3161. Block Placement Queries

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3161. Block Placement Queries

Description

There exists an infinite number line, with its origin at 0 and extending towards the positive x-axis.

You are given a 2D array queries, which contains two types of queries:

  • For a query of type 1, queries[i] = [1, x]. Build an obstacle at distance x from the origin. It is guaranteed that there is no obstacle at distance x when the query is asked.
  • For a query of type 2, queries[i] = [2, x, sz]. Check if it is possible to place a block of size sz anywhere in the range [0, x] on the line, such that the block entirely lies in the range [0, x]. A block cannot ** be placed if it intersects with any obstacle, but it may touch it. Note that you do not** actually place the block. Queries are separate.

Return a boolean array results, where results[i] is true if you can place the block specified in the i^th query of type 2, and false otherwise.

Example 1:

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Input: queries = [[1,2],[2,3,3],[2,3,1],[2,2,2]]

Output: [false,true,true]

Explanation:

For query 0, place an obstacle at x = 2. A block of size at most 2 can be placed before x = 3.

Example 2:

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Input: queries = [[1,7],[2,7,6],[1,2],[2,7,5],[2,7,6]]

Output: [true,true,false]

Explanation:

  • Place an obstacle at x = 7 for query 0. A block of size at most 7 can be placed before x = 7.
  • Place an obstacle at x = 2 for query 2. Now, a block of size at most 5 can be placed before x = 7, and a block of size at most 2 before x = 2.

Constraints:

  • 1 <= queries.length <= 15 * 10^4
  • 2 <= queries[i].length <= 3
  • 1 <= queries[i][0] <= 2
  • 1 <= x, sz <= min(5 * 10^4, 3 * queries.length)
  • The input is generated such that for queries of type 1, no obstacle exists at distance x when the query is asked.
  • The input is generated such that there is at least one query of type 2.

Hints/Notes

  • Biweekly Contest 131
  • segment tree

Solution

Language: C++

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class Solution {
public:
    vector<int> mx;

    vector<bool> getResults(vector<vector<int>>& queries) {
        int boundary = 0;
        for (auto query : queries) {
            boundary = max(boundary, query[1]);
        }
        boundary++;
        mx = vector<int>(boundary * 4, 0);
        set<int> obstacles{0, boundary};
        vector<bool> res;
        for (auto q : queries) {
            int x = q[1];
            auto it = obstacles.lower_bound(x);
            int pre = *prev(it); // x 左侧最近障碍物的位置
            if (q[0] == 1) {
                int nxt = *it;
                obstacles.insert(x);
                update(1, 0, boundary, x, x - pre);
                update(1, 0, boundary, nxt, nxt - x);
            } else {
                int max_gap = max(query(1, 0, boundary, pre), x - pre);
                res.push_back(max_gap >= q[2]);
            }
        }
        return res;
    }

    void update(int o, int l, int r, int i, int val) {
        if (l == r) {
            mx[o] = val;
            return;
        }
        int m = (r - l) / 2 + l;
        if (i <= m) {
            update(o * 2, l, m, i, val);
        } else {
            update(o * 2 + 1, m + 1, r, i, val);
        }
        mx[o] = max(mx[o * 2], mx[o * 2 + 1]);
    }

    int query(int o, int l, int r, int R) {
        if (r <= R) {
            return mx[o];
        }
        int m = (r - l) / 2 + l;
        if (R <= m) {
            return query(o * 2, l, m, R);
        }
        return max(mx[o * 2], query(o * 2 + 1, m + 1, r, R));
    }
};