CPMagical CP Library

Problem solution · C++

Set Intersection Size At Least Two

Set Intersection Size At Least Two: 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
36 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 Set Intersection Size At Least Two, 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

  • 36 lines of C++ from the credited upstream file 757.cpp.
  • The implementation visibly relies on sequence storage.
  • 1 loop block 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 codeSet Intersection Size At Least Two · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int intersectionSizeTwo(vector<vector<int>>& intervals) {    int ans = 0;    int mx = -1;    int secondMax = -1;     ranges::sort(intervals, ranges::less{}, [](const vector<int>& interval) {      const int start = interval[0];      const int end = interval[1];      return pair<int, int>{end, -start};    });     for (const vector<int>& interval : intervals) {      const int start = interval[0];      const int end = interval[1];      // The maximum and the second maximum still satisfy.      if (mx >= start && secondMax >= start)        continue;      if (mx >= start) {        // The maximum still satisfy.        secondMax = mx;        mx = end;  // Add `end` to the set.        ans += 1;      } else {        // The maximum and the second maximum can't satisfy.        mx = end;             // Add `end` to the set.        secondMax = end - 1;  // Add `end - 1` to the set.        ans += 2;      }    }     return ans;  }};

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 ↗