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Problem solution · C++

Closest Room

Closest Room: 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
40 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 Closest Room, 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

  • 40 lines of C++ from the credited upstream file 1847.cpp.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 3 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 codeClosest Room · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  vector<int> closestRoom(vector<vector<int>>& rooms,                          vector<vector<int>>& queries) {    vector<int> ans(queries.size());    set<int> roomIds;     for (int i = 0; i < queries.size(); ++i)      queries[i].push_back(i);     ranges::sort(rooms, ranges::greater{},                 [](const vector<int>& room) { return room[1]; });    ranges::sort(queries, ranges::greater{},                 [](const vector<int>& query) { return query[1]; });     int i = 0;  // rooms' index    for (const vector<int>& query : queries) {      while (i < rooms.size() && rooms[i][1] >= query[1])        roomIds.insert(rooms[i++][0]);      ans[query[2]] = searchClosestRoomId(roomIds, query[0]);    }     return ans;  }  private:  int searchClosestRoomId(set<int>& roomIds, int preferred) {    const auto it = roomIds.lower_bound(preferred);    const int id1 = it == roomIds.cbegin() ? -1 : *(prev(it));    const int id2 = it == roomIds.cend() ? -1 : *it;    if (id1 == -1)      return id2;    if (id2 == -1)      return id1;    if (abs(preferred - id1) <= abs(preferred - id2))      return id1;    return id2;  }};

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