Find the Largest Area of Square Inside Two Rectangles

• Initialize a variable maxSide to 0.
• Use a nested loop to iterate through all unique pairs of rectangles (i, j).
• For each pair, calculate the coordinates of their intersection rectangle.
• If the intersection is valid (has positive width and height), calculate the side s of the largest square that can fit inside (s = min(width, height)).
• Update maxSide = max(maxSide, s).
• Use a third nested loop to iterate through all unique triples of rectangles (i, j, k).
• For each triple, calculate their common intersection.
• If the intersection is valid, calculate the side s of the largest square and update maxSide accordingly.
• After all loops complete, return the area, which is (long)maxSide * maxSide.

This approach is a brute-force method that considers more combinations than necessary. The algorithm proceeds as follows:

1. Initialize a variable maxSide to 0, which will store the side length of the largest square found so far.
2. First, it iterates through all unique pairs of rectangles (i, j). For each pair, it computes their intersection. The intersection of two rectangles R_i = (x1i, y1i, x2i, y2i) and R_j = (x1j, y1j, x2j, y2j) is a new rectangle with its bottom-left corner at (max(x1i, x1j), max(y1i, y1j)) and its top-right corner at (min(x2i, x2j), min(y2i, y2j)). If this intersection is valid (i.e., its width and height are positive), we find the side of the largest square it can contain, which is min(width, height). We update maxSide with this value if it's larger.
3. Next, the algorithm iterates through all unique triples of rectangles (i, j, k). It finds their common intersection, for instance, by first intersecting R_i and R_j, and then intersecting the resulting rectangle with R_k. Similar to the pairwise case, it calculates the largest square in this triple intersection and updates maxSide.
4. The final result is the area of the square with the side maxSide, which is maxSide * maxSide.

This method is fundamentally flawed in terms of efficiency because the work done checking triples is redundant. Any square that fits in the intersection of three rectangles R_i, R_j, R_k also fits in the intersection of any pair from that triple (e.g., R_i and R_j). Thus, the maximum square side will always be found by considering only pairs.

1// Snippet demonstrating the inefficient triple loop logic2long maxSide = 0;3int n = bottomLeft.length;4// O(N^2) part for pairs (essential)5for (int i = 0; i < n; i++) {6    for (int j = i + 1; j < n; j++) {7        // ... pairwise intersection calculation ...8        // maxSide = max(maxSide, side_from_pair_ij);9    }10}11 12// O(N^3) part for triples (inefficient and redundant)13for (int i = 0; i < n; i++) {14    for (int j = i + 1; j < n; j++) {15        for (int k = j + 1; k < n; k++) {16            // Find intersection of i and j17            long ix1 = Math.max(bottomLeft[i][0], bottomLeft[j][0]);18            long iy1 = Math.max(bottomLeft[i][1], bottomLeft[j][1]);19            long ix2 = Math.min(topRight[i][0], topRight[j][0]);20            long iy2 = Math.min(topRight[i][1], topRight[j][1]);21 22            // Intersect with k23            ix1 = Math.max(ix1, bottomLeft[k][0]);24            iy1 = Math.max(iy1, bottomLeft[k][1]);25            ix2 = Math.min(ix2, topRight[k][0]);26            iy2 = Math.min(iy2, topRight[k][1]);27 28            if (ix2 > ix1 && iy2 > iy1) {29                long side = Math.min(ix2 - ix1, iy2 - iy1);30                maxSide = Math.max(maxSide, side);31            }32        }33    }34}35// return maxSide * maxSide;

• Initialize a variable maxSide to 0 to keep track of the maximum possible side of a square.
• Iterate through each unique pair of rectangles using two nested loops. Let the outer loop run for i from 0 to n-1 and the inner loop for j from i+1 to n-1.
• For each pair of rectangles i and j, calculate the coordinates of their intersection.
  • intersectX1 = max(bottomLeft[i][0], bottomLeft[j][0])
  • intersectY1 = max(bottomLeft[i][1], bottomLeft[j][1])
  • intersectX2 = min(topRight[i][0], topRight[j][0])
  • intersectY2 = min(topRight[i][1], topRight[j][1])
• Check if the intersection is a valid rectangle (i.e., if intersectX2 > intersectX1 and intersectY2 > intersectY1).
• If a valid intersection exists, calculate the side length of the largest square that can fit inside it: side = min(intersectX2 - intersectX1, intersectY2 - intersectY1).
• Update maxSide with the maximum value found so far: maxSide = max(maxSide, side).
• After iterating through all pairs, the maximum area is (long)maxSide * maxSide. Return this value.

The most efficient way to solve this problem is to realize that we only need to consider pairs of rectangles. The intersection of three or more rectangles will always be a sub-region of the intersection of any two of those rectangles. Therefore, it cannot contain a larger square.

The algorithm is as follows:

1. Initialize a variable, maxSide, to 0. This will store the side length of the largest square found.
2. Iterate through all unique pairs of rectangles. A common way to do this is with a nested loop structure where the outer loop variable i goes from 0 to n-2 and the inner loop variable j goes from i+1 to n-1.
3. For each pair of rectangles, rect_i and rect_j, calculate their intersection. The intersection is also a rectangle whose boundaries are determined by the maximum of the bottom-left coordinates and the minimum of the top-right coordinates.
4. If the calculated intersection has a positive width and height, it's a valid overlapping region. The side length of the largest square that can be inscribed in this intersection rectangle is the minimum of its width and height.
5. Compare this side length with maxSide and update maxSide if the new side is larger.
6. After checking all pairs, maxSide will hold the maximum possible side length. The result is the area, calculated as (long)maxSide * maxSide to prevent potential integer overflow since coordinates can be large.

1class Solution {2    public long largestSquareArea(int[][] bottomLeft, int[][] topRight) {3        int n = bottomLeft.length;4        long maxSide = 0;5 6        for (int i = 0; i < n; i++) {7            for (int j = i + 1; j < n; j++) {8                // Calculate the intersection of rectangle i and rectangle j.9                // Bottom-left corner of the intersection10                long intersectX1 = Math.max(bottomLeft[i][0], bottomLeft[j][0]);11                long intersectY1 = Math.max(bottomLeft[i][1], bottomLeft[j][1]);12 13                // Top-right corner of the intersection14                long intersectX2 = Math.min(topRight[i][0], topRight[j][0]);15                long intersectY2 = Math.min(topRight[i][1], topRight[j][1]);16 17                // Check if there is a valid intersection18                if (intersectX2 > intersectX1 && intersectY2 > intersectY1) {19                    // The side of the largest square in the intersection is the minimum of its width and height.20                    long side = Math.min(intersectX2 - intersectX1, intersectY2 - intersectY1);21                    // Update the maximum side found so far.22                    maxSide = Math.max(maxSide, side);23                }24            }25        }26 27        // The result is the area of the square with the maximum side.28        return maxSide * maxSide;29    }30}