Add solution #1725

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,502 LeetCode solutions in JavaScript
+# 1,503 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1329,6 +1329,7 @@
 1721|[Swapping Nodes in a Linked List](./solutions/1721-swapping-nodes-in-a-linked-list.js)|Medium|
 1722|[Minimize Hamming Distance After Swap Operations](./solutions/1722-minimize-hamming-distance-after-swap-operations.js)|Medium|
 1723|[Find Minimum Time to Finish All Jobs](./solutions/1723-find-minimum-time-to-finish-all-jobs.js)|Hard|
+1725|[Number Of Rectangles That Can Form The Largest Square](./solutions/1725-number-of-rectangles-that-can-form-the-largest-square.js)|Easy|
 1726|[Tuple with Same Product](./solutions/1726-tuple-with-same-product.js)|Medium|
 1732|[Find the Highest Altitude](./solutions/1732-find-the-highest-altitude.js)|Easy|
 1748|[Sum of Unique Elements](./solutions/1748-sum-of-unique-elements.js)|Easy|

solutions/1725-number-of-rectangles-that-can-form-the-largest-square.js

@@ -0,0 +1,38 @@
+/**
+ * 1725. Number Of Rectangles That Can Form The Largest Square
+ * https://leetcode.com/problems/number-of-rectangles-that-can-form-the-largest-square/
+ * Difficulty: Easy
+ *
+ * You are given an array rectangles where rectangles[i] = [li, wi] represents the ith rectangle
+ * of length li and width wi.
+ *
+ * You can cut the ith rectangle to form a square with a side length of k if both k <= li and
+ * k <= wi. For example, if you have a rectangle [4,6], you can cut it to get a square with a
+ * side length of at most 4.
+ *
+ * Let maxLen be the side length of the largest square you can obtain from any of the given
+ * rectangles.
+ *
+ * Return the number of rectangles that can make a square with a side length of maxLen.
+ */
+
+/**
+ * @param {number[][]} rectangles
+ * @return {number}
+ */
+var countGoodRectangles = function(rectangles) {
+  let maxSide = 0;
+  let result = 0;
+
+  for (const [length, width] of rectangles) {
+    const side = Math.min(length, width);
+    if (side > maxSide) {
+      maxSide = side;
+      result = 1;
+    } else if (side === maxSide) {
+      result++;
+    }
+  }
+
+  return result;
+};