LeetCode #1751 — HARD

Maximum Number of Events That Can Be Attended II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an array of events where events[i] = [startDay_i, endDay_i, value_i]. The i^th event starts at startDay_iand ends at endDay_i, and if you attend this event, you will receive a value of value_i. You are also given an integer k which represents the maximum number of events you can attend.

You can only attend one event at a time. If you choose to attend an event, you must attend the entire event. Note that the end day is inclusive: that is, you cannot attend two events where one of them starts and the other ends on the same day.

Return the maximum sum of values that you can receive by attending events.

Example 1:

Input: events = [[1,2,4],[3,4,3],[2,3,1]], k = 2
Output: 7
Explanation: Choose the green events, 0 and 1 (0-indexed) for a total value of 4 + 3 = 7.

Example 2:

Input: events = [[1,2,4],[3,4,3],[2,3,10]], k = 2
Output: 10
Explanation: Choose event 2 for a total value of 10.
Notice that you cannot attend any other event as they overlap, and that you do not have to attend k events.

Example 3:

Input: events = [[1,1,1],[2,2,2],[3,3,3],[4,4,4]], k = 3
Output: 9
Explanation: Although the events do not overlap, you can only attend 3 events. Pick the highest valued three.

Constraints:

1 <= k <= events.length
1 <= k * events.length <= 10⁶
1 <= startDay_i <= endDay_i <= 10⁹
1 <= value_i <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given an array of events where events[i] = [startDayi, endDayi, valuei]. The ith event starts at startDayi and ends at endDayi, and if you attend this event, you will receive a value of valuei. You are also given an integer k which represents the maximum number of events you can attend. You can only attend one event at a time. If you choose to attend an event, you must attend the entire event. Note that the end day is inclusive: that is, you cannot attend two events where one of them starts and the other ends on the same day. Return the maximum sum of values that you can receive by attending events.

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

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

Example 2

[[1,2,4],[3,4,3],[2,3,10]]
2

Example 3

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

Core Insight

What unlocks the optimal approach

Sort the events by its startTime.
For every event, you can either choose it and consider the next event available, or you can ignore it. You can efficiently find the next event that is available using binary search.

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

Largest constraint values

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 #1751: Maximum Number of Events That Can Be Attended II
class Solution {
    private int[][] events;
    private int[][] f;
    private int n;

    public int maxValue(int[][] events, int k) {
        Arrays.sort(events, (a, b) -> a[0] - b[0]);
        this.events = events;
        n = events.length;
        f = new int[n][k + 1];
        return dfs(0, k);
    }

    private int dfs(int i, int k) {
        if (i >= n || k <= 0) {
            return 0;
        }
        if (f[i][k] != 0) {
            return f[i][k];
        }
        int j = search(events, events[i][1], i + 1);
        int ans = Math.max(dfs(i + 1, k), dfs(j, k - 1) + events[i][2]);
        return f[i][k] = ans;
    }

    private int search(int[][] events, int x, int lo) {
        int l = lo, r = n;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (events[mid][0] > x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

// Accepted solution for LeetCode #1751: Maximum Number of Events That Can Be Attended II
func maxValue(events [][]int, k int) int {
	sort.Slice(events, func(i, j int) bool { return events[i][0] < events[j][0] })
	n := len(events)
	f := make([][]int, n)
	for i := range f {
		f[i] = make([]int, k+1)
	}
	var dfs func(i, k int) int
	dfs = func(i, k int) int {
		if i >= n || k <= 0 {
			return 0
		}
		if f[i][k] > 0 {
			return f[i][k]
		}
		j := sort.Search(n, func(h int) bool { return events[h][0] > events[i][1] })
		ans := max(dfs(i+1, k), dfs(j, k-1)+events[i][2])
		f[i][k] = ans
		return ans
	}
	return dfs(0, k)
}

# Accepted solution for LeetCode #1751: Maximum Number of Events That Can Be Attended II
class Solution:
    def maxValue(self, events: List[List[int]], k: int) -> int:
        @cache
        def dfs(i: int, k: int) -> int:
            if i >= len(events):
                return 0
            _, ed, val = events[i]
            ans = dfs(i + 1, k)
            if k:
                j = bisect_right(events, ed, lo=i + 1, key=lambda x: x[0])
                ans = max(ans, dfs(j, k - 1) + val)
            return ans

        events.sort()
        return dfs(0, k)

// Accepted solution for LeetCode #1751: Maximum Number of Events That Can Be Attended II
impl Solution {
    pub fn max_value(mut events: Vec<Vec<i32>>, k: i32) -> i32 {
        events.sort_by_key(|e| e[0]);
        let n = events.len();
        let mut f = vec![vec![0; (k + 1) as usize]; n];

        fn dfs(i: usize, k: i32, events: &Vec<Vec<i32>>, f: &mut Vec<Vec<i32>>, n: usize) -> i32 {
            if i >= n || k <= 0 {
                return 0;
            }
            if f[i][k as usize] != 0 {
                return f[i][k as usize];
            }
            let j = search(events, events[i][1], i + 1, n);
            let ans = dfs(i + 1, k, events, f, n).max(dfs(j, k - 1, events, f, n) + events[i][2]);
            f[i][k as usize] = ans;
            ans
        }

        fn search(events: &Vec<Vec<i32>>, x: i32, lo: usize, n: usize) -> usize {
            let mut l = lo;
            let mut r = n;
            while l < r {
                let mid = (l + r) / 2;
                if events[mid][0] > x {
                    r = mid;
                } else {
                    l = mid + 1;
                }
            }
            l
        }

        dfs(0, k, &events, &mut f, n)
    }
}

// Accepted solution for LeetCode #1751: Maximum Number of Events That Can Be Attended II
function maxValue(events: number[][], k: number): number {
    events.sort((a, b) => a[0] - b[0]);
    const n = events.length;
    const f: number[][] = Array.from({ length: n }, () => Array(k + 1).fill(0));

    const dfs = (i: number, k: number): number => {
        if (i >= n || k <= 0) {
            return 0;
        }
        if (f[i][k] > 0) {
            return f[i][k];
        }

        const ed = events[i][1],
            val = events[i][2];

        let left = i + 1,
            right = n;
        while (left < right) {
            const mid = (left + right) >> 1;
            if (events[mid][0] > ed) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        const p = left;

        f[i][k] = Math.max(dfs(i + 1, k), dfs(p, k - 1) + val);
        return f[i][k];
    };

    return dfs(0, k);
}

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 + n × k)

Space

O(n × k)

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.