Best Time to Buy and Sell Stock IV

Question

leetcode: Best Time to Buy and Sell Stock IV | LeetCode OJ
lintcode: (393) Best Time to Buy and Sell Stock IV

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.

Example
Given prices = [4,4,6,1,1,4,2,5], and k = 2, return 6.

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

Challenge
O(nk) time.
copy

题解1

卖股票系列中最难的一道，较易实现的方法为使用动态规划，动规的实现又分为大约3大类方法，这里先介绍一种最为朴素的方法，过不了大量数据，会 TLE.

最多允许 k 次交易，由于一次增加收益的交易至少需要两天，故当 k >= n/2时，此题退化为卖股票的第二道题，即允许任意多次交易。当 k < n/2 时，使用动规来求解，动规的几个要素如下：

f[i][j] 代表第 i 天为止交易 k 次获得的最大收益，那么将问题分解为前 x 天交易 k-1 次，第 x+1 天至第 i 天交易一次两个子问题，于是动态方程如下：

f[i][j] = max(f[x][j - 1] + profit(x + 1, i))
copy

简便起见，初始化二维矩阵为0，下标尽可能从1开始，便于理解。

Python

class Solution:
    """
    @param k: an integer
    @param prices: a list of integer
    @return: an integer which is maximum profit
    """
    def maxProfit(self, k, prices):
        if prices is None or len(prices) <= 1 or k <= 0:
            return 0

        n = len(prices)
        # k >= prices.length / 2 ==> multiple transactions Stock II
        if k >= n / 2:
            profit_max = 0
            for i in xrange(1, n):
                diff = prices[i] - prices[i - 1]
                if diff > 0:
                    profit_max += diff
            return profit_max

        f = [[0 for i in xrange(k + 1)] for j in xrange(n + 1)]
        for j in xrange(1, k + 1):
            for i in xrange(1, n + 1):
                for x in xrange(0, i + 1):
                    f[i][j] = max(f[i][j], f[x][j - 1] + self.profit(prices, x + 1, i))

        return f[n][k]

    # calculate the profit of prices(l, u)
    def profit(self, prices, l, u):
        if l >= u:
            return 0
        valley = 2**31 - 1
        profit_max = 0
        for price in prices[l - 1:u]:
            profit_max = max(profit_max, price - valley)
            valley = min(valley, price)
        return profit_max
copy

C++

class Solution {
public:
    /**
     * @param k: An integer
     * @param prices: Given an integer array
     * @return: Maximum profit
     */
    int maxProfit(int k, vector<int> &prices) {
        if (prices.size() <= 1 || k <= 0) return 0;

        int n = prices.size();
        // k >= prices.length / 2 ==> multiple transactions Stock II
        if (k >= n / 2) {
            int profit_max = 0;
            for (int i = 1; i < n; ++i) {
                int diff = prices[i] - prices[i - 1];
                if (diff > 0) {
                    profit_max += diff;
                }
            }
            return profit_max;
        }

        vector<vector<int> > f = vector<vector<int> >(n + 1, vector<int>(k + 1, 0));
        for (int j = 1; j <= k; ++j) {
            for (int i = 1; i <= n; ++i) {
                for (int x = 0; x <= i; ++x) {
                    f[i][j] = max(f[i][j], f[x][j - 1] + profit(prices, x + 1, i));
                }
            }
        }

        return f[n][k];
    }

private:
    int profit(vector<int> &prices, int l, int u) {
        if (l >= u) return 0;

        int valley = INT_MAX;
        int profit_max = 0;
        for (int i = l - 1; i < u; ++i) {
            profit_max = max(profit_max, prices[i] - valley);
            valley = min(valley, prices[i]);
        }

        return profit_max;
    }
};
copy

Java

class Solution {
    /**
     * @param k: An integer
     * @param prices: Given an integer array
     * @return: Maximum profit
     */
    public int maxProfit(int k, int[] prices) {
        if (prices == null || prices.length <= 1 || k <= 0) return 0;

        int n = prices.length;
        if (k >= n / 2) {
            int profit_max = 0;
            for (int i = 1; i < n; i++) {
                if (prices[i] - prices[i - 1] > 0) {
                    profit_max += prices[i] - prices[i - 1];
                }
            }
            return profit_max;
        }

        int[][] f = new int[n + 1][k + 1];
        for (int j = 1; j <= k; j++) {
            for (int i = 1; i <= n; i++) {
                for (int x = 0; x <= i; x++) {
                    f[i][j] = Math.max(f[i][j], f[x][j - 1] + profit(prices, x + 1, i));
                }
            }
        }

        return f[n][k];
    }

    private int profit(int[] prices, int l, int u) {
        if (l >= u) return 0;

        int valley = Integer.MAX_VALUE;
        int profit_max = 0;
        for (int i = l - 1; i < u; i++) {
            profit_max = Math.max(profit_max, prices[i] - valley);
            valley = Math.min(valley, prices[i]);
        }
        return profit_max;
    }
};
copy

源码分析

注意 Python 中的多维数组初始化方式，不可简单使用[[0] * k] * n], 具体原因是因为 Python 中的对象引用方式。可以优化的地方是 profit 方法及最内存循环。

复杂度分析

三重循环，时间复杂度近似为 $O(n^2 \cdot k)$ , 使用了 f 二维数组，空间复杂度为 $O(n \cdot k)$ .