feat: add solution 3000. Maximum Area of Longest Diagonal Rectangle

Author: tzuyi0817

solutions/3000. Maximum Area of Longest Diagonal Rectangle/README.md

@@ -0,0 +1,74 @@
+# [3000. Maximum Area of Longest Diagonal Rectangle](https://leetcode.com/problems/maximum-area-of-longest-diagonal-rectangle)
+
+## Description
+
+<div class="elfjS" data-track-load="description_content"><p>You are given a 2D <strong>0-indexed </strong>integer array <code>dimensions</code>.</p>
+
+<p>For all indices <code>i</code>, <code>0 &lt;= i &lt; dimensions.length</code>, <code>dimensions[i][0]</code> represents the length and <code>dimensions[i][1]</code> represents the width of the rectangle<span style="font-size: 13.3333px;"> <code>i</code></span>.</p>
+
+<p>Return <em>the <strong>area</strong> of the rectangle having the <strong>longest</strong> diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the <strong>maximum</strong> area.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre><strong>Input:</strong> dimensions = [[9,3],[8,6]]
+<strong>Output:</strong> 48
+<strong>Explanation:</strong> 
+For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈ 9.487.
+For index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10.
+So, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre><strong>Input:</strong> dimensions = [[3,4],[4,3]]
+<strong>Output:</strong> 12
+<strong>Explanation:</strong> Length of diagonal is the same for both which is 5, so maximum area = 12.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= dimensions.length &lt;= 100</code></li>
+	<li><code><font face="monospace">dimensions[i].length == 2</font></code></li>
+	<li><code><font face="monospace">1 &lt;= dimensions[i][0], dimensions[i][1] &lt;= 100</font></code></li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+## Solutions
+
+**Solution: `Brute Force`**
+
+- Time complexity: <em>O(n)</em>
+- Space complexity: <em>O(1)</em>
+
+<p>&nbsp;</p>
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number[][]} dimensions
+ * @return {number}
+ */
+const areaOfMaxDiagonal = function (dimensions) {
+  let maxDiagonal = 0;
+  let result = 0;
+
+  for (const [length, width] of dimensions) {
+    const diagonal = Math.hypot(length, width);
+
+    if (diagonal < maxDiagonal) continue;
+
+    const area = length * width;
+
+    result = diagonal === maxDiagonal ? Math.max(area, result) : area;
+    maxDiagonal = diagonal;
+  }
+
+  return result;
+};
+```

solutions/3000. Maximum Area of Longest Diagonal Rectangle/solution.js

@@ -0,0 +1,21 @@
+/**
+ * @param {number[][]} dimensions
+ * @return {number}
+ */
+const areaOfMaxDiagonal = function (dimensions) {
+  let maxDiagonal = 0;
+  let result = 0;
+
+  for (const [length, width] of dimensions) {
+    const diagonal = Math.hypot(length, width);
+
+    if (diagonal < maxDiagonal) continue;
+
+    const area = length * width;
+
+    result = diagonal === maxDiagonal ? Math.max(area, result) : area;
+    maxDiagonal = diagonal;
+  }
+
+  return result;
+};