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Multiple States VI

Description

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).

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/description/

Example I:

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 II:

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 III:

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

Questions

  • LeetCode - 123. Best Time to Buy and Sell Stock III

Transform Function

before[i] = Math.max(before[i - 1], prices[i] - min);
after[i] = Math.max(after[i + 1], max - prices[i]);
res = Math.max(before[i] + after[i], res);

Template I

class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null || prices.length <= 1) return 0;
        int n = prices.length;
        int[] before = new int[n];
        int min = prices[0];
        for(int i = 1; i < n; i++) {
            min = Math.min(min, prices[i]);
            before[i] = Math.max(before[i - 1], prices[i] - min);
        }
        int[] after = new int[n];
        int max = prices[n - 1];
        for(int i = n - 2; i >= 0; i--) {
            max = Math.max(prices[i], max);
            after[i] = Math.max(after[i + 1], max - prices[i]);
        }
        int res = 0;
        for(int i = 0; i < n; i++) res = Math.max(before[i] + after[i], res);
        return res;
    }

}

Optimized

hold_one[i] = Math.max(-prices[i], hold_one[i - 1])
hold_one_sell_one = Math.max(hold_one[i - 1] + prices[i], hold_one_sell_one[i - 1])
hold_two_sell_one = Math.max(hold_one_sell_one[i - 1] - prices[i], hold_two_sell_one[i - 1])
hold_two_sell_two = Math.max(hold_two_sell_one[i - 1] + prices[i], hold_two_sell_two[i - 1])

Template II

class Solution {
    public int maxProfit(int[] prices) {
         if(prices == null || prices.length == 0) return 0;
         int n = prices.length;
         int[] holdOne = new int[n + 1];
         int[] holdOneSellOne = new int[n + 1];
         int[] holdTwoSellOne = new int[n + 1];
         int[] holdTwoSellTwo = new int[n + 1];
         holdOne[0] = Integer.MIN_VALUE;
         holdTwoSellOne[0] = Integer.MIN_VALUE;
         for(int i = 1; i <= n; i++) {
             holdOne[i] = Math.max(holdOne[i - 1], -prices[i - 1]);
             holdOneSellOne[i] = Math.max(holdOneSellOne[i - 1], holdOne[i - 1] + prices[i - 1]);
             holdTwoSellOne[i] = Math.max(holdTwoSellOne[i - 1], holdOneSellOne[i - 1] - prices[i - 1]);
             holdTwoSellTwo[i] = Math.max(holdTwoSellTwo[i - 1], holdTwoSellOne[i - 1] + prices[i - 1]);
         }

         return holdTwoSellTwo[n];
     }
}

Optimized

int hold_one = Integer.MIN_VALUE;
int hold_one_sell_one = 0;
int hold_two_sell_one = Integer.MIN_VALUE;
int hold_two_sell_two = 0;

Template III

class Solution {
    public int maxProfit(int[] prices) {
        int hold_one = Integer.MIN_VALUE, hold_one_sell_one = 0;
        int hold_two_sell_one = Integer.MIN_VALUE;
        int hold_two_sell_two = 0;
        for(int p : prices) {
            hold_one = Math.max(hold_one, -p);
            hold_one_sell_one = Math.max(hold_one + p, hold_one_sell_one);
            hold_two_sell_one = Math.max(hold_one_sell_one - p, hold_two_sell_one);
            hold_two_sell_two = Math.max(hold_two_sell_one + p, hold_two_sell_two);
        }
        return hold_two_sell_two;
    }
}