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Problem solution · C++

Maximum Number of Events That Can Be Attended II

Maximum Number of Events That Can Be Attended II: a C++ solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Binary search
Source
walkccc LeetCode Solutions
Length
29 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

Binary search

For Maximum Number of Events That Can Be Attended II, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 29 lines of C++ from the credited upstream file 1751.cpp.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

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 codeMaximum Number of Events That Can Be Attended II · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int maxValue(vector<vector<int>>& events, int k) {    vector<vector<int>> mem(events.size(), vector<int>(k + 1, -1));    ranges::sort(events);    return maxValue(events, 0, k, mem);  }  private:  // Returns the maximum sum of values that you can receive by attending  // events[i..n), where k is the maximum number of attendancevents.  int maxValue(const vector<vector<int>>& events, int i, int k,               vector<vector<int>>& mem) {    if (k == 0 || i == events.size())      return 0;    if (mem[i][k] != -1)      return mem[i][k];     // Binary search `events` to find the first index j    // s.t. events[j][0] > events[i][1].    const auto it = upper_bound(        events.begin() + i, events.end(), events[i][1],        [](int end, const vector<int>& event) { return event[0] > end; });    const int j = distance(events.begin(), it);    return mem[i][k] = max(events[i][2] + maxValue(events, j, k - 1, mem),                           maxValue(events, i + 1, k, mem));  }};

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