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Problem solution · C++

Range Module

Range Module: 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
52 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 Range Module, 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

  • 52 lines of C++ from the credited upstream file 715-2.cpp.
  • The implementation visibly relies on ordered lookup.
  • 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 codeRange Module · C++C++
Use this to learn the idea, then write your own version.
class RangeModule { public:  void addRange(int left, int right) {    const auto [l, r] = getOverlapRanges(left, right);    if (l == r) {            // There's no overlap.      ranges[left] = right;  // Add a new range.      return;    }     auto last = r;    const int newLeft = min(l->first, left);    const int newRight = max((--last)->second, right);    ranges.erase(l, r);    ranges[newLeft] = newRight;  // Add a new range.  }   bool queryRange(int left, int right) {    auto it = ranges.upper_bound(left);    return it != ranges.begin() && (--it)->second >= right;  }   void removeRange(int left, int right) {    const auto [l, r] = getOverlapRanges(left, right);    if (l == r)  // There's no overlap.      return;     auto last = r;    const int newLeft = min(l->first, left);    const int newRight = max((--last)->second, right);    ranges.erase(l, r);    // Add new ranges if needed.    if (newLeft < left)      ranges[newLeft] = left;    if (right < newRight)      ranges[right] = newRight;  }  private:  using IT = map<int, int>::iterator;  map<int, int> ranges;   pair<IT, IT> getOverlapRanges(int left, int right) {    // Point to the first element with `second` >= `left`.    IT l = ranges.upper_bound(left);    // Point to the first element with `first` > `right`.    IT r = ranges.upper_bound(right);    if (l != ranges.begin() && (--l)->second < left)      ++l;    return {l, r};  }};

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