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Problem solution · C++

Two Best Non-Overlapping Events

Two Best Non-Overlapping Events: a C++ solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sorting and greedy selection
Source
walkccc LeetCode Solutions
Length
35 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

Sorting and greedy selection

For Two Best Non-Overlapping Events, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 35 lines of C++ from the credited upstream file 2054.cpp.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

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 codeTwo Best Non-Overlapping Events · C++C++
Use this to learn the idea, then write your own version.
struct Event {  int time;  int value;  bool isStart;}; class Solution { public:  int maxTwoEvents(vector<vector<int>>& events) {    int ans = 0;    int maxValue = 0;    vector<Event> evts;     for (const vector<int>& event : events) {      const int start = event[0];      const int end = event[1];      const int value = event[2];      evts.emplace_back(start, value, true);      evts.emplace_back(end + 1, value, false);    }     ranges::sort(evts, ranges::less{}, [](const Event& evt) {      return pair<int, bool>{evt.time, evt.isStart};    });     for (const auto& [_, value, isStart] : evts)      if (isStart)        ans = max(ans, value + maxValue);      else        maxValue = max(maxValue, value);     return ans;  }};

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