CPMagical CP Library

Problem solution · C++

Booking Concert Tickets in Groups

Booking Concert Tickets in Groups: a C++ solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Segment tree or range structure
Source
walkccc LeetCode Solutions
Length
130 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Segment tree or range structure

For Booking Concert Tickets in Groups, the implementation stores interval information in a range-query data structure so updates and queries avoid rescanning the full input.

  1. Choose the aggregate stored for each interval or prefix.
  2. Build or initialize the structure from the input.
  3. Apply updates and combine the affected nodes to answer each query.

Code notes

  • 130 lines of C++ from the credited upstream file 2286.cpp.
  • The implementation visibly relies on sequence storage.
  • 1 loop block detected.

Complexity

Count the build once, then multiply the logarithmic update or query path by the number of operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeBooking Concert Tickets in Groups · C++C++
Use this to learn the idea, then write your own version.
struct SegmentTreeNode {  int lo;  int hi;  std::unique_ptr<SegmentTreeNode> left;  std::unique_ptr<SegmentTreeNode> right;  int mx;  long sum;  SegmentTreeNode(int lo, int hi, std::unique_ptr<SegmentTreeNode>&& left,                  std::unique_ptr<SegmentTreeNode>&& right, int mx, long sum)      : lo(lo),        hi(hi),        left(std::move(left)),        right(std::move(right)),        mx(mx),        sum(sum) {}}; class SegmentTree { public:  explicit SegmentTree(int n, int m) : m(m), root(std::move(build(0, n - 1))) {}   vector<int> maxRange(int k, int maxRow) {    return maxRange(root, k, maxRow);  }   long sumRange(int maxRow) {    return sumRange(root, 0, maxRow);  }   // Substracts k from the seats row.  void substract(int row, int k) {    substract(root, row, k);  }  private:  const int m;  std::unique_ptr<SegmentTreeNode> root;   std::unique_ptr<SegmentTreeNode> build(int l, int r) {    if (l == r)      return make_unique<SegmentTreeNode>(l, r, nullptr, nullptr, m, m);    const int mid = (l + r) / 2;    std::unique_ptr<SegmentTreeNode> left = build(l, mid);    std::unique_ptr<SegmentTreeNode> right = build(mid + 1, r);    return make_unique<SegmentTreeNode>(l, r, std::move(left), std::move(right),                                        max(left->mx, right->mx),                                        left->sum + right->sum);  }   vector<int> maxRange(std::unique_ptr<SegmentTreeNode>& root, int k,                       int maxRow) {    if (root->lo == root->hi) {      if (root->sum < k || root->lo > maxRow)        return {};      return {root->lo, m - static_cast<int>(root->sum)};  // {row, col}    }    // Greedily search the left subtree    if (root->left->mx >= k)      return maxRange(root->left, k, maxRow);    return maxRange(root->right, k, maxRow);  }   long sumRange(std::unique_ptr<SegmentTreeNode>& root, int i, int j) {    if (root->lo == i && root->hi == j)      return root->sum;    const int mid = (root->lo + root->hi) / 2;    if (j <= mid)      return sumRange(root->left, i, j);    if (i > mid)      return sumRange(root->right, i, j);    return sumRange(root->left, i, mid) + sumRange(root->right, mid + 1, j);  }   void substract(std::unique_ptr<SegmentTreeNode>& root, int row, int k) {    if (root == nullptr)      return;    if (root->lo == root->hi && root->hi == row) {      root->mx -= k;      root->sum -= k;      return;    }    const int mid = (root->lo + root->hi) / 2;    if (row <= mid)      substract(root->left, row, k);    else      substract(root->right, row, k);    root->mx = max(root->left->mx, root->right->mx);    root->sum = root->left->sum + root->right->sum;  }}; class BookMyShow { public:  BookMyShow(int n, int m) : tree(n, m), seats(n, m) {}   vector<int> gather(int k, int maxRow) {    const vector<int> res = tree.maxRange(k, maxRow);    if (res.size() == 2) {      const int row = res[0];      tree.substract(row, k);      seats[row] -= k;    }    return res;  }   bool scatter(int k, int maxRow) {    if (tree.sumRange(maxRow) < k)      return false;     while (k > 0)      if (seats[minVacantRow] >= k) {        tree.substract(minVacantRow, k);        seats[minVacantRow] -= k;        k = 0;      } else {        tree.substract(minVacantRow, seats[minVacantRow]);        k -= seats[minVacantRow];        seats[minVacantRow] = 0;        ++minVacantRow;      }     return true;  }  private:  SegmentTree tree;  vector<int> seats;  // the remaining seats at each row  int minVacantRow = 0;};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