123. Best Time to Buy and Sell Stock III - Hard

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: [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: [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: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

 

M1: dp (optimal), time = O(n), space = O(1)

class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        if(n < 2) {
            return 0;
        }
        
        int[] local = new int[3];   // local[i]: the max profit in first i days, stock must be sold on day i
        int[] global = new int[3];  // global[i]: the max profit in first i days
        
        for(int i = 1; i < n; i++) {
            int diff = prices[i] - prices[i - 1];
            for(int j = 2; j >= 1; j--) {
        // local[i-1][j]+diff是把本来在第i-1天卖出的第j次交易改在第i天卖出
        // 因为是连续的，因此不会多出一次交易                
        local[j] = Math.max(global[j - 1] + Math.max(0, diff), local[j] + diff);
                global[j] = Math.max(global[j], local[j]);
            }
        }
        return global[2];
    }
}

 

M2: dp, time = O(n), space = O(n)

class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        if(n < 2) {
            return 0;
        }
        // mustSell[i][j]: the max profit in first i days with at most j transactions, stock must be sold on day i
        // globalBest[i][j]: the max profit in first i days with at most j transactions
        int[][] mustSell = new int[n][3];
        int[][] globalBest = new int[n][3];
        
        for(int i = 1; i < n; i++) {
            int diff = prices[i] - prices[i - 1];
            mustSell[i][0] = 0;
            for(int j = 1; j <= 2; j++) {
                mustSell[i][j] = Math.max(globalBest[i - 1][j - 1] + diff, mustSell[i - 1][j] + diff);
                globalBest[i][j] = Math.max(globalBest[i - 1][j], mustSell[i][j]);
            }
        }
        return globalBest[n - 1][2];
    }
}