Add solution #502

Author: JoshCrozier

README.md

@@ -164,6 +164,7 @@
 492|[Construct the Rectangle](./0492-construct-the-rectangle.js)|Easy|
 500|[Keyboard Row](./0500-keyboard-row.js)|Easy|
 501|[Find Mode in Binary Search Tree](./0501-find-mode-in-binary-search-tree.js)|Easy|
+502|[IPO](./0502-ipo.js)|Hard|
 506|[Relative Ranks](./0506-relative-ranks.js)|Easy|
 509|[Fibonacci Number](./0509-fibonacci-number.js)|Easy|
 520|[Detect Capital](./0520-detect-capital.js)|Easy|

solutions/0502-ipo.js

@@ -0,0 +1,52 @@
+/**
+ * 502. IPO
+ * https://leetcode.com/problems/ipo/
+ * Difficulty: Hard
+ *
+ * Suppose LeetCode will start its IPO soon. In order to sell a good price of
+ * its shares to Venture Capital, LeetCode would like to work on some projects
+ * to increase its capital before the IPO. Since it has limited resources, it
+ * can only finish at most k distinct projects before the IPO. Help LeetCode
+ * design the best way to maximize its total capital after finishing at most
+ * k distinct projects.
+ *
+ * You are given n projects where the ith project has a pure profit profits[i]
+ * and a minimum capital of capital[i] is needed to start it.
+ *
+ * Initially, you have w capital. When you finish a project, you will obtain
+ * its pure profit and the profit will be added to your total capital.
+ *
+ * Pick a list of at most k distinct projects from given projects to maximize
+ * your final capital, and return the final maximized capital.
+ *
+ * The answer is guaranteed to fit in a 32-bit signed integer.
+ */
+
+/**
+ * @param {number} k
+ * @param {number} w
+ * @param {number[]} profits
+ * @param {number[]} capital
+ * @return {number}
+ */
+var findMaximizedCapital = function(k, w, profits, capital) {
+  const queue = new MinPriorityQueue();
+  const descendingQueue = new MaxPriorityQueue();
+
+  for (let i = 0; i < capital.length; i++) {
+    queue.enqueue([capital[i], profits[i]], capital[i]);
+  }
+
+  for (let i = 0; i < k; i++) {
+    while (!queue.isEmpty() && queue.front().element[0] <= w) {
+      const element = queue.dequeue().element;
+      descendingQueue.enqueue(element, element[1]);
+    }
+    if (descendingQueue.isEmpty()) {
+      return w;
+    }
+    w += descendingQueue.dequeue().element[1];
+  }
+
+  return w;
+};