Problem Description

You are given a circle and an axis-aligned rectangle. The circle is defined by its radius and the coordinates of its center (xCenter, yCenter). The rectangle is defined by the coordinates of its bottom-left corner (x1, y1) and its top-right corner (x2, y2). Your task is to determine if there exists at least one point that is inside both the given circle and rectangle. The problem is asking you to return true if such a point exists (meaning the circle and rectangle overlap), and false otherwise.

Intuition

To solve this problem, let's first understand the scenarios where a circle and a rectangle can overlap:

1. Any corner of the rectangle is inside the circle.
2. Any part of the circle's circumference intersects the rectangle.
3. The circle is entirely inside the rectangle, or vice versa.

However, checking every point on the circumference directly would be impractical. The key insight here is that the closest point on the rectangle to the circle's center can indicate whether they overlap without checking every point. If this closest point is also inside the circle, then they overlap.

The solution code uses a helper function f to find the closest point on the rectangle to a given point (xCenter, yCenter). The function f calculates the minimum distance along the x or y dimension from the rectangle to the center of the circle. If the center of the circle is within the bounds of the rectangle along one axis, then the minimum distance is 0 for that axis. Otherwise, f returns the distance to the closest edge of the rectangle along that axis.

Once we calculate the minimum distances a and b for the x and y axes, the point on the rectangle closest to the circle's center is at (xCenter + a, yCenter + b). If that point is within the circle (meaning the sum of squares a * a + b * b is less than or equal to radius * radius), then we have an overlap, and the function returns true. If the closest point is outside the circle, the function returns false.

This approach effectively checks whether the rectangle and the circle overlap by considering only the closest point on the rectangle to the circle's center, greatly simplifying the problem.

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Solution Approach

The solution employs a straightforward geometric approach, converting the overlap problem into one of measuring distances and comparing them to the circle's radius. Here's how it's implemented:

1. Function Declaration: The Solution class contains a method checkOverlap which takes as its arguments the circle's radius and its center coordinates, along with the coordinates of the rectangle's bottom-left and top-right corners.

2. Distance Function (f): The method uses a nested helper function f, which is designed to calculate the shortest one-dimensional distance from a point (either xCenter or yCenter) to an interval defined by two bounds (either x1, x2 or y1, y2). The bounds represent the sides of the rectangle along one axis.

   • It works as follows: if the center coordinate (for one dimension) is between the rectangle's edge coordinates, then the circle's center is directly above or alongside the edge (when projected onto that axis), and thus the shortest distance is 0.
   • If the circle's center is outside the rectangle along that axis, the function returns the distance to the nearest edge.

3. Finding the Closest Point: The checkOverlap method then calls f for both the x and y dimensions to find the x-distance (a) and y-distance (b) from the circle’s center to the rectangle's closest edge or point.

4. Checking Overlap: After the closest point is identified by its x and y distances to the circle's center, we check if this point is inside the circle. The point lies inside if the sum of the squares of the distances a * a + b * b does not exceed the square of the radius of the circle radius * radius.

5. Returning the Result: Finally, the checkOverlap method returns true if the closest point is inside the circle (hence, an overlap) or false otherwise.

This algorithm is particularly effective because it reduces a 2D problem into a comparative evaluation of calculated 1D distances. No looping or complex structures are necessary, making the solution very efficient in terms of time complexity, which is O(1), as the calculations are done in a fixed number of steps regardless of input size.

Example Walkthrough

Let's go through a small example to illustrate the solution approach.

Assume we are given the following inputs:

• Radius of the circle (radius): 2
• Coordinates of the circle's center (xCenter, yCenter): (2, 2)
• Coordinates of the bottom-left corner of the rectangle (x1, y1): (3, 1)
• Coordinates of the top-right corner of the rectangle (x2, y2): (5, 3)

Now, we need to determine if at least one point on the rectangle is also inside the circle.

Following the solution approach:

1. We start by preparing to invoke the checkOverlap method from the Solution class, which will take our given inputs.

2. Next, we use the helper function f to calculate the closest x-distance (a) from the circle's center to the rectangle's closest edge.

   • In this case, the xCenter is 2, and the x bounds of the rectangle are 3 and 5. Since the xCenter is less than x1, the shortest x-distance to the rectangle is x1 - xCenter, which is 3 - 2 = 1.

3. We apply the same logic to find the closest y-distance (b) from the circle's center to the rectangle's closest edge.

   • The yCenter is 2, and the y bounds are 1 and 3. Since the yCenter is within these bounds, the circle's center is directly alongside the rectangle on the y-axis, so the shortest y-distance is 0.

4. With the closest x-distance (a) as 1 and the closest y-distance (b) as 0, we now determine whether the closest point on the rectangle to the circle's center is inside the circle.

   • We calculate a * a + b * b which equals 1 * 1 + 0 * 0 = 1.
   • The square of the circle's radius is radius * radius or 2 * 2 = 4.

5. Since 1 (the sum of squares of the distances) is less than 4 (the square of the radius), the closest point on the rectangle to the circle's center is within the circle. Therefore, there is an overlap.

6. The checkOverlap method would return true for these inputs, indicating that the rectangle and the circle do indeed overlap.

This example shows that you do not need to check every point on the rectangle or circle to see if there is an overlap. Instead, by cleverly using the closest point and comparing distance squared, we can efficiently solve the problem with a constant time algorithm.

Solution Implementation

Time and Space Complexity

Time Complexity: O(1)

The checkOverlap function consists of a nested helper f function. The operations inside the f function are simple arithmetic and conditional operations. These operations take constant time. Since there are no loops or recursive calls, the overall time complexity of the function is O(1), which means it executes in constant time regardless of the size of the input parameters.

Space Complexity: O(1)

The space complexity is also O(1) because the function only uses a fixed amount of additional space. The helper function f does not utilize additional memory that scales with the input size, and no extra data structures are employed. The variables a and b are the only additional memory used, and their space requirement does not depend on the size of the input.

Learn more about how to find time and space complexity quickly using problem constraints.