Create 3000-maximum-area-of-longest-diagonal-rectangle.java

Author: sanajitjana

Easy/3000-maximum-area-of-longest-diagonal-rectangle.java

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+/**
+ * Problem: 2679. Sum in a Matrix (variation: Area of Max Diagonal)
+ * Difficulty: Easy
+ * URL: https://leetcode.com/problems/area-of-maximum-diagonal/
+ * 
+ * Approach:
+ * - For each rectangle (length, width), compute the squared diagonal = l² + w².
+ * - Also compute its area = l * w.
+ * - Track the maximum diagonal squared.
+ *   - If a rectangle has a larger diagonal, update max diagonal and max area.
+ *   - If diagonal is equal to current max, take the larger area.
+ * - Using squared diagonal avoids floating-point issues from Math.sqrt.
+ * 
+ * Time Complexity: O(n)   — one pass through all rectangles
+ * Space Complexity: O(1) — only a few variables stored
+ */
+
+class Solution {
+    public int areaOfMaxDiagonal(int[][] dimensions) {
+        int maxDiaSq = 0;  // store squared diagonal
+        int maxArea = 0;
+
+        for (int[] arr : dimensions) {
+            int diaSq = arr[0] * arr[0] + arr[1] * arr[1];
+            int area = arr[0] * arr[1];
+
+            if (diaSq > maxDiaSq) {
+                maxDiaSq = diaSq;
+                maxArea = area;
+            } else if (diaSq == maxDiaSq && area > maxArea) {
+                maxArea = area;
+            }
+        }
+
+        return maxArea;
+    }
+}