123. Best Time to Buy and Sell Stock III.md

123. Best Time to Buy and Sell Stock III

Question

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.

Thinking:

• Method1:DP

class Solution {
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length <= 1) return 0;
        int[] local = new int[3];
        int[] global = new int[3];
        for(int i = 0; i < prices.length - 1; i++){
            int diff = prices[i + 1] - prices[i];
            for(int j = 2; j >= 1; j--){
                local[j] = Math.max(global[j - 1] + Math.max(diff, 0), local[j] + diff);
                global[j] = Math.max(global[j], local[j]);
            }
        }
        return global[2];
    }
}

Second time

1. (Wrong)I tried my best to solve this question but cannot pass one case: [1,2,4,2,5,7,2,4,9,0], maybe I can use recursion to solve this question, but not a good idea.

class Solution {
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length == 0) return 0;
        int first = 0, second = 0, pre = -1, temp = 0;;
        for(int i = 1; i < prices.length; i++){
            if(prices[i] < prices[i - 1]){
                temp = pre == -1 ? prices[i - 1] - prices[0] : prices[i - 1] - prices[pre];
                pre = i;
                if(temp >= first){
                    second = first;
                    first = temp;
                }else if(temp > second) second = temp;
            }
        }
        if(pre == -1) return prices[prices.length - 1] - prices[0];
        if(pre != prices.length - 1){
            temp = prices[prices.length - 1] - prices[pre];
            if(temp >= first){
                second = first;
                first = temp;
            }else if(temp > second) second = temp;
        }
        return first + second;
    }
}

Amazon Session

• Method 1: DP

   class Solution {
   	public int maxProfit(int[] prices) {
   		if(prices == null || prices.length <= 1) return 0;
   		int len = prices.length;
   		int[][] buys = new int[len + 1][3];
   		int[][] sells = new int[len + 1][3];
   		buys[1][1] = -prices[0];
   		buys[1][2] = Integer.MIN_VALUE;
   		for(int i = 2; i <= len; i++){
   			for(int j = 1; j <= 2; j++){
   				buys[i][j] = Math.max(buys[i - 1][j], sells[i - 1][j - 1] - prices[i - 1]);
   				sells[i][j] = Math.max(sells[i - 1][j], buys[i - 1][j] + prices[i - 1]);
   			}
   		}       
   		return Math.max(sells[len][0], Math.max(sells[len][1], sells[len][2]));
   	}
   }