Add solution #1691

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,478 LeetCode solutions in JavaScript
+# 1,479 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1303,6 +1303,7 @@
 1688|[Count of Matches in Tournament](./solutions/1688-count-of-matches-in-tournament.js)|Easy|
 1689|[Partitioning Into Minimum Number Of Deci-Binary Numbers](./solutions/1689-partitioning-into-minimum-number-of-deci-binary-numbers.js)|Medium|
 1690|[Stone Game VII](./solutions/1690-stone-game-vii.js)|Medium|
+1691|[Maximum Height by Stacking Cuboids](./solutions/1691-maximum-height-by-stacking-cuboids.js)|Hard|
 1716|[Calculate Money in Leetcode Bank](./solutions/1716-calculate-money-in-leetcode-bank.js)|Easy|
 1718|[Construct the Lexicographically Largest Valid Sequence](./solutions/1718-construct-the-lexicographically-largest-valid-sequence.js)|Medium|
 1726|[Tuple with Same Product](./solutions/1726-tuple-with-same-product.js)|Medium|

solutions/1691-maximum-height-by-stacking-cuboids.js

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+/**
+ * 1691. Maximum Height by Stacking Cuboids
+ * https://leetcode.com/problems/maximum-height-by-stacking-cuboids/
+ * Difficulty: Hard
+ *
+ * Given n cuboids where the dimensions of the ith cuboid is cuboids[i] = [widthi, lengthi, heighti]
+ * (0-indexed). Choose a subset of cuboids and place them on each other.
+ *
+ * You can place cuboid i on cuboid j if widthi <= widthj and lengthi <= lengthj and
+ * heighti <= heightj. You can rearrange any cuboid's dimensions by rotating it to put it on
+ * another cuboid.
+ *
+ * Return the maximum height of the stacked cuboids.
+ */
+
+/**
+ * @param {number[][]} cuboids
+ * @return {number}
+ */
+var maxHeight = function(cuboids) {
+  const sortedCuboids = cuboids.map(dim => dim.sort((a, b) => a - b))
+    .sort((a, b) => a[0] - b[0] || a[1] - b[1] || a[2] - b[2]);
+  const n = sortedCuboids.length;
+  const maxHeights = new Array(n).fill(0);
+  let result = 0;
+
+  for (let i = 0; i < n; i++) {
+    maxHeights[i] = sortedCuboids[i][2];
+    for (let j = 0; j < i; j++) {
+      if (sortedCuboids[j][0] <= sortedCuboids[i][0]
+          && sortedCuboids[j][1] <= sortedCuboids[i][1]
+          && sortedCuboids[j][2] <= sortedCuboids[i][2]) {
+        maxHeights[i] = Math.max(maxHeights[i], maxHeights[j] + sortedCuboids[i][2]);
+      }
+    }
+    result = Math.max(result, maxHeights[i]);
+  }
+
+  return result;
+};