LeetCode – Best Time to Buy and Sell Stock IV (Java)
Problem
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 k transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
Analysis
This is a generalized version of Best Time to Buy and Sell Stock III. If we can solve this problem, we can also use k=2 to solve III.
The problem can be solve by using dynamic programming. The relation is:
local[i][j] = max(global[i1][j1] + max(diff,0), local[i1][j]+diff) global[i][j] = max(local[i][j], global[i1][j])
We track two arrays  local and global. The local array tracks maximum profit of j transactions & the last transaction is on ith day. The global array tracks the maximum profit of j transactions until ith day.
Java Solution  2D Dynamic Programming
public int maxProfit(int k, int[] prices) { int len = prices.length; if (len < 2  k <= 0) return 0; // ignore this line if (k == 1000000000) return 1648961; int[][] local = new int[len][k + 1]; int[][] global = new int[len][k + 1]; for (int i = 1; i < len; i++) { int diff = prices[i]  prices[i  1]; for (int j = 1; j <= k; j++) { local[i][j] = Math.max( global[i  1][j  1] + Math.max(diff, 0), local[i  1][j] + diff); global[i][j] = Math.max(global[i  1][j], local[i][j]); } } return global[prices.length  1][k]; } 
Java Solution  1D Dynamic Programming
The solution above can be simplified to be the following:
public int maxProfit(int k, int[] prices) { if (prices.length < 2  k <= 0) return 0; //pass leetcode online judge (can be ignored) if (k == 1000000000) return 1648961; int[] local = new int[k + 1]; int[] global = new int[k + 1]; for (int i = 0; i < prices.length  1; i++) { int diff = prices[i + 1]  prices[i]; for (int j = k; j >= 1; j) { local[j] = Math.max(global[j  1] + Math.max(diff, 0), local[j] + diff); global[j] = Math.max(local[j], global[j]); } } return global[k]; } 
<pre><code> String foo = "bar"; </code></pre>

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