Sweep line (event counting)

O(n log n)

Turn each interval into a start event (+1) and an end event (-1), sort all events by position, then sweep through adding deltas to a running counter to find the busiest moment.

Signals

maximum number of overlapping meetings/intervals at onceminimum rooms/resources neededbusiest moment in timehow many intervals cover a given pointevents happening at the same time

Template

function maxOverlap(intervals) {
    const events = [];
    for (const [start, end] of intervals) {
        events.push([start, 1]);
        events.push([end, -1]);
    }
    events.sort((a, b) => a[0] - b[0] || a[1] - b[1]);
    let active = 0;
    let best = 0;
    for (const [, delta] of events) {
        active += delta;
        best = Math.max(best, active);
    }
    return best;
}

Looks similar, isn't

Intervals: merge & insert: Merge outputs combined, non-overlapping ranges as the final answer. A sweep line does not combine ranges; it counts how many are simultaneously active, which is what you need for 'minimum meeting rooms', not a merged schedule.

n up to 1e5 intervals -> O(n log n): 2n events sorted by position, then one linear sweep.

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