LeetCode #3440 — MEDIUM

Reschedule Meetings for Maximum Free Time II

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

The Problem

Problem Statement

You are given an integer eventTime denoting the duration of an event. You are also given two integer arrays startTime and endTime, each of length n.

These represent the start and end times of n non-overlapping meetings that occur during the event between time t = 0 and time t = eventTime, where the i^th meeting occurs during the time [startTime[i], endTime[i]].

You can reschedule at most one meeting by moving its start time while maintaining the same duration, such that the meetings remain non-overlapping, to maximize the longest continuous period of free time during the event.

Return the maximum amount of free time possible after rearranging the meetings.

Note that the meetings can not be rescheduled to a time outside the event and they should remain non-overlapping.

Note: In this version, it is valid for the relative ordering of the meetings to change after rescheduling one meeting.

Example 1:

Input: eventTime = 5, startTime = [1,3], endTime = [2,5]

Output: 2

Explanation:

Reschedule the meeting at [1, 2] to [2, 3], leaving no meetings during the time [0, 2].

Example 2:

Input: eventTime = 10, startTime = [0,7,9], endTime = [1,8,10]

Output: 7

Explanation:

Reschedule the meeting at [0, 1] to [8, 9], leaving no meetings during the time [0, 7].

Example 3:

Input: eventTime = 10, startTime = [0,3,7,9], endTime = [1,4,8,10]

Output: 6

Explanation:

Reschedule the meeting at [3, 4] to [8, 9], leaving no meetings during the time [1, 7].

Example 4:

Input: eventTime = 5, startTime = [0,1,2,3,4], endTime = [1,2,3,4,5]

Output: 0

Explanation:

There is no time during the event not occupied by meetings.

Constraints:

1 <= eventTime <= 10⁹
n == startTime.length == endTime.length
2 <= n <= 10⁵
0 <= startTime[i] < endTime[i] <= eventTime
endTime[i] <= startTime[i + 1] where i lies in the range [0, n - 2].

Step 01

Brute Force Baseline

Problem summary: You are given an integer eventTime denoting the duration of an event. You are also given two integer arrays startTime and endTime, each of length n. These represent the start and end times of n non-overlapping meetings that occur during the event between time t = 0 and time t = eventTime, where the ith meeting occurs during the time [startTime[i], endTime[i]]. You can reschedule at most one meeting by moving its start time while maintaining the same duration, such that the meetings remain non-overlapping, to maximize the longest continuous period of free time during the event. Return the maximum amount of free time possible after rearranging the meetings. Note that the meetings can not be rescheduled to a time outside the event and they should remain non-overlapping. Note: In this version, it is valid for the relative ordering of the meetings to change after rescheduling one meeting.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

5
[1,3]
[2,5]

Example 2

10
[0,7,9]
[1,8,10]

Example 3

10
[0,3,7,9]
[1,4,8,10]

Step 02

Core Insight

What unlocks the optimal approach

If we reschedule a meeting earlier or later, we need to find a gap of length at least <code>endTime[i] - startTime[i]</code>. Try maintaining the gaps in some sorted data structure.

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 #3440: Reschedule Meetings for Maximum Free Time II
class Solution {
    public int maxFreeTime(int eventTime, int[] startTime, int[] endTime) {
        int n = startTime.length;
        int[] pre = new int[n];
        int[] suf = new int[n];

        pre[0] = startTime[0];
        suf[n - 1] = eventTime - endTime[n - 1];

        for (int i = 1; i < n; i++) {
            pre[i] = Math.max(pre[i - 1], startTime[i] - endTime[i - 1]);
        }

        for (int i = n - 2; i >= 0; i--) {
            suf[i] = Math.max(suf[i + 1], startTime[i + 1] - endTime[i]);
        }

        int ans = 0;
        for (int i = 0; i < n; i++) {
            int l = (i == 0) ? 0 : endTime[i - 1];
            int r = (i == n - 1) ? eventTime : startTime[i + 1];
            int w = endTime[i] - startTime[i];
            ans = Math.max(ans, r - l - w);

            if (i > 0 && pre[i - 1] >= w) {
                ans = Math.max(ans, r - l);
            } else if (i + 1 < n && suf[i + 1] >= w) {
                ans = Math.max(ans, r - l);
            }
        }

        return ans;
    }
}

// Accepted solution for LeetCode #3440: Reschedule Meetings for Maximum Free Time II
func maxFreeTime(eventTime int, startTime []int, endTime []int) int {
	n := len(startTime)
	pre := make([]int, n)
	suf := make([]int, n)

