Find Free Meeting Slots

A calendar service stores busy blocks as [start, end) in minutes from midnight. Blocks may overlap and are unsorted.

Given busy, the working day dayStart/dayEnd, and minLength, return all free slots [start, end) within the working day whose length is at least minLength, in chronological order. Busy blocks may extend outside the working day.

Example 1

Input: busy = [[540,600],[570,630],[720,780]], dayStart = 540, dayEnd = 1020, minLength = 30
Output: [[630,720],[780,1020]]
Explanation: 09:00–10:30 busy (merged), lunch 12:00–13:00.

Solution and explanation

Show the solution

Time O(n log n) · Space O(n)

Sort by start, keep cursor = the earliest time not known to be busy (starting at dayStart). For each block, if it starts after the cursor, the gap [cursor, min(start, dayEnd)) is free — keep it if long enough; then move the cursor to max(cursor, end). Finally, the tail [cursor, dayEnd) is free. Clamping to the working day handles blocks that spill outside it.

Reference solution

function freeSlots(busy, dayStart, dayEnd, minLength) {
  const blocks = [...busy].sort((a, b) => a[0] - b[0]);
  const out = [];
  let cursor = dayStart;
  for (const [s, e] of blocks) {
    const gapEnd = Math.min(s, dayEnd);
    if (gapEnd - cursor >= minLength) out.push([cursor, gapEnd]);
    cursor = Math.max(cursor, e);
    if (cursor >= dayEnd) break;
  }
  if (dayEnd - cursor >= minLength) out.push([cursor, dayEnd]);
  return out;
}

function freeSlots(busy: number[][], dayStart: number, dayEnd: number, minLength: number): number[][] {
  const out: number[][] = [];
  let cursor = dayStart;
  for (const [s, e] of [...busy].sort((a, b) => a[0] - b[0])) {
    const gapEnd = Math.min(s, dayEnd);
    if (gapEnd - cursor >= minLength) out.push([cursor, gapEnd]);
    cursor = Math.max(cursor, e);
  }
  if (dayEnd - cursor >= minLength) out.push([cursor, dayEnd]);
  return out;
}

def free_slots(busy: list[list[int]], day_start: int, day_end: int, min_length: int) -> list[list[int]]:
    out: list[list[int]] = []
    cursor = day_start
    for s, e in sorted(busy):
        gap_end = min(s, day_end)
        if gap_end - cursor >= min_length:
            out.append([cursor, gap_end])
        cursor = max(cursor, e)
    if day_end - cursor >= min_length:
        out.append([cursor, day_end])
    return out

<?php

function freeSlots(array $busy, int $dayStart, int $dayEnd, int $minLength): array
{
    usort($busy, fn($a, $b) => $a[0] <=> $b[0]);
    $out = [];
    $cursor = $dayStart;
    foreach ($busy as [$s, $e]) {
        $gapEnd = min($s, $dayEnd);
        if ($gapEnd - $cursor >= $minLength) {
            $out[] = [$cursor, $gapEnd];
        }
        $cursor = max($cursor, $e);
    }
    if ($dayEnd - $cursor >= $minLength) {
        $out[] = [$cursor, $dayEnd];
    }
    return $out;
}

class Solution {
    public int[][] freeSlots(int[][] busy, int dayStart, int dayEnd, int minLength) {
        int[][] blocks = busy.clone();
        Arrays.sort(blocks, (a, b) -> Integer.compare(a[0], b[0]));
        List<int[]> out = new ArrayList<>();
        int cursor = dayStart;
        for (int[] b : blocks) {
            int gapEnd = Math.min(b[0], dayEnd);
            if (gapEnd - cursor >= minLength) out.add(new int[]{cursor, gapEnd});
            cursor = Math.max(cursor, b[1]);
        }
        if (dayEnd - cursor >= minLength) out.add(new int[]{cursor, dayEnd});
        return out.toArray(new int[0][]);
    }
}

Follow-up questions

Find slots free for all attendees (merge everyone's busy blocks first).
Handle time zones and recurring events.

Example 1

Hints

Write and run your solution

Solution and explanation

Reference solution

Follow-up questions

Example 1

Hints

Write and run your solution

Solution and explanation

Reference solution

Follow-up questions

More Intervals problems