CPMagical CP Library

Problem solution · C++

Circle and Rectangle Overlapping

Circle and Rectangle Overlapping: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
22 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

Direct simulation

For Circle and Rectangle Overlapping, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 22 lines of C++ from the credited upstream file 1401.cpp.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeCircle and Rectangle Overlapping · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  bool checkOverlap(int radius, int x_center, int y_center, int x1, int y1,                    int x2, int y2) {    auto clamp = [&](int center, int mn, int mx) {      return max(mn, min(mx, center));    };     // the closest point to the circle within the rectangle    int closestX = clamp(x_center, x1, x2);    int closestY = clamp(y_center, y1, y2);     // the distance between the circle's center and its closest point    int distanceX = x_center - closestX;    int distanceY = y_center - closestY;     // If the distance < the circle's radius, an intersection occurs.    return (distanceX * distanceX) + (distanceY * distanceY) <=           (radius * radius);  }};

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