Best Time to Buy and Sell Stock

Say you have an array for which the ith element is the price of a given stock on day i. If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

难度：easy

Solution: Maximum Subarray

这个问题有个通用的解法：Kadane's Algorithm

public class Solution {
    public int maxProfit(int[] prices) {
        int maxRevenue = 0;
        if(prices.length == 0) {
            return maxRevenue;
        }
        int buyPrice = prices[0];
        for(int i = 1; i < prices.length; ++i) {
            maxRevenue = Math.max(maxRevenue, prices[i] - buyPrice);
            buyPrice = Math.min(buyPrice, prices[i]);
        }
        return maxRevenue;
    }
}

Best Time to Buy and Sell Stock II

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 as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

难度：easy

public class Solution {
    public int maxProfit(int[] prices) {
        int maxRevenue = 0;
        for(int i = 1; i < prices.length; ++i) {
            maxRevenue = Math.max(maxRevenue, maxRevenue + prices[i] - prices[i - 1]);
        }
        return maxRevenue;
    }
}

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).

难度：hard

Solution: Dynamic Programming

Reference

public class Solution {
    public int maxProfit(int[] prices) {
        int k = 2, maxRevenue = 0;
        if(prices.length == 0) {
            return maxRevenue;
        }
        int[][] dp = new int[k + 1][prices.length];
        for(int i = 1; i <= k; ++i) {
            int preMax = dp[i - 1][0] - prices[0];
            for(int j = 1; j < prices.length; ++j) {
                dp[i][j] = Math.max(dp[i][j - 1], preMax + prices[j]);
                preMax = Math.max(preMax, dp[i - 1][j] - prices[j]);
                maxRevenue = Math.max(maxRevenue, dp[i][j]);
            }
        }
        return maxRevenue;
    }
}

Best Time to Buy and Sell Stock IV

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 k transactions. Note: You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Solution: Dynamic Programming

class Solution {
    public int maxProfit(int k, int[] prices) {
        int m = prices.length;
        int maxRevenue = 0;
        if(m == 0) {
            return maxRevenue;
        }

        if(k >= m / 2) {
            for(int i = 1; i < m; ++i) {
                if(prices[i] > prices[i - 1]) {
                    maxRevenue += (prices[i] - prices[i - 1]);
                }
            }
        } else {
            int[][] dp = new int[k + 1][prices.length];
            for(int i = 1; i <= k; ++i) {
                int preMax = dp[i - 1][0] - prices[0];
                for(int j = 1; j < prices.length; ++j) {
                    dp[i][j] = Math.max(dp[i][j - 1], preMax + prices[j]);
                    preMax = Math.max(preMax, dp[i - 1][j] - prices[j]);
                    maxRevenue = Math.max(maxRevenue, dp[i][j]);
                }
            }
        }
        return maxRevenue;
    }
}

Best Time to Buy and Sell Stock with Cooldown

Medium

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 as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions: You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)

Solution: Dynamic Programming + State Machine

State Machine is a good way of solving DP problem with complex choices.

class Solution {
    public int maxProfit(int[] prices) {
        if(prices.length <= 1) {
            return 0;
        }
        int m = prices.length;
        int afterSell = prices[1] - prices[0], beforeBuy = 0, afterBuy = Math.max(-prices[1], -prices[0]);
        for(int i = 2; i < m; ++i) {
            int tempAS = afterSell;
            int tempBB = beforeBuy;
            int tempAB = afterBuy;
            afterSell = prices[i] + tempAB;
            beforeBuy = Math.max(tempAS, tempBB);
            afterBuy = Math.max(tempAB, tempBB - prices[i]);
        }

        return Math.max(beforeBuy, afterSell);
    }
}