Add solution #1199

JoshCrozier · JoshCrozier · commit 79c9a2db8f90 · 2025-06-28T00:31:36.000-05:00
diff --git a/README.md b/README.md
@@ -1117,6 +1117,7 @@
 1196|[How Many Apples Can You Put into the Basket](./solutions/1196-how-many-apples-can-you-put-into-the-basket.js)|Easy|
 1197|[Minimum Knight Moves](./solutions/1197-minimum-knight-moves.js)|Medium|
 1198|[Find Smallest Common Element in All Rows](./solutions/1198-find-smallest-common-element-in-all-rows.js)|Medium|
+1199|[Minimum Time to Build Blocks](./solutions/1199-minimum-time-to-build-blocks.js)|Hard|
 1200|[Minimum Absolute Difference](./solutions/1200-minimum-absolute-difference.js)|Easy|
 1201|[Ugly Number III](./solutions/1201-ugly-number-iii.js)|Medium|
 1202|[Smallest String With Swaps](./solutions/1202-smallest-string-with-swaps.js)|Medium|
diff --git a/solutions/1199-minimum-time-to-build-blocks.js b/solutions/1199-minimum-time-to-build-blocks.js
@@ -0,0 +1,42 @@
+/**
+ * 1199. Minimum Time to Build Blocks
+ * https://leetcode.com/problems/minimum-time-to-build-blocks/
+ * Difficulty: Hard
+ *
+ * You are given a list of blocks, where blocks[i] = t means that the i-th block needs
+ * t units of time to be built. A block can only be built by exactly one worker.
+ *
+ * A worker can either split into two workers (number of workers increases by one) or
+ * build a block then go home. Both decisions cost some time.
+ *
+ * The time cost of spliting one worker into two workers is given as an integer split.
+ * Note that if two workers split at the same time, they split in parallel so the cost
+ * would be split.
+ *
+ * Output the minimum time needed to build all blocks.
+ *
+ * Initially, there is only one worker.
+ */
+
+/**
+ * @param {number[]} blocks
+ * @param {number} split
+ * @return {number}
+ */
+var minBuildTime = function(blocks, split) {
+  const minHeap = new PriorityQueue((a, b) => a - b);
+
+  for (const block of blocks) {
+    minHeap.enqueue(block);
+  }
+
+  while (minHeap.size() > 1) {
+    const first = minHeap.dequeue();
+    const second = minHeap.dequeue();
+
+    const mergedTime = split + Math.max(first, second);
+    minHeap.enqueue(mergedTime);
+  }
+
+  return minHeap.dequeue();
+};
