1274. Number of Ships in a Rectangle - Detailed Explanation

Problem Statement

You are given a rectangular region in the sea defined by two points: the bottom‐left coordinate and the top‐right coordinate. There are some “ships” in this region, but you do not know their exact positions. You are provided an API:

• hasShips(topRight, bottomLeft):
  Returns true if there is at least one ship in the rectangle defined by the two coordinates, otherwise returns false.

Your task is to count how many ships are present in the given rectangle.
Note:

• Ships are located at distinct grid points.
• There is a limited number of ships (for instance, at most 10).
• The rectangle can be very large, so you cannot check every cell individually.

Example 1

• Input:
  bottomLeft = [0, 0], topRight = [4, 4]
  Assume the ships are at coordinates (1, 2), (3, 3), and (4, 1).
• Output:
  3
• Explanation:
  The rectangle from (0, 0) to (4, 4) contains exactly 3 ships.

Example 2

• Input:
  bottomLeft = [1, 1], topRight = [2, 2]
  Assume a ship exists at (2, 2).
• Output:
  1
• Explanation:
  The region covers 4 cells, but only one cell (2,2) has a ship.

Example 3

• Input:
  bottomLeft = [0, 0], topRight = [0, 0]
  Assume there is no ship at (0, 0).
• Output:
  0
• Explanation:
  The rectangle is just a single point with no ship.

Constraints

• The coordinates are given as integers.
• The given rectangle can be very large.
• The number of ships is limited (e.g., ≤ 10).
• The API hasShips is available to check if a subrectangle contains at least one ship.

Hints and Intuition

• Hint 1:
  You cannot check every single cell because the grid may be huge. Instead, use the API to rule out large areas without ships.

• Hint 2:
  Use a divide and conquer strategy. If the current rectangle does not contain any ship (using hasShips), return 0 immediately. If it does and the rectangle represents a single cell, return 1. Otherwise, split the rectangle into four smaller parts and count the ships in each.

Detailed Approach

A. Why Not Brute Force?

• Brute Force Approach:
  Checking every cell one by one in the given rectangle is not feasible due to the potential huge coordinate space.
• Optimal Idea:
  Instead of scanning every cell, use the API to quickly eliminate regions with no ships and only drill down into areas where ships may be present.

B. Divide and Conquer (Recursive Subdivision)

1. Check the Region:
   Use hasShips(topRight, bottomLeft) to determine if there is at least one ship in the current rectangle.
   • If false, return 0.
2. Base Case:
   If the rectangle represents a single point (i.e. bottomLeft equals topRight), then return 1 (since hasShips returned true).
3. Subdivision:
   • Calculate midpoints:
     • midX = (bottomLeft[0] + topRight[0]) // 2
     • midY = (bottomLeft[1] + topRight[1]) // 2
   • Divide the rectangle into four smaller rectangles:
     1. Bottom-Left Subrectangle:
        from bottomLeft to [midX, midY]
     2. Bottom-Right Subrectangle:
        from [midX + 1, bottomLeft[1]] to [topRight[0], midY]
        (if midX + 1 <= topRight[0])
     3. Top-Left Subrectangle:
        from [bottomLeft[0], midY + 1] to [midX, topRight[1]]
        (if midY + 1 <= topRight[1])
     4. Top-Right Subrectangle:
        from [midX + 1, midY + 1] to topRight
        (if both conditions hold)
4. Recursive Counting:
   Recursively count the number of ships in each subrectangle and sum up the results.

Step-by-Step Walkthrough with Visual Example

Consider Example 1:

• Region: BottomLeft = (0, 0), TopRight = (4, 4)
• Assumed Ship Locations: (1, 2), (3, 3), (4, 1)

Step 1:

• Check the entire rectangle using hasShips((4, 4), (0, 0)) → returns true.

Step 2:

• Since the rectangle is not a single cell, calculate midpoints:
  midX = (0 + 4) // 2 = 2 and midY = (0 + 4) // 2 = 2.

Step 3:

• Divide the region into four subrectangles:
  1. Subrectangle 1: (0,0) to (2,2)
     • Check: If any ship exists (ship at (1,2) is in this region).
  2. Subrectangle 2: (3,0) to (4,2)
     • Check: (Ship at (4,1) falls in this subrectangle.)
  3. Subrectangle 3: (0,3) to (2,4)
     • Check: Likely empty.
  4. Subrectangle 4: (3,3) to (4,4)
     • Check: (Ship at (3,3) falls here.)

Step 4:

• Recursively apply the same process on each subrectangle.
• For subrectangles that reduce to a single point and return true, count 1.
• For those where hasShips returns false, count 0.

Final Count:

• Sum the counts from the four regions → 3 ships.

5. Python Implementation

In Python, we assume a helper object (e.g., sea) that provides the method hasShips(topRight, bottomLeft). For testing purposes, you can simulate the API.