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: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
Example 1:
Input: prices = [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
Example 2:
Input: prices = [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.
Example 3:
Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
Example 4:
Input: prices = [1]
Output: 0
Constraints:
1 <= prices.length <= 105
0 <= prices[i] <= 105
Solution:
class Solution {
public int maxProfit(int[] prices) {
int n = prices.length;
// (i, j, k) = max profit on day i with or w/ stock, with at most k txns
int[][][] dp = new int[n][2][3];
dp[0][1][0] = Integer.MIN_VALUE;
dp[0][1][1] = -prices[0];
dp[0][1][2] = -prices[0];
int max = 0;
for (int i = 1; i < n; i ++) {
for (int k = 2; k > 0; k --) {
dp[i][0][k] = Math.max(dp[i - 1][0][k], dp[i - 1][1][k] + prices[i]);
dp[i][1][k] = Math.max(dp[i - 1][1][k], dp[i - 1][0][k - 1] - prices[i]);
max = Math.max(max, dp[i][0][k]);
}
}
return max;
}
}