LeetCode-in-Java

3161. Block Placement Queries

Hard

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:

1. 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.
2. 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 ith query of type 2, and false otherwise.

Example 1:

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:

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 * 104
• 2 <= queries[i].length <= 3
• 1 <= queries[i][0] <= 2
• 1 <= x, sz <= min(5 * 104, 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.

Solution

import java.util.Arrays;
import java.util.List;

public class Solution {
    public List<Boolean> getResults(int[][] queries) {
        int m = queries.length;
        int[] pos = new int[m + 1];
        int size = 0;
        pos[size++] = 0;
        int max = 0;
        for (int[] q : queries) {
            max = Math.max(max, q[1]);
            if (q[0] == 1) {
                pos[size++] = q[1];
            }
        }
        Arrays.sort(pos, 0, size);
        max++;
        UnionFind left = new UnionFind(max + 1);
        UnionFind right = new UnionFind(max + 1);
        BIT bit = new BIT(max);
        initializePositions(size, pos, bit, left, right, max);
        return List.of(getBooleans(queries, m, size, left, right, bit));
    }

    private void initializePositions(
            int size, int[] pos, BIT bit, UnionFind left, UnionFind right, int max) {
        for (int i = 1; i < size; i++) {
            int pre = pos[i - 1];
            int cur = pos[i];
            bit.update(cur, cur - pre);
            for (int j = pre + 1; j < cur; j++) {
                left.parent[j] = pre;
                right.parent[j] = cur;
            }
        }
        for (int j = pos[size - 1] + 1; j < max; j++) {
            left.parent[j] = pos[size - 1];
            right.parent[j] = max;
        }
    }

    private Boolean[] getBooleans(
            int[][] queries, int m, int size, UnionFind left, UnionFind right, BIT bit) {
        Boolean[] ans = new Boolean[m - size + 1];
        int index = ans.length - 1;
        for (int i = m - 1; i >= 0; i--) {
            int[] q = queries[i];
            int x = q[1];
            int pre = left.find(x - 1);
            if (q[0] == 1) {
                int next = right.find(x + 1);
                left.parent[x] = pre;
                right.parent[x] = next;
                bit.update(next, next - pre);
            } else {
                int maxGap = Math.max(bit.query(pre), x - pre);
                ans[index--] = maxGap >= q[2];
            }
        }
        return ans;
    }

    private static final class BIT {
        int n;
        int[] tree;

        public BIT(int n) {
            this.n = n;
            tree = new int[n];
        }

        public void update(int i, int v) {
            while (i < n) {
                tree[i] = Math.max(tree[i], v);
                i += i & -i;
            }
        }

        public int query(int i) {
            int result = 0;
            while (i > 0) {
                result = Math.max(result, tree[i]);
                i &= i - 1;
            }
            return result;
        }
    }

    private static final class UnionFind {
        private final int[] parent;

        public UnionFind(int n) {
            parent = new int[n];
            for (int i = 1; i < n; i++) {
                parent[i] = i;
            }
        }

        public int find(int x) {
            if (parent[x] != x) {
                parent[x] = find(parent[x]);
            }
            return parent[x];
        }
    }
}

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