Add solution #3047

Author: JoshCrozier

README.md

@@ -2149,6 +2149,7 @@
 3043|[Find the Length of the Longest Common Prefix](./solutions/3043-find-the-length-of-the-longest-common-prefix.js)|Medium|
 3044|[Most Frequent Prime](./solutions/3044-most-frequent-prime.js)|Medium|
 3046|[Split the Array](./solutions/3046-split-the-array.js)|Easy|
+3047|[Find the Largest Area of Square Inside Two Rectangles](./solutions/3047-find-the-largest-area-of-square-inside-two-rectangles.js)|Medium|
 3066|[Minimum Operations to Exceed Threshold Value II](./solutions/3066-minimum-operations-to-exceed-threshold-value-ii.js)|Medium|
 3068|[Find the Maximum Sum of Node Values](./solutions/3068-find-the-maximum-sum-of-node-values.js)|Hard|
 3105|[Longest Strictly Increasing or Strictly Decreasing Subarray](./solutions/3105-longest-strictly-increasing-or-strictly-decreasing-subarray.js)|Easy|

solutions/3047-find-the-largest-area-of-square-inside-two-rectangles.js

@@ -0,0 +1,39 @@
+/**
+ * 3047. Find the Largest Area of Square Inside Two Rectangles
+ * https://leetcode.com/problems/find-the-largest-area-of-square-inside-two-rectangles/
+ * Difficulty: Medium
+ *
+ * There exist n rectangles in a 2D plane with edges parallel to the x and y axis. You
+ * are given two 2D integer arrays bottomLeft and topRight where bottomLeft[i] = [a_i, b_i]
+ * and topRight[i] = [c_i, d_i] represent the bottom-left and top-right coordinates of the
+ * ith rectangle, respectively.
+ *
+ * You need to find the maximum area of a square that can fit inside the intersecting
+ * region of at least two rectangles. Return 0 if such a square does not exist.
+ */
+
+/**
+ * @param {number[][]} bottomLeft
+ * @param {number[][]} topRight
+ * @return {number}
+ */
+var largestSquareArea = function(bottomLeft, topRight) {
+  const n = bottomLeft.length;
+  let maxSide = 0;
+
+  for (let i = 0; i < n; i++) {
+    for (let j = i + 1; j < n; j++) {
+      const xLeft = Math.max(bottomLeft[i][0], bottomLeft[j][0]);
+      const yBottom = Math.max(bottomLeft[i][1], bottomLeft[j][1]);
+      const xRight = Math.min(topRight[i][0], topRight[j][0]);
+      const yTop = Math.min(topRight[i][1], topRight[j][1]);
+
+      if (xLeft < xRight && yBottom < yTop) {
+        const side = Math.min(xRight - xLeft, yTop - yBottom);
+        maxSide = Math.max(maxSide, side);
+      }
+    }
+  }
+
+  return maxSide * maxSide;
+};