Code Random Recording Algorithm Training Camp 50th Day|Special Topics on Stocks (2)

Table of contents

LeeCode 123. Best Time to Buy and Sell Stocks III

LeeCode 188. Best Time to Buy and Sell Stocks IV

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LeeCode 123. Best Time to Buy and Sell Stocks III

123. The Best Time to Buy and Sell Stocks III - LeetCode

Moving back to five parts :

1. Determine the meaning of the dp array and the subscript: 5 possible states in a day: 0. No operation; 1. Holding stocks for the first time; 2. Not holding stocks for the first time; 3. Holding stocks for the second time; 4. Do not hold stocks for the second time; in dp[i][j], i represents the i-th day, j represents the five states [0 - 4], and dp[i][j] represents the remaining state j of the i-th day maximum cash.

2.确定递推公式：dp[i][1] = max(dp[i-1][0] - prices[i], dp[i - 1][1]); dp[i][2] = max(dp[i - 1][1] + prices[i], dp[i - 1][2]); dp[i][3] = max(dp[i - 1][3], dp[i - 1][2] - prices[i]); dp[i][4] = max(dp[i - 1][4], dp[i - 1][3] + prices[i]);

3. How to initialize the dp array: dp[0][0] = 0; dp[0][1] = -prices[0]; dp[0][2] = 0; dp[0][3] = - prices[0]; dp[0][4] = 0;

4. Determine the traversal order: traverse from front to back;

5. Example of recursive dp array;

code:

class Solution {
public:
    int maxProfit(vector<int>& prices) {
    	if (prices.size() == 0) return 0;
		vector<vector<int>> dp(prices.size(), vector<int>(5, 0));
		dp[0][1] = -prices[0];
		dp[0][3] = -prices[0];
		for (int i = 1; i < prices.size(); i++) {
			dp[i][0] = dp[i - 1][0];
			dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] - prices[i]);
			dp[i][2] = max(dp[i - 1][2], dp[i - 1][1] + prices[i]);
			dp[i][3] = max(dp[i - 1][3], dp[i - 1][2] - prices[i]);
			dp[i][4] = max(dp[i - 1][4], dp[i - 1][3] + prices[i]);
		} 
        return dp[prices.size() - 1][4];
    }
};

Time complexity: O(n) Space complexity: O(5n) 

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LeeCode 188. Best Time to Buy and Sell Stocks IV

188. The Best Time to Buy and Sell Stocks IV - LeetCode

Moving back to five parts :

1. Determine the meaning of the dp array and the subscript: define a two-dimensional dp array, dp[i][j]: the state of the i-th day is j, and the remaining maximum cash;

2. Determine the recursive formula:

​for (int j = 0; j < 2 * k - 1; j += 2) {
    dp[i][j + 1] = max(dp[i - 1][j + 1], dp[i - 1][j] - prices[i]);
    dp[i][j + 2] = max(dp[i - 1][j + 2], dp[i - 1][j + 1] + prices[i]);
}

3. How to initialize the dp array:

for (int j = 1; j < 2 * k; j += 2) {
    dp[0][j] = -prices[0];
}

4. Determine the traversal order: traverse from front to back;

5. Example of recursive dp array;

code:

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
    	if (prices.size() == 0) return 0;
    	vector<vector<int>> dp(prices.size(), vector<int>(2 * k + 1, 0));
    	for (int j = 1; j < 2 * k; j += 2) {
    		dp[0][j] = -prices[0];
		}
        for (int i = 1; i < prices.size(); i++) {
        	for (int j = 0; j < 2 * k - 1; j += 2) {
        		dp[i][j + 1] = max(dp[i - 1][j + 1], dp[i - 1][j] - prices[i]);
        		dp[i][j + 2] = max(dp[i - 1][j + 2], dp[i - 1][j + 1] + prices[i]);
			}
		}
		return dp[prices.size() - 1][2 * k];
    }
};

Time complexity: O(n*k) Space complexity: O(n*k)