LeetCode 123. Best Time to Buy and Sell Stock III

123. Best Time to Buy and Sell Stock III

Difficulty: Hard

Say you have an array for which the i^th 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.

Solution

Language: Java

分段DP，以i为分界线

public class Solution {
    public int maxProfit(int[] prices) {
        if (prices.length < 2) return 0;
        
        int n = prices.length;
        int[] preProfit = new int[n];
        int[] postProfit = new int[n];
        
        int curMin = prices[0];
        for (int i = 1; i < n; i++) {
            curMin = Math.min(curMin, prices[i]);
            preProfit[i] = Math.max(preProfit[i - 1], prices[i] - curMin);
        }
        
        int curMax = prices[n - 1];
        for (int i = n - 2; i >= 0; i--) {
            curMax = Math.max(curMax, prices[i]);
            postProfit[i] = Math.max(postProfit[i + 1], curMax - prices[i]);
        }
        
        int maxProfit = 0;
        for (int i = 0; i < n; i++) {
            maxProfit = Math.max(maxProfit, preProfit[i] + postProfit[i]);
        }
        
        return  maxProfit;
    }
}

简化DP

class Solution {
      public int maxProfit(int[] prices) {
            int hold1 = Integer.MIN_VALUE, hold2 = Integer.MIN_VALUE;
            int release1 = 0, release2 = 0;
            for(int i:prices){                                             // Assume we only have 0 money at first
                  hold1      = Math.max(hold1,      -i);               // The maximum if we've just buy  1st stock so far. 
                  release1 = Math.max(release1, hold1+i);       // The maximum if we've just sold 1nd stock so far.
                  hold2      = Math.max(hold2,      release1-i);   // The maximum if we've just buy  2nd stock so far.
                  release2 = Math.max(release2, hold2+i);       // The maximum if we've just sold 2nd stock so far.
           }
            return release2; ///Since release1 is initiated as 0, so release2 will always higher than release1.
     }
}