LeetCode – Best Time to Buy and Sell Stock III (Java)
Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete at most two transactions.
Note:
A transaction is a buy & a sell. You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
Analysis
Comparing to I and II, III limits the number of transactions to 2. This can be solve by "devide and conquer". We use left[i] to track the maximum profit for transactions before i, and use right[i] to track the maximum profit for transactions after i. You can use the following example to understand the Java solution:
Prices: 1 4 5 7 6 3 2 9 left = [0, 3, 4, 6, 6, 6, 6, 8] right= [8, 7, 7, 7, 7, 7, 7, 0]
The maximum profit = 13
Java Solution
public int maxProfit(int[] prices) { if (prices == null  prices.length < 2) { return 0; } //highest profit in 0 ... i int[] left = new int[prices.length]; int[] right = new int[prices.length]; // DP from left to right left[0] = 0; int min = prices[0]; for (int i = 1; i < prices.length; i++) { min = Math.min(min, prices[i]); left[i] = Math.max(left[i  1], prices[i]  min); } // DP from right to left right[prices.length  1] = 0; int max = prices[prices.length  1]; for (int i = prices.length  2; i >= 0; i) { max = Math.max(max, prices[i]); right[i] = Math.max(right[i + 1], max  prices[i]); } int profit = 0; for (int i = 0; i < prices.length; i++) { profit = Math.max(profit, left[i] + right[i]); } return profit; } 
<pre><code> String foo = "bar"; </code></pre>

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