Best Time to Buy and Sell Stock IV
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).
Solution:
public class Solution {
public int maxProfit(int k, int[] prices) {
int n = prices.length;
if (k >= n) {
return solveMaxProfit(prices);
}
// 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.
int[][] local = new int[n][k + 1];
int[][] global = new int[n][k + 1];
for (int i = 1; i < n; 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[n - 1][k];
}
private int solveMaxProfit(int[] prices) {
int profit = 0;
for (int i = 1; i < prices.length; i++) {
// as long as there is a price gap, we gain a profit.
if (prices[i] > prices[i - 1]) {
profit += prices[i] - prices[i - 1];
}
}
return profit;
}
}