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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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struct IndexedQuery {
  int queryIndex;
  int query;
};

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);

    ranges::sort(logs, ranges::less{},
                 [](const vector<int>& log) { return log[1]; });

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

    // For each query, we care about logs[i..j].
    for (const auto& [queryIndex, query] : getIndexedQueries(queries)) {
      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[queryIndex] = n - servers;
    }

    return ans;
  }

 private:
  vector<IndexedQuery> getIndexedQueries(const vector<int>& queries) {
    vector<IndexedQuery> indexedQueries;
    for (int i = 0; i < queries.size(); ++i)
      indexedQueries.push_back({i, queries[i]});
    ranges::sort(indexedQueries,
                 [](const IndexedQuery& a, const IndexedQuery& b) {
      return a.query < b.query;
    });
    return indexedQueries;
  }
};

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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];

    Arrays.sort(logs, (a, b) -> Integer.compare(a[1], b[1]));

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

    // For each query, we care about logs[i..j].
    for (IndexedQuery indexedQuery : getIndexedQueries(queries)) {
      final int queryIndex = indexedQuery.queryIndex;
      final int query = indexedQuery.query;
      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[queryIndex] = n - servers;
    }

    return ans;
  }

  private record IndexedQuery(int queryIndex, int query){};

  private IndexedQuery[] getIndexedQueries(int[] queries) {
    IndexedQuery[] indexedQueries = new IndexedQuery[queries.length];
    for (int i = 0; i < queries.length; ++i)
      indexedQueries[i] = new IndexedQuery(i, queries[i]);
    Arrays.sort(indexedQueries, (a, b) -> Integer.compare(a.query, b.query));
    return indexedQueries;
  }
}

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from dataclasses import dataclass


@dataclass(frozen=True)
class IndexedQuery:
  queryIndex: int
  query: int

  def __iter__(self):
    yield self.queryIndex
    yield self.query


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)

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

    i = 0
    j = 0
    servers = 0

    # For each query, we care about logs[i..j].
    for queryIndex, query in sorted([IndexedQuery(i, query)
                                     for i, query in enumerate(queries)],
                                    key=lambda x: x.query):
      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[queryIndex] = n - servers

    return ans