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3275. K-th Nearest Obstacle Queries 👍

  • Time: $O(q\log k)$
  • Space: $O(q)$
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class Solution {
 public:
  vector<int> resultsArray(vector<vector<int>>& queries, int k) {
    vector<int> ans;
    priority_queue<int> maxHeap;

    for (const vector<int>& query : queries) {
      const int x = query[0];
      const int y = query[1];
      maxHeap.push(abs(x) + abs(y));
      if (maxHeap.size() > k)
        maxHeap.pop();
      ans.push_back(maxHeap.size() == k ? maxHeap.top() : -1);
    }

    return ans;
  }
};

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class Solution {
  public int[] resultsArray(int[][] queries, int k) {
    int[] ans = new int[queries.length];
    Queue<Integer> maxHeap = new PriorityQueue<>(Comparator.reverseOrder());

    for (int i = 0; i < queries.length; ++i) {
      final int x = queries[i][0];
      final int y = queries[i][1];
      maxHeap.offer(Math.abs(x) + Math.abs(y));
      if (maxHeap.size() > k)
        maxHeap.poll();
      ans[i] = maxHeap.size() == k ? maxHeap.peek() : -1;
    }

    return ans;
  }
}

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class Solution:
  def resultsArray(self, queries: list[list[int]], k: int) -> list[int]:
    ans = []
    maxHeap = []

    for x, y in queries:
      heapq.heappush(maxHeap, -(abs(x) + abs(y)))
      if len(maxHeap) > k:
        heapq.heappop(maxHeap)
      ans.append(-maxHeap[0] if len(maxHeap) == k else -1)

    return ans