[LeetCode] Problem 123 - Best Time to Buy and Sell Stock III

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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

No.1

Input: [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.

No.2

Input: [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.

No.3

Input: [7,6,4,3,1]

Output: 0

Explanation: In this case, no transaction is done, i.e. max profit = 0.

Code

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// T[i][j] = max{T[i][j - 1], max{prices[j] - prices[m] + T[i - 1][m]}}, 0 <= m <= j - 1
// => T[i][j] = max{T[i][j - 1], prices[j] + maxDiff}
//    maxDiff = max{maxDiff, T[i - 1][j] - prices[j]}
public int maxProfit(int[] prices) {
    int n = prices.length;
    
    if (n == 0)
        return 0;
    
    int[] T = new int[n];
    int[] prev = new int[n];

    for (int k = 1; k <= 2; k++) {
        int maxDiff = -prices[0];

        for (int i = 1; i < n; i++) {
            T[i] = Math.max(T[i - 1], prices[i] + maxDiff);
            maxDiff = Math.max(maxDiff, prev[i] - prices[i]);
            prev[i] = T[i];
        }
    }

    return T[n - 1];
}