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Leetcode 3000. Maximum Area of Longest Diagonal Rectangle

Problem Statement

The problem requires determining the maximum area of a rectangle that can be inscribed inside a grid such that the rectangle’s longest diagonal is parallel to one of the grid’s diagonals.

Here is the expected definition for the problem:

Given a grid of size `m x n`, return the maximum area of a rectangle that can be inscribed inside the grid such that one of its longest diagonals is parallel to one of the grid's diagonals.

Clarifying Questions

1. Grid Boundaries: Can the rectangle touch the grid boundaries?
   • Assumption: Yes, the rectangle can touch the grid boundaries.
2. Shape Constraints: Does the rectangle need to be strictly contained within the grid?
   • Assumption: Yes, the rectangle must fit completely inside the grid without crossing the borders.
3. Input Constraints: What are the typical constraints on m and n (the dimensions of the grid)?
   • Assumption: Typical constraints would be 1 <= m, n <= 10^5.
4. Output: Should we return the area as an integer?
   • Assumption: Yes.

Strategy

1. Understanding the Diagonal Condition: For the rectangle’s longest diagonal to be parallel to one of the grid’s diagonals:
   • If the rectangle’s diagonal is parallel to the grid’s diagonals, the sides of the rectangle are rotated at an angle of 45 degrees in relation to the sides of the grid.
2. Maximum Rectangle Size: When a rectangle is rotated:
   • With diagonals of length d by d:
   • A square inscribed in a rectangle would have its maximal diagonal inside the bounds of another square (translating the square diagonally).
3. Area Calculation:
   • For an inscribed rectangle with diagonal d, the area would be maximized when the sides are floor(m/√2) and floor(n/√2) respectively.
4. Optimal Approach: Derived from inscribing rectangles, we seek maximal dimensional combinations of grid sub-sections.

Code

Here’s an approach to calculate the maximum area:

public class Solution {
    public int maxInscribedRectangleArea(int m, int n) {
        // Maximum possible diagonal length in the rotated grid
        double maxSide = Math.min(m, n) / Math.sqrt(2);
        
        // Maximum sides of the rectangle inscribed
        int maxLength = (int) maxSide;
        
        // Return the area as an integer
        return maxLength * maxLength;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.maxInscribedRectangleArea(3, 6)); // Example case
    }
}

Time Complexity

• Time Analysis: The approach runs in O(1) time as it computes basic mathematical operations directly influenced by the input values m and n.

• Space Analysis: The space complexity is also O(1) since no additional space allocations are required apart from a few variables.

This solution efficiently calculates the maximum area of a rectangle meeting the specified diagonal condition within the problem constraints.

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