LeetCode #2054 — MEDIUM

Two Best Non-Overlapping Events

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

The Problem

Problem Statement

You are given a 0-indexed 2D integer array of events where events[i] = [startTime_i, endTime_i, value_i]. The i^th event starts at startTime_iand ends at endTime_i, and if you attend this event, you will receive a value of value_i. You can choose at most two non-overlapping events to attend such that the sum of their values is maximized.

Return this maximum sum.

Note that the start time and end time is inclusive: that is, you cannot attend two events where one of them starts and the other ends at the same time. More specifically, if you attend an event with end time t, the next event must start at or after t + 1.

Example 1:

Input: events = [[1,3,2],[4,5,2],[2,4,3]]
Output: 4
Explanation: Choose the green events, 0 and 1 for a sum of 2 + 2 = 4.

Example 2:

Input: events = [[1,3,2],[4,5,2],[1,5,5]]
Output: 5
Explanation: Choose event 2 for a sum of 5.

Example 3:

Input: events = [[1,5,3],[1,5,1],[6,6,5]]
Output: 8
Explanation: Choose events 0 and 2 for a sum of 3 + 5 = 8.

Constraints:

2 <= events.length <= 10⁵
events[i].length == 3
1 <= startTime_i <= endTime_i <= 10⁹
1 <= value_i <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed 2D integer array of events where events[i] = [startTimei, endTimei, valuei]. The ith event starts at startTimei and ends at endTimei, and if you attend this event, you will receive a value of valuei. You can choose at most two non-overlapping events to attend such that the sum of their values is maximized. Return this maximum sum. Note that the start time and end time is inclusive: that is, you cannot attend two events where one of them starts and the other ends at the same time. More specifically, if you attend an event with end time t, the next event must start at or after t + 1.

Baseline thinking

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

Pattern signal: Array · Binary Search · Dynamic Programming

Example 1

[[1,3,2],[4,5,2],[2,4,3]]

Example 2

[[1,3,2],[4,5,2],[1,5,5]]

Example 3

[[1,5,3],[1,5,1],[6,6,5]]

Core Insight

What unlocks the optimal approach

How can sorting the events on the basis of their start times help? How about end times?
How can we quickly get the maximum score of an interval not intersecting with the interval we chose?

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 #2054: Two Best Non-Overlapping Events
class Solution {
    public int maxTwoEvents(int[][] events) {
        Arrays.sort(events, (a, b) -> a[0] - b[0]);
        int n = events.length;
        int[] f = new int[n + 1];
        for (int i = n - 1; i >= 0; --i) {
            f[i] = Math.max(f[i + 1], events[i][2]);
        }
        int ans = 0;
        for (int[] e : events) {
            int v = e[2];
            int left = 0, right = n;
            while (left < right) {
                int mid = (left + right) >> 1;
                if (events[mid][0] > e[1]) {
                    right = mid;
                } else {
                    left = mid + 1;
                }
            }
            if (left < n) {
                v += f[left];
            }
            ans = Math.max(ans, v);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2054: Two Best Non-Overlapping Events
func maxTwoEvents(events [][]int) int {
	sort.Slice(events, func(i, j int) bool {
		return events[i][0] < events[j][0]
	})
	n := len(events)
	f := make([]int, n+1)
	for i := n - 1; i >= 0; i-- {
		f[i] = max(f[i+1], events[i][2])
	}
	ans := 0
	for _, e := range events {
		v := e[2]
		left, right := 0, n
		for left < right {
			mid := (left + right) >> 1
			if events[mid][0] > e[1] {
				right = mid
			} else {
				left = mid + 1
			}
		}
		if left < n {
			v += f[left]
		}
		ans = max(ans, v)
	}
	return ans
}

# Accepted solution for LeetCode #2054: Two Best Non-Overlapping Events
class Solution:
    def maxTwoEvents(self, events: List[List[int]]) -> int:
        events.sort()
        n = len(events)
        f = [events[-1][2]] * n
        for i in range(n - 2, -1, -1):
            f[i] = max(f[i + 1], events[i][2])
        ans = 0
        for _, e, v in events:
            idx = bisect_right(events, e, key=lambda x: x[0])
            if idx < n:
                v += f[idx]
            ans = max(ans, v)
        return ans

// Accepted solution for LeetCode #2054: Two Best Non-Overlapping Events
impl Solution {
    pub fn max_two_events(mut events: Vec<Vec<i32>>) -> i32 {
        events.sort_by(|a, b| a[0].cmp(&b[0]));

        let n: usize = events.len();
        let mut f: Vec<i32> = vec![0; n + 1];

        for i in (0..n).rev() {
            f[i] = f[i + 1].max(events[i][2]);
        }

        let mut ans: i32 = 0;

        for e in &events {
            let mut v: i32 = e[2];

            let mut left: usize = 0;
            let mut right: usize = n;
            while left < right {
                let mid = (left + right) >> 1;
                if events[mid][0] > e[1] {
                    right = mid;
                } else {
                    left = mid + 1;
                }
            }

            if left < n {
                v += f[left];
            }

            ans = ans.max(v);
        }

        ans
    }
}

// Accepted solution for LeetCode #2054: Two Best Non-Overlapping Events
function maxTwoEvents(events: number[][]): number {
    events.sort((a, b) => a[0] - b[0]);
    const n = events.length;
    const f: number[] = Array(n + 1).fill(0);
    for (let i = n - 1; ~i; --i) {
        f[i] = Math.max(f[i + 1], events[i][2]);
    }
    let ans = 0;
    for (const [_, end, v] of events) {
        let [left, right] = [0, n];
        while (left < right) {
            const mid = (left + right) >> 1;
            if (events[mid][0] > end) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        const t = left < n ? f[left] : 0;
        ans = Math.max(ans, t + v);
    }
    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(n × log n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.