2286. Booking Concert Tickets in Groups

Problem

A concert hall has n rows numbered from 0 to n - 1, each with m seats, numbered from 0 to m - 1. You need to design a ticketing system that can allocate seats in the following cases:

• If a group of k spectators can sit together in a row.

• If every member of a group of k spectators can get a seat. They may or may not sit together.

Note that the spectators are very picky. Hence:

• They will book seats only if each member of their group can get a seat with row number less than or equal to maxRow. maxRow can vary from group to group.

• In case there are multiple rows to choose from, the row with the smallest number is chosen. If there are multiple seats to choose in the same row, the seat with the smallest number is chosen.

Implement the BookMyShow class:

• BookMyShow(int n, int m) Initializes the object with n as number of rows and m as number of seats per row.

• int[] gather(int k, int maxRow) Returns an array of length 2 denoting the row and seat number (respectively) of the first seat being allocated to the k members of the group, who must sit together. In other words, it returns the smallest possible r and c such that all [c, c + k - 1] seats are valid and empty in row r, and r <= maxRow. Returns [] in case it is not possible to allocate seats to the group.

• boolean scatter(int k, int maxRow) Returns true if all k members of the group can be allocated seats in rows 0 to maxRow, who may or may not sit together. If the seats can be allocated, it allocates k seats to the group with the smallest row numbers, and the smallest possible seat numbers in each row. Otherwise, returns false.

  Example 1:

Input
["BookMyShow", "gather", "gather", "scatter", "scatter"]
[[2, 5], [4, 0], [2, 0], [5, 1], [5, 1]]
Output
[null, [0, 0], [], true, false]

Explanation
BookMyShow bms = new BookMyShow(2, 5); // There are 2 rows with 5 seats each 
bms.gather(4, 0); // return [0, 0]
                  // The group books seats [0, 3] of row 0. 
bms.gather(2, 0); // return []
                  // There is only 1 seat left in row 0,
                  // so it is not possible to book 2 consecutive seats. 
bms.scatter(5, 1); // return True
                   // The group books seat 4 of row 0 and seats [0, 3] of row 1. 
bms.scatter(5, 1); // return False
                   // There is only one seat left in the hall.

  Constraints:

• 1 <= n <= 5 * 10^4

• 1 <= m, k <= 10^9

• 0 <= maxRow <= n - 1

• At most 5 * 10^4 calls in total will be made to gather and scatter.

Solution

class BookMyShow {
    private final int n;
    private final int m;
    // max number of seats in a row for some segment of the rows
    private final int[] max;
    // total number of seats for some segment of the rows
    private final long[] total;
    // number of rows with zero free places on the left and on the right
    // using this to quickly skip already zero rows
    // actual nodes are placed in [1,this.n], the first and last element only shows there the first
    // non-zero row
    private final int[] numZerosRight;
    private final int[] numZerosLeft;

    public BookMyShow(int n, int m) {
        // make n to be a power of 2 (for simplicity)
        this.n = nextPow2(n);
        this.m = m;
        // segment tree for max number of seats in a row
        this.max = new int[this.n * 2 - 1];
        // total number of seats for a segment of the rows
        this.total = new long[this.n * 2 - 1];
        this.numZerosRight = new int[this.n + 2];
        this.numZerosLeft = new int[this.n + 2];
        // initialize max and total, for max we firstly set values to m
        // segments of size 1 are placed starting from this.n - 1
        Arrays.fill(max, this.n - 1, this.n + n - 1, m);
        Arrays.fill(total, this.n - 1, this.n + n - 1, m);
        // calculate values of max and total for segments based on values of their children
        for (int i = this.n - 2, i1 = i * 2 + 1, i2 = i * 2 + 2; i >= 0; i--, i1 -= 2, i2 -= 2) {
            max[i] = Math.max(max[i1], max[i2]);
            total[i] = total[i1] + total[i2];
        }
    }

    public int[] gather(int k, int maxRow) {
        // find most left row with enough free places
        int mostLeft = mostLeft(0, 0, n, k, maxRow + 1);
        if (mostLeft == -1) {
            return new int[0];
        }
        // get corresponding segment tree node
        int v = n - 1 + mostLeft;
        int[] ans = {mostLeft, m - max[v]};
        // update max and total for this node
        max[v] -= k;
        total[v] -= k;
        // until this is a root of segment tree we update its parent
        while (v != 0) {
            v = (v - 1) / 2;
            max[v] = Math.max(max[v * 2 + 1], max[v * 2 + 2]);
            total[v] = total[v * 2 + 1] + total[v * 2 + 2];
        }
        return ans;
    }

    private int mostLeft(int v, int l, int r, int k, int qr) {
        if (l >= qr || max[v] < k) {
            return -1;
        }
        if (l == r - 1) {
            return l;
        }
        int mid = (l + r) / 2;
        int left = mostLeft(v * 2 + 1, l, mid, k, qr);
        if (left != -1) {
            return left;
        }
        return mostLeft(v * 2 + 2, mid, r, k, qr);
    }

    public boolean scatter(int k, int maxRow) {
        // find total number of free places in the rows [0; maxRow+1)
        long sum = total(0, 0, n, maxRow + 1);
        if (sum < k) {
            return false;
        }
        int i = 0;
        // to don't update parent for both of its children we use a queue
        Deque<Integer> deque = new ArrayDeque<>();
        while (k != 0) {
            i = i + numZerosRight[i] + 1;
            int v = n - 1 + i - 1;
            int spent = Math.min(k, max[v]);
            k -= spent;
            max[v] -= spent;
            total[v] -= spent;
            if (max[v] == 0) {
                // update numZerosRight and numZerosLeft
                numZerosRight[i - numZerosLeft[i] - 1] += numZerosRight[i] + 1;
                numZerosLeft[i + numZerosRight[i] + 1] += numZerosLeft[i] + 1;
            }
            if (v != 0) {
                v = (v - 1) / 2;
                // if we already have the parent node in the queue we don't need to update it
                if (deque.isEmpty() || deque.peekLast() != v) {
                    deque.addLast(v);
                }
            }
        }
        // update max and total
        while (!deque.isEmpty()) {
            int v = deque.pollFirst();
            max[v] = Math.max(max[v * 2 + 1], max[v * 2 + 2]);
            total[v] = total[v * 2 + 1] + total[v * 2 + 2];
            if (v != 0) {
                v = (v - 1) / 2;
                // if we already have the parent node in the queue we don't need to update it
                if (deque.isEmpty() || deque.peekLast() != v) {
                    deque.addLast(v);
                }
            }
        }
        return true;
    }

    // find sum of [ql, qr)
    private long total(int v, int l, int r, int qr) {
        if (l >= qr) {
            return 0;
        }
        if (r <= qr) {
            return total[v];
        }
        int mid = (l + r) / 2;
        return total(v * 2 + 1, l, mid, qr) + total(v * 2 + 2, mid, r, qr);
    }

    private static int nextPow2(int n) {
        if ((n & (n - 1)) == 0) {
            return n;
        }
        return Integer.highestOneBit(n) << 1;
    }
}

/**
 * Your BookMyShow object will be instantiated and called as such:
 * BookMyShow obj = new BookMyShow(n, m);
 * int[] param_1 = obj.gather(k,maxRow);
 * boolean param_2 = obj.scatter(k,maxRow);
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).