Update solution for Problem 3193: Area of Maximum Diagonal Rectangle - correct algorithm using Pythagorean theorem to find longest diagonal

Author: TechnoBlogger14o3

Medium/2025-08-26-3193-MaximumAreaOfLongestDiagonalRectangle/explanation.md

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+# Area of Maximum Diagonal Rectangle - Problem #3193
+
+## Problem Description
+Given a 2D array `dimensions` where `dimensions[i] = [lengthi, widthi]`, return the area of the rectangle having the longest diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the maximum area.
+
+## Examples
+```
+Input: dimensions = [[9,3],[8,6]]
+Output: 48
+Explanation: 
+For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9² + 3²) = sqrt(90) ≈ 9.487.
+For index = 1, length = 8 and width = 6. Diagonal length = sqrt(8² + 6²) = sqrt(100) = 10.
+So, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48.
+
+Input: dimensions = [[3,4],[4,3]]
+Output: 12
+Explanation: Length of diagonal is the same for both which is 5, so maximum area = 12.
+```
+
+## Approach
+This is a geometry problem that requires finding the rectangle with the longest diagonal. The key insight is to calculate the diagonal length of each rectangle using the Pythagorean theorem and track the one with the longest diagonal.
+
+## Algorithm
+1. **Iterate through rectangles**: Process each rectangle in the dimensions array
+2. **Calculate diagonal length**: Use Pythagorean theorem: sqrt(length² + width²)
+3. **Track longest diagonal**: Keep track of the rectangle with the longest diagonal
+4. **Handle ties**: If diagonals are equal, keep the rectangle with larger area
+5. **Return area**: Return the area of the rectangle with the longest diagonal
+
+## Key Implementation Details
+- **Pair Generation**: Use nested loops to generate all possible pairs
+- **Orientation Handling**: For each pair, try both length×width and width×length combinations
+- **Area Calculation**: Area = length × width
+- **Maximum Tracking**: Update maximum area when a larger valid rectangle is found
+
+## Mathematical Intuition
+The maximum area rectangle can be formed by:
+- Using the largest length and width from different dimension pairs
+- Considering both orientations (length×width vs width×length)
+- Finding the optimal combination that maximizes the area
+
+## Time Complexity
+- **Time**: O(n²) where n is the number of dimensions
+- **Space**: O(1) - only using constant extra space
+
+## Example Walkthrough
+For input `[[3,4],[5,5],[5,3],[1,1],[4,4]]`:
+- Pair [3,4] and [5,5]: Try [3,5] and [5,3] → areas 15 and 15
+- Pair [3,4] and [5,3]: Try [3,5] and [5,3] → areas 15 and 15
+- Pair [5,5] and [4,4]: Try [5,4] and [4,5] → areas 20 and 20
+- Maximum area: 20
+
+## Key Insights
+- **Pair Combinations**: Need to consider all possible pairs of dimensions
+- **Orientation Flexibility**: Each dimension can be used as length or width
+- **Area Maximization**: Find the combination that gives the largest area
+- **Efficient Pairing**: Use nested loops to generate all pairs
+
+## Alternative Approaches
+- **Sorting**: Sort dimensions by area and try largest combinations first
+- **Greedy**: Always use the largest available dimensions
+- **Dynamic Programming**: Use DP to find optimal combinations
+- **Binary Search**: Binary search on the answer area
+
+## Edge Cases
+- **Empty Array**: Return 0
+- **Single Element**: Return 0 (need at least two dimensions)
+- **All Same Dimensions**: Handle when all dimensions are identical
+- **Zero Dimensions**: Handle when length or width is 0
+
+## Applications
+- **Geometry Problems**: Rectangle area optimization
+- **Design Applications**: Optimal layout design
+- **Resource Allocation**: Maximize resource utilization
+- **Game Development**: Optimal game board dimensions
+- **UI Design**: Optimal component sizing
+
+## Optimization Opportunities
+- **Early Termination**: Stop when remaining pairs can't improve the answer
+- **Sorting**: Sort dimensions to try larger combinations first
+- **Memory Access**: Optimize array access patterns
+- **Branch Prediction**: Minimize conditional branches
+
+## Related Problems
+- **Maximum Rectangle**: Find largest rectangle in a grid
+- **Rectangle Area**: Calculate area of overlapping rectangles
+- **Optimal Rectangle**: Find rectangle with specific properties
+- **Geometry Optimization**: Various geometric optimization problems

Medium/2025-08-26-3193-MaximumAreaOfLongestDiagonalRectangle/solution.java

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+class Solution {
+    public int areaOfMaxDiagonal(int[][] dimensions) {
+        if (dimensions == null || dimensions.length == 0) return 0;
+        
+        double maxDiagonal = 0;
+        int maxArea = 0;
+        
+        for (int[] rect : dimensions) {
+            int length = rect[0];
+            int width = rect[1];
+            
+            // Calculate diagonal length using Pythagorean theorem
+            double diagonal = Math.sqrt(length * length + width * width);
+            
+            // If this diagonal is longer, update both diagonal and area
+            if (diagonal > maxDiagonal) {
+                maxDiagonal = diagonal;
+                maxArea = length * width;
+            }
+            // If diagonal is equal, keep the rectangle with larger area
+            else if (Math.abs(diagonal - maxDiagonal) < 1e-9) {
+                maxArea = Math.max(maxArea, length * width);
+            }
+        }
+        
+        return maxArea;
+    }
+}

Medium/2025-08-26-3193-MaximumAreaOfLongestDiagonalRectangle/solution.js

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+/**
+ * @param {number[][]} dimensions
+ * @return {number}
+ */
+var areaOfMaxDiagonal = function(dimensions) {
+    if (!dimensions || dimensions.length === 0) return 0;
+    
+    let maxDiagonal = 0;
+    let maxArea = 0;
+    
+    for (const rect of dimensions) {
+        const length = rect[0];
+        const width = rect[1];
+        
+        // Calculate diagonal length using Pythagorean theorem
+        const diagonal = Math.sqrt(length * length + width * width);
+        
+        // If this diagonal is longer, update both diagonal and area
+        if (diagonal > maxDiagonal) {
+            maxDiagonal = diagonal;
+            maxArea = length * width;
+        }
+        // If diagonal is equal, keep the rectangle with larger area
+        else if (Math.abs(diagonal - maxDiagonal) < 1e-9) {
+            maxArea = Math.max(maxArea, length * width);
+        }
+    }
+    
+    return maxArea;
+};

Medium/2025-08-26-3193-MaximumAreaOfLongestDiagonalRectangle/solution.py

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+from typing import List
+import math
+
+class Solution:
+    def areaOfMaxDiagonal(self, dimensions: List[List[int]]) -> int:
+        if not dimensions:
+            return 0
+        
+        max_diagonal = 0
+        max_area = 0
+        
+        for rect in dimensions:
+            length = rect[0]
+            width = rect[1]
+            
+            # Calculate diagonal length using Pythagorean theorem
+            diagonal = math.sqrt(length * length + width * width)
+            
+            # If this diagonal is longer, update both diagonal and area
+            if diagonal > max_diagonal:
+                max_diagonal = diagonal
+                max_area = length * width
+            # If diagonal is equal, keep the rectangle with larger area
+            elif abs(diagonal - max_diagonal) < 1e-9:
+                max_area = max(max_area, length * width)
+        
+        return max_area