There is an infinite 2D plane.
\n\nYou are given a positive integer k. You are also given a 2D array queries, which contains the following queries:
queries[i] = [x, y]: Build an obstacle at coordinate (x, y) in the plane. It is guaranteed that there is no obstacle at this coordinate when this query is made.After each query, you need to find the distance of the kth nearest obstacle from the origin.
Return an integer array results where results[i] denotes the kth nearest obstacle after query i, or results[i] == -1 if there are less than k obstacles.
Note that initially there are no obstacles anywhere.
\n\nThe distance of an obstacle at coordinate (x, y) from the origin is given by |x| + |y|.
\n
Example 1:
\n\nInput: queries = [[1,2],[3,4],[2,3],[-3,0]], k = 2
\n\nOutput: [-1,7,5,3]
\n\nExplanation:
\n\nqueries[0], there are less than 2 obstacles.queries[1], there are obstacles at distances 3 and 7.queries[2], there are obstacles at distances 3, 5, and 7.queries[3], there are obstacles at distances 3, 3, 5, and 7.Example 2:
\n\nInput: queries = [[5,5],[4,4],[3,3]], k = 1
\n\nOutput: [10,8,6]
\n\nExplanation:
\n\nqueries[0], there is an obstacle at distance 10.queries[1], there are obstacles at distances 8 and 10.queries[2], there are obstacles at distances 6, 8, and 10.\n
Constraints:
\n\n1 <= queries.length <= 2 * 105queries[i] are unique.-109 <= queries[i][0], queries[i][1] <= 1091 <= k <= 105k obstacles. Can the k + 1th obstacle ever be the answer to any query?",
"Maintain a max heap of size k, thus heap will contain minimum element at the top in that queue.",
"Remove top element and insert new element from input array if current max is larger than this."
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