123. Best Time to Buy and Sell Stock III

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 (ie, you must sell the stock before you buy again).

Solution 1

public static int maxProfit2(int[] prices) {if (prices == null || prices.length == 0) return 0;    int lenght = prices.length;    int[] leftProfit = new int[lenght];    int leftMaxProfit = 0;    int leftMin = prices[0];    for (int i=0; i<lenght; i++) {        if (prices[i] < leftMin) leftMin = prices[i];        if (prices[i] - leftMin > leftMaxProfit) leftMaxProfit = prices[i]-leftMin;        leftProfit[i] = leftMaxProfit;    }    int maxProfit = 0;    int rightMaxProfit = 0;    int rightMax = prices[lenght-1];    for (int i=lenght-1; i>=0; i--) {        if (prices[i] > rightMax) rightMax = prices[i];        if (rightMax - prices[i] > rightMaxProfit) rightMaxProfit = rightMax - prices[i];        int currentProfit = rightMaxProfit + (i>0 ? leftProfit[i-1] : 0);        if (currentProfit > maxProfit) {            maxProfit = currentProfit;        }    }    return maxProfit;}

Solution 2 DP

public static int maxProfit3(int[] prices) {    if(prices == null || prices.length == 0) return 0;    int[] dp = new int[prices.length];//k == 0, dp[i] = 0    int K = 2, tmpMax = 0;      for(int k =1; k<=K; k++) {        tmpMax = dp[0] - prices[0];        dp[0] = 0;        for(int i = 1; i<prices.length; i++) {            tmpMax = Math.max(tmpMax, dp[i] - prices[i]);            dp[i] = Math.max(dp[i-1], prices[i]+tmpMax);        }    }    return dp[prices.length -1];}

Solution 3

Best solution O(N)

First assume that we have no money, so buy1 means that we have to borrow money from others,  we want to borrow less so that we have to make our balance as max as we can(because this is negative).


sell1 means we decide to sell the stock, after selling it we have price[i] money and we have to give back the money we owed, so we have price[i] - |buy1| = prices[i ] + buy1, we want to make this max.


buy2 means we want to buy another stock, we already have sell1 money, so after buying stock2 we have buy2 = sell1 - price[i] money left, we want more money left, so we make it max


sell2 means we want to sell stock2, we can have price[i] money after selling it, and we have buy2 money left before, so sell2 = buy2 + prices[i], we make this max.

public int maxProfit(int[] prices) {         int sell1 = 0, sell2 = 0, buy1 = Integer.MIN_VALUE, buy2 = Integer.MIN_VALUE;        for(int i = 0; i < prices.length; i++){            buy1 = Math.max(buy1, -prices[i]);            sell1 = Math.max(sell1, buy1 + prices[i]);            buy2 = Math.max(buy2, sell1 - prices[i]);            sell2 = Math.max(sell2, buy2 + prices[i]);        }        return sell2;    }