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2747. Count Zero Request Servers 👍

  • Time: $O(\texttt{sort}(n) + \texttt{sort}(q) + q\log n)$
  • Space: $O(n + q)$
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class Solution {
 public:
  vector<int> countServers(int n, vector<vector<int>>& logs, int x,
                           vector<int>& queries) {
    vector<int> ans(queries.size());
    vector<int> count(n + 1);
    vector<pair<int, int>> queryAndIndexes;

    for (int i = 0; i < queries.size(); ++i)
      queryAndIndexes.emplace_back(queries[i], i);

    sort(
        logs.begin(), logs.end(),
        [](const vector<int>& a, const vector<int>& b) { return a[1] < b[1]; });
    sort(queryAndIndexes.begin(), queryAndIndexes.end());

    int i = 0;
    int j = 0;
    int servers = 0;

    // For each query, we care about logs[i..j].
    for (const auto& [query, index] : queryAndIndexes) {
      for (; j < logs.size() && logs[j][1] <= query; ++j)
        if (++count[logs[j][0]] == 1)
          ++servers;
      for (; i < logs.size() && logs[i][1] < query - x; ++i)
        if (--count[logs[i][0]] == 0)
          --servers;
      ans[index] = n - servers;
    }

    return ans;
  }
};

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class Solution {
  public int[] countServers(int n, int[][] logs, int x, int[] queries) {
    int[] ans = new int[queries.length];
    int[] count = new int[n + 1];
    Pair<Integer, Integer>[] queryAndIndexes = new Pair[queries.length];

    for (int i = 0; i < queries.length; ++i)
      queryAndIndexes[i] = new Pair<>(queries[i], i);

    Arrays.sort(logs, (a, b) -> Integer.compare(a[1], b[1]));
    Arrays.sort(queryAndIndexes, Comparator.comparingInt(Pair::getKey));

    int i = 0;
    int j = 0;
    int servers = 0;

    // For each query, we care about logs[i..j].
    for (Pair<Integer, Integer> queryAndIndex : queryAndIndexes) {
      final int query = queryAndIndex.getKey();
      final int index = queryAndIndex.getValue();
      for (; j < logs.length && logs[j][1] <= query; ++j)
        if (++count[logs[j][0]] == 1)
          ++servers;
      for (; i < logs.length && logs[i][1] < query - x; ++i)
        if (--count[logs[i][0]] == 0)
          --servers;
      ans[index] = n - servers;
    }

    return ans;
  }
}

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class Solution:
  def countServers(self, n: int, logs: List[List[int]], x: int, queries: List[int]) -> List[int]:
    ans = [0] * len(queries)
    count = [0] * (n + 1)
    queryAndIndexes = sorted([(query, i) for i, query in enumerate(queries)])

    logs.sort(key=lambda log: log[1])

    i = 0
    j = 0
    servers = 0

    # For each query, we care about logs[i..j].
    for query, index in queryAndIndexes:
      while j < len(logs) and logs[j][1] <= query:
        count[logs[j][0]] += 1
        if count[logs[j][0]] == 1:
          servers += 1
        j += 1
      while i < len(logs) and logs[i][1] < query - x:
        count[logs[i][0]] -= 1
        if count[logs[i][0]] == 0:
          servers -= 1
        i += 1
      ans[index] = n - servers

    return ans