	pre[0] = startTime[0]
	suf[n-1] = eventTime - endTime[n-1]

	for i := 1; i < n; i++ {
		pre[i] = max(pre[i-1], startTime[i]-endTime[i-1])
	}

	for i := n - 2; i >= 0; i-- {
		suf[i] = max(suf[i+1], startTime[i+1]-endTime[i])
	}

	ans := 0
	for i := 0; i < n; i++ {
		l := 0
		if i > 0 {
			l = endTime[i-1]
		}
		r := eventTime
		if i < n-1 {
			r = startTime[i+1]
		}
		w := endTime[i] - startTime[i]
		ans = max(ans, r-l-w)

		if i > 0 && pre[i-1] >= w {
			ans = max(ans, r-l)
		} else if i+1 < n && suf[i+1] >= w {
			ans = max(ans, r-l)
		}
	}

	return ans
}

# Accepted solution for LeetCode #3440: Reschedule Meetings for Maximum Free Time II
class Solution:
    def maxFreeTime(
        self, eventTime: int, startTime: List[int], endTime: List[int]
    ) -> int:
        n = len(startTime)
        pre = [0] * n
        suf = [0] * n
        pre[0] = startTime[0]
        suf[n - 1] = eventTime - endTime[-1]
        for i in range(1, n):
            pre[i] = max(pre[i - 1], startTime[i] - endTime[i - 1])
        for i in range(n - 2, -1, -1):
            suf[i] = max(suf[i + 1], startTime[i + 1] - endTime[i])
        ans = 0
        for i in range(n):
            l = 0 if i == 0 else endTime[i - 1]
            r = eventTime if i == n - 1 else startTime[i + 1]
            w = endTime[i] - startTime[i]
            ans = max(ans, r - l - w)
            if i and pre[i - 1] >= w:
                ans = max(ans, r - l)
            elif i + 1 < n and suf[i + 1] >= w:
                ans = max(ans, r - l)
        return ans

// Accepted solution for LeetCode #3440: Reschedule Meetings for Maximum Free Time II
impl Solution {
    pub fn max_free_time(event_time: i32, start_time: Vec<i32>, end_time: Vec<i32>) -> i32 {
        let n = start_time.len();
        let mut pre = vec![0; n];
        let mut suf = vec![0; n];

        pre[0] = start_time[0];
        suf[n - 1] = event_time - end_time[n - 1];

        for i in 1..n {
            pre[i] = pre[i - 1].max(start_time[i] - end_time[i - 1]);
        }

        for i in (0..n - 1).rev() {
            suf[i] = suf[i + 1].max(start_time[i + 1] - end_time[i]);
        }

        let mut ans = 0;
        for i in 0..n {
            let l = if i == 0 { 0 } else { end_time[i - 1] };
            let r = if i == n - 1 {
                event_time
            } else {
                start_time[i + 1]
            };
            let w = end_time[i] - start_time[i];
            ans = ans.max(r - l - w);

            if i > 0 && pre[i - 1] >= w {
                ans = ans.max(r - l);
            } else if i + 1 < n && suf[i + 1] >= w {
                ans = ans.max(r - l);
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #3440: Reschedule Meetings for Maximum Free Time II
function maxFreeTime(eventTime: number, startTime: number[], endTime: number[]): number {
    const n = startTime.length;
    const pre: number[] = Array(n).fill(0);
    const suf: number[] = Array(n).fill(0);

    pre[0] = startTime[0];
    suf[n - 1] = eventTime - endTime[n - 1];

    for (let i = 1; i < n; i++) {
        pre[i] = Math.max(pre[i - 1], startTime[i] - endTime[i - 1]);
    }

    for (let i = n - 2; i >= 0; i--) {
        suf[i] = Math.max(suf[i + 1], startTime[i + 1] - endTime[i]);
    }

    let ans = 0;
    for (let i = 0; i < n; i++) {
        const l = i === 0 ? 0 : endTime[i - 1];
        const r = i === n - 1 ? eventTime : startTime[i + 1];
        const w = endTime[i] - startTime[i];

        ans = Math.max(ans, r - l - w);

        if (i > 0 && pre[i - 1] >= w) {
            ans = Math.max(ans, r - l);
        } else if (i + 1 < n && suf[i + 1] >= w) {
            ans = Math.max(ans, r - l);
        }
    }

    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)

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.