Alert Using Same Key-Card Three or More Times in a One Hour Period LeetCode solution in Java

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class Solution {
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  public List<String> alertNames(String[] keyName, String[] keyTime) {
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    Map<String, PriorityQueue<Integer>> map = new HashMap<>();
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    // for every entry in keyName and keyTime, add that time to a priorityqueue for that name
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    for (int i = 0; i < keyName.length; i++) {
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      PriorityQueue<Integer> pq = map.getOrDefault(keyName[i], new PriorityQueue<Integer>());
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      // convert the time to an integer (0- 2359 inclusive) for easy comparisons
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      pq.add(
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          Integer.parseInt(keyTime[i].substring(0, 2)) * 100
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              + Integer.parseInt(keyTime[i].substring(3)));
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      map.put(keyName[i], pq);
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    }
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    // Generate the "answer" list
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    List<String> ans = new ArrayList<>();
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    for (String s : map.keySet()) {
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      // For each name in the map, determine if that name used the keycard within 1 hour
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      PriorityQueue<Integer> pq = map.get(s);
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      if (active(pq)) {
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        ans.add(s);
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      }
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    }
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    // Sort the names alphabetically
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    Collections.sort(ans);
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    return ans;
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  }
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  // Greedy function to determine if there were 3 uses within an hour
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  private boolean active(PriorityQueue<Integer> pq) {
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    // If there are two or less entries, the user could not have entered 3 times, return false
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    if (pq.size() < 3) return false;
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    // Create rolling data
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    // Using PriorityQueues, the lowest number is removed first by default
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    int a = pq.poll();
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    int b = pq.poll();
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    int c = pq.poll();
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    // Test if two entrances are within 1 hour (100 in integer)
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    if (c - a <= 100) return true;
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    while (pq.size() > 0) {
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      a = b;
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      b = c;
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      c = pq.poll();
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      if (c - a <= 100) return true;
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    }
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    // If the full Queue has been checked, return false
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    return false;
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  }
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}