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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:

  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.

Solutions

Solution 1: Binary Indexed Tree + Ordered Set

Thinking

Obstacles are inserted only, and a query asks whether a block of size \(sz\) fits in \([0,x]\). Scanning gaps online is \(O(q^2)\).

Reversed time turns insertions into deletions, which only enlarge gaps. A Fenwick tree can store prefix maxima of gap lengths keyed by the right endpoint.

Load every obstacle plus sentinels, write gaps, then go backward: a query checks the prefix maximum up to \(pre\) and the tail \(x-pre\); a deletion updates the successor gap to \(nxt-pre\). Reverse the collected answers.

Obstacles are only inserted, so we can process the queries offline in reverse and turn "insert an obstacle" into "delete an obstacle". After a deletion the adjacent gap only grows, and a Fenwick tree can maintain prefix maxima.

Put all obstacles into an ordered set, and add sentinels \(0\) and \(m+1\), where \(m\) is the maximum coordinate. For every pair of neighboring obstacles \(x_1, x_2\), update index \(x_2\) in the Fenwick tree with the gap \(x_2 - x_1\).

Then scan the queries from back to front:

  • Type \(2\): find the last obstacle \(pre \le x\). The block can be placed if the maximum gap in \([0, pre]\) or the tail gap \((pre, x]\) is at least \(sz\).
  • Type \(1\): delete obstacle \(x\), and update the gap at its successor \(nxt\) to \(nxt - pre\).

The time complexity is \(O(q \times \log m)\), and the space complexity is \(O(m)\), where \(q\) is the number of queries and \(m\) is the maximum coordinate.

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class BinaryIndexedTree:
    def __init__(self, n: int):
        self.n = n
        self.c = [0] * (n + 1)

    def update(self, x: int, v: int):
        while x <= self.n:
            self.c[x] = max(self.c[x], v)
            x += x & -x

    def query(self, x: int) -> int:
        mx = 0
        while x:
            mx = max(mx, self.c[x])
            x -= x & -x
        return mx


class Solution:
    def getResults(self, queries: List[List[int]]) -> List[bool]:
        m = max(q[1] for q in queries)
        sl = SortedList([0, m + 1])
        for q in queries:
            if q[0] == 1:
                sl.add(q[1])
        tree = BinaryIndexedTree(m + 1)
        for x1, x2 in pairwise(sl):
            tree.update(x2, x2 - x1)
        ans = []
        for q in reversed(queries):
            x = q[1]
            if q[0] == 1:
                i = sl.index(x)
                tree.update(sl[i + 1], sl[i + 1] - sl[i - 1])
                sl.remove(x)
            else:
                i = sl.bisect_right(x)
                pre = sl[i - 1]
                ans.append(tree.query(pre) >= q[2] or x - pre >= q[2])
        return ans[::-1]

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class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
    }

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

    public int query(int x) {
        int mx = 0;
        while (x > 0) {
            mx = Math.max(mx, c[x]);
            x -= x & -x;
        }
        return mx;
    }
}

class Solution {
    public List<Boolean> getResults(int[][] queries) {
        int m = 0;
        for (int[] q : queries) {
            m = Math.max(m, q[1]);
        }
        TreeSet<Integer> ts = new TreeSet<>();
        ts.add(0);
        ts.add(m + 1);
        for (int[] q : queries) {
            if (q[0] == 1) {
                ts.add(q[1]);
            }
        }
        BinaryIndexedTree tree = new BinaryIndexedTree(m + 1);
        int pre = 0;
        for (int x : ts) {
            if (x > 0) {
                tree.update(x, x - pre);
            }
            pre = x;
        }
        List<Boolean> ans = new ArrayList<>();
        for (int i = queries.length - 1; i >= 0; --i) {
            int[] q = queries[i];
            int x = q[1];
            if (q[0] == 1) {
                int nxt = ts.higher(x);
                tree.update(nxt, nxt - ts.lower(x));
                ts.remove(x);
            } else {
                int p = ts.floor(x);
                ans.add(tree.query(p) >= q[2] || x - p >= q[2]);
            }
        }
        Collections.reverse(ans);
        return ans;
    }
}

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class BinaryIndexedTree {
private:
    int n;
    vector<int> c;

public:
    BinaryIndexedTree(int n) {
        this->n = n;
        c.resize(n + 1);
    }

    void update(int x, int v) {
        while (x <= n) {
            c[x] = max(c[x], v);
            x += x & -x;
        }
    }

    int query(int x) {
        int mx = 0;
        while (x > 0) {
            mx = max(mx, c[x]);
            x -= x & -x;
        }
        return mx;
    }
};

class Solution {
public:
    vector<bool> getResults(vector<vector<int>>& queries) {
        int m = 0;
        for (auto& q : queries) {
            m = max(m, q[1]);
        }
        set<int> ts{0, m + 1};
        for (auto& q : queries) {
            if (q[0] == 1) {
                ts.insert(q[1]);
            }
        }
        BinaryIndexedTree tree(m + 1);
        int pre = 0;
        for (int x : ts) {
            if (x) {
                tree.update(x, x - pre);
            }
            pre = x;
        }
        vector<bool> ans;
        for (int i = queries.size() - 1; i >= 0; --i) {
            int x = queries[i][1];
            if (queries[i][0] == 1) {
                auto it = ts.find(x);
                tree.update(*next(it), *next(it) - *prev(it));
                ts.erase(it);
            } else {
                auto it = prev(ts.upper_bound(x));
                ans.push_back(tree.query(*it) >= queries[i][2] || x - *it >= queries[i][2]);
            }
        }
        ranges::reverse(ans);
        return ans;
    }
};

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func getResults(queries [][]int) []bool {
    m := 0
    for _, q := range queries {
        m = max(m, q[1])
    }
    st := redblacktree.New[int, struct{}]()
    st.Put(0, struct{}{})
    st.Put(m+1, struct{}{})
    for _, q := range queries {
        if q[0] == 1 {
            st.Put(q[1], struct{}{})
        }
    }
    tree := newBinaryIndexedTree(m + 1)
    it := st.Iterator()
    it.Next()
    pre := it.Key()
    for it.Next() {
        x := it.Key()
        tree.update(x, x-pre)
        pre = x
    }
    ans := []bool{}
    for i := len(queries) - 1; i >= 0; i-- {
        q := queries[i]
        x := q[1]
        if q[0] == 1 {
            nxt, _ := st.Ceiling(x + 1)
            p, _ := st.Floor(x - 1)
            st.Remove(x)
            tree.update(nxt.Key, nxt.Key-p.Key)
        } else {
            node, _ := st.Floor(x)
            p := node.Key
            ans = append(ans, tree.query(p) >= q[2] || x-p >= q[2])
        }
    }
    slices.Reverse(ans)
    return ans
}

type binaryIndexedTree struct {
    n int
    c []int
}

func newBinaryIndexedTree(n int) *binaryIndexedTree {
    return &binaryIndexedTree{n: n, c: make([]int, n+1)}
}

func (t *binaryIndexedTree) update(x, v int) {
    for x <= t.n {
        t.c[x] = max(t.c[x], v)
        x += x & -x
    }
}

func (t *binaryIndexedTree) query(x int) int {
    mx := 0
    for x > 0 {
        mx = max(mx, t.c[x])
        x -= x & -x
    }
    return mx
}

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