.IntSlice[0] < s {
			heap.Pop(pq)
		}
		for _, e := range g[s] {
			heap.Push(pq, e)
		}
		if pq.Len() > 0 {
			heap.Pop(pq)
			ans++
		}
	}
	return
}

type hp struct{ sort.IntSlice }

func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	n := len(h.IntSlice)
	v := h.IntSlice[n-1]
	h.IntSlice = h.IntSlice[:n-1]
	return v
}
func (h *hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }

# Accepted solution for LeetCode #1353: Maximum Number of Events That Can Be Attended
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

// Accepted solution for LeetCode #1353: Maximum Number of Events That Can Be Attended
use std::cmp::Reverse;
use std::collections::{BinaryHeap, HashMap};

impl Solution {
    pub fn max_events(events: Vec<Vec<i32>>) -> i32 {
        let mut g: HashMap<i32, Vec<i32>> = HashMap::new();
        let mut l = i32::MAX;
        let mut r = 0;

        for event in events {
            let s = event[0];
            let e = event[1];
            g.entry(s).or_default().push(e);
            l = l.min(s);
            r = r.max(e);
        }

        let mut pq = BinaryHeap::new();
        let mut ans = 0;

        for s in l..=r {
            while let Some(&Reverse(top)) = pq.peek() {
                if top < s {
                    pq.pop();
                } else {
                    break;
                }
            }
            if let Some(ends) = g.get(&s) {
                for &e in ends {
                    pq.push(Reverse(e));
                }
            }
            if pq.pop().is_some() {
                ans += 1;
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #1353: Maximum Number of Events That Can Be Attended
function maxEvents(events: number[][]): number {
    const g: Map<number, number[]> = new Map();
    let l = Infinity,
        r = 0;
    for (const [s, e] of events) {
        if (!g.has(s)) g.set(s, []);
        g.get(s)!.push(e);
        l = Math.min(l, s);
        r = Math.max(r, e);
    }

    const pq = new MinPriorityQueue<number>();
    let ans = 0;
    for (let s = l; s <= r; s++) {
        while (!pq.isEmpty() && pq.front() < s) {
            pq.dequeue();
        }
        for (const e of g.get(s) || []) {
            pq.enqueue(e);
        }
        if (!pq.isEmpty()) {
            pq.dequeue();
            ans++;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(M × log n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.