CPMagical CP Library

Problem solution · C++

Block Placement Queries

Block Placement Queries: 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
71 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 Block Placement Queries, 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

  • 71 lines of C++ from the credited upstream file 3161.cpp.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 5 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 codeBlock Placement Queries · C++C++
Use this to learn the idea, then write your own version.
class FenwickTree { public:  FenwickTree(int n) : vals(n + 1) {}   void maximize(int i, int val) {    while (i < vals.size()) {      vals[i] = max(vals[i], val);      i += lowbit(i);    }  }   int get(int i) const {    int res = 0;    while (i > 0) {      res = max(res, vals[i]);      i -= lowbit(i);    }    return res;  }  private:  vector<int> vals;   static int lowbit(int i) {    return i & -i;  }}; class Solution { public:  vector<bool> getResults(vector<vector<int>>& queries) {    const int n = min(50000, static_cast<int>(queries.size()) * 3);    vector<bool> ans;    FenwickTree tree(n + 1);    set<int> obstacles{0, n};  // sentinel values     for (const vector<int>& query : queries) {      const int type = query[0];      if (type == 1) {        const int x = query[1];        obstacles.insert(x);      }    }     for (auto it = obstacles.begin(); std::next(it) != obstacles.end(); ++it) {      const int x1 = *it;      const int x2 = *std::next(it);      tree.maximize(x2, x2 - x1);    }     for (int i = queries.size() - 1; i >= 0; --i) {      const int type = queries[i][0];      const int x = queries[i][1];      if (type == 1) {        const auto it = obstacles.find(x);        if (next(it) != obstacles.end())  // x is not the last element.          tree.maximize(*next(it), *next(it) - *prev(it));        obstacles.erase(it);      } else {        const int sz = queries[i][2];        const auto it = obstacles.upper_bound(x);        const int prev = *std::prev(it);        ans.push_back(tree.get(prev) >= sz || x - prev >= sz);      }    }     ranges::reverse(ans);    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 ↗