In real-world scheduling scenarios, there's often flexibility in meeting times. LeetCode 3440 explores a variation where you're allowed to reschedule one meeting to potentially maximize your continuous free time during a larger event. Unlike its predecessor, the relative order of meetings can now be changed. This introduces an interesting optimization problem that mixes greedy scanning with intelligent rescheduling.
🧠 Problem Summary
You are given:
- An integer
eventTime, representing the full span of the event (fromt = 0tot = eventTime). - Two arrays:
startTime[]andendTime[], where each pair defines the start and end times ofnnon-overlapping meetings.
You can reschedule at most one meeting (by changing its start time but keeping its duration the same), as long as:
- The new time doesn't exceed event bounds.
- Meetings stay non-overlapping.
Goal: Maximize the longest continuous free time after such rescheduling.
🔧 Intuition
The main idea is to:
- Traverse from left to right, tracking the largest gap before each meeting.
- Traverse from right to left, tracking the largest gap after each meeting.
- At each step, determine the maximum free block achievable by shifting a meeting into a known gap.
The meeting durations stay fixed. By checking if we can fit a meeting into a larger previous or subsequent gap, we determine the best possible outcome.
💪 C++ Code
#include <ranges>
class Solution {
public:
int maxFreeTime(int eventTime, vector<int>& startTime, vector<int>& endTime) {
int n = startTime.size();
int max_gap_before = 0;
int last_end = 0;
int max_free_time = 0;
for (int i = 0; i < n; ++i) {
int meeting_time = endTime[i] - startTime[i];
int next_start = (i == n - 1) ? eventTime : startTime[i + 1];
int free_time = next_start - last_end;
if (meeting_time > max_gap_before) free_time -= meeting_time;
max_free_time = max(max_free_time, free_time);
max_gap_before = max(max_gap_before, startTime[i] - last_end);
last_end = endTime[i];
}
int max_gap_after = 0;
int last_start = eventTime;
for (int i = n - 1; i >= 0; --i) {
int meeting_time = endTime[i] - startTime[i];
int prev_end = (i == 0) ? 0 : endTime[i - 1];
int free_time = last_start - prev_end;
if (meeting_time <= max_gap_after)
max_free_time = max(max_free_time, free_time);
max_gap_after = max(max_gap_after, last_start - endTime[i]);
last_start = startTime[i];
}
return max_free_time;
}
};
🐍 Python Code
class Solution:
def maxFreeTime(self, eventTime: int, startTime: List[int], endTime: List[int]) -> int:
n = len(startTime)
max_gap_before, max_free_time, last_end = 0, 0, 0
for i in range(n):
meeting_time = endTime[i] - startTime[i]
next_start = eventTime if i == n - 1 else startTime[i + 1]
free_time = next_start - last_end
if meeting_time > max_gap_before:
free_time -= meeting_time
max_free_time = max(max_free_time, free_time)
max_gap_before = max(max_gap_before, startTime[i] - last_end)
last_end = endTime[i]
max_gap_after, last_start = 0, eventTime
for i in reversed(range(n)):
meeting_time = endTime[i] - startTime[i]
prev_end = 0 if i == 0 else endTime[i - 1]
free_time = last_start - prev_end
if meeting_time <= max_gap_after:
max_free_time = max(max_free_time, free_time)
max_gap_after = max(max_gap_after, last_start - endTime[i])
last_start = startTime[i]
return max_free_time
🔌 JavaScript Code
var maxFreeTime = function(eventTime, startTime, endTime) {
let n = startTime.length;
let maxGapBefore = 0, maxFreeTime = 0, lastEnd = 0;
for (let i = 0; i < n; i++) {
let duration = endTime[i] - startTime[i];
let nextStart = (i === n - 1) ? eventTime : startTime[i + 1];
let freeTime = nextStart - lastEnd;
if (duration > maxGapBefore) freeTime -= duration;
maxFreeTime = Math.max(maxFreeTime, freeTime);
maxGapBefore = Math.max(maxGapBefore, startTime[i] - lastEnd);
lastEnd = endTime[i];
}
let maxGapAfter = 0, lastStart = eventTime;
for (let i = n - 1; i >= 0; i--) {
let duration = endTime[i] - startTime[i];
let prevEnd = (i === 0) ? 0 : endTime[i - 1];
let freeTime = lastStart - prevEnd;
if (duration <= maxGapAfter)
maxFreeTime = Math.max(maxFreeTime, freeTime);
maxGapAfter = Math.max(maxGapAfter, lastStart - endTime[i]);
lastStart = startTime[i];
}
return maxFreeTime;
};
📝 Key Takeaways
- Forward and reverse traversal helps us identify optimal gaps.
- Keep track of the largest gaps before and after each position.
- The ability to break order adds complexity but increases optimization potential.
- Avoid brute force. This solution runs in O(n) with no heap or segment trees.
✅ Final Thoughts
This variant of the rescheduling problem showcases the power of greedy planning and gap analysis. Real-life calendar apps, conference schedulers, or even logistics timelines could benefit from such logic. It's not always about cramming more into the schedule — sometimes it's about creating meaningful free time.
Let this pattern guide your approach in future scheduling problems!
Top comments (3)
Well Explained
Thanks Anna
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