Find All People With Secret LeetCode solution in JavaScript

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var findAllPeople = function (n, meetings, firstPerson) {
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  const timeToMeeting = mapSortedTimeToMeetings(meetings);
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  const peopleThatCurrentlyHaveSecret = new Set([0, firstPerson]);
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  for (const peopleInMeetings of timeToMeeting.values()) {
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    const personToMeetingsWithPeople =
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      mapPeopleToPeopleTheyAreHavingMeetingsWith(peopleInMeetings);
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    let peopleInMeetingsWithSecret = findPeopleThatHaveTheSecret(
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      peopleInMeetings,
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      peopleThatCurrentlyHaveSecret
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    );
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    // BFS algorithm
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    while (peopleInMeetingsWithSecret.size > 0) {
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      const nextPeopleInMeetingsWithSecret = new Set();
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      for (const attendee of peopleInMeetingsWithSecret) {
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        for (const personInMeetingWithAttendee of personToMeetingsWithPeople[
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          attendee
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        ]) {
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          // only add new people that have the secret otherwise there will be an
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          // infinite loop
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          if (!peopleThatCurrentlyHaveSecret.has(personInMeetingWithAttendee)) {
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            nextPeopleInMeetingsWithSecret.add(personInMeetingWithAttendee);
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            peopleThatCurrentlyHaveSecret.add(personInMeetingWithAttendee);
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          }
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        }
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      }
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      peopleInMeetingsWithSecret = nextPeopleInMeetingsWithSecret;
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    }
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  }
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  return [...peopleThatCurrentlyHaveSecret];
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};
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// groups all the meetings by time
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// keys (time) is sorted in ascending order
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function mapSortedTimeToMeetings(meetings) {
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  meetings.sort((a, b) => a[2] - b[2]);
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  const timeToMeeting = new Map();
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  for (const [person1, person2, time] of meetings) {
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    if (!timeToMeeting.has(time)) {
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      timeToMeeting.set(time, []);
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    }
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    timeToMeeting.get(time).push([person1, person2]);
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  }
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  return timeToMeeting;
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}
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// creates an adjacency list of people and people they are having meetings with
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function mapPeopleToPeopleTheyAreHavingMeetingsWith(peopleInMeetings) {
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  const personToMeetingsWithPeople = {};
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  for (const [person1, person2] of peopleInMeetings) {
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    if (!personToMeetingsWithPeople[person1]) {
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      personToMeetingsWithPeople[person1] = [];
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    }
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    if (!personToMeetingsWithPeople[person2]) {
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      personToMeetingsWithPeople[person2] = [];
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    }
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    personToMeetingsWithPeople[person1].push(person2);
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    personToMeetingsWithPeople[person2].push(person1);
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  }
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  return personToMeetingsWithPeople;
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}
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// finds all the people that are in meetings that have the secret
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// set data structue is used so that people are not duplicated
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function findPeopleThatHaveTheSecret(
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  peopleInMeetings,
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  peopleThatCurrentlyHaveSecret
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) {
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  const peopleInMeetingsWithSecret = new Set();
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  for (const peopleInMeeting of peopleInMeetings) {
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    for (const person of peopleInMeeting) {
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      if (peopleThatCurrentlyHaveSecret.has(person)) {
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        peopleInMeetingsWithSecret.add(person);
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      }
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    }
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  }
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  return peopleInMeetingsWithSecret;
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}