Add solution #188

Author: JoshCrozier

README.md

@@ -175,6 +175,7 @@
 174|[Dungeon Game](./0174-dungeon-game.js)|Hard|
 179|[Largest Number](./0179-largest-number.js)|Medium|
 187|[Repeated DNA Sequences](./0187-repeated-dna-sequences.js)|Medium|
+188|[Best Time to Buy and Sell Stock IV](./0188-best-time-to-buy-and-sell-stock-iv.js)|Hard|
 189|[Rotate Array](./0189-rotate-array.js)|Medium|
 190|[Reverse Bits](./0190-reverse-bits.js)|Easy|
 191|[Number of 1 Bits](./0191-number-of-1-bits.js)|Easy|

solutions/0188-best-time-to-buy-and-sell-stock-iv.js

@@ -0,0 +1,42 @@
+/**
+ * 188. Best Time to Buy and Sell Stock IV
+ * https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/
+ * Difficulty: Hard
+ *
+ * You are given an integer array prices where prices[i] is the price of a given stock on
+ * the ith day, and an integer k.
+ *
+ * Find the maximum profit you can achieve. You may complete at most k transactions: i.e.
+ * you may buy at most k times and sell at most k times.
+ *
+ * Note: You may not engage in multiple transactions simultaneously (i.e., you must sell
+ * the stock before you buy again).
+ */
+
+/**
+ * @param {number} k
+ * @param {number[]} prices
+ * @return {number}
+ */
+var maxProfit = function(k, prices) {
+  if (prices.length < 2 || k === 0) return 0;
+
+  if (k >= prices.length / 2) {
+    let profit = 0;
+    for (let i = 1; i < prices.length; i++) {
+      profit += prices[i] > prices[i - 1] ? (prices[i] - prices[i - 1]) : 0;
+    }
+    return profit;
+  }
+
+  const buy = new Array(k + 1).fill(-Infinity);
+  const sell = new Array(k + 1).fill(0);
+  for (let i = 0; i < prices.length; i++) {
+    for (let j = k; j >= 1; j--) {
+      sell[j] = Math.max(sell[j], buy[j] + prices[i]);
+      buy[j] = Math.max(buy[j], sell[j - 1] - prices[i]);
+    }
+  }
+
+  return sell[k];
+};