Description Link to heading

123.best-time-to-buy-and-sell-stock-iii

Solution Link to heading

The key point is to find what dp should denotes and the recursion formula:

dp[i] denotes only considering first i days, then by day i, there are five possible cases:

  • no operation, written as dp[i][0];
  • buy stock once in first i days, written as dp[i][1]:
    • dp[i][1] = max(dp[i - 1][0] - prices[i - 1], dp[i - 1][1])
  • sell stock once in first i days, written as dp[i][2]:
    • dp[i][2] = max(dp[i - 1][1] + prices[i - 1], dp[i - 1][2])
  • by stock twice in first i days, written as dp[i][3]:
    • dp[i][3] = max(dp[i - 1][2] - prices[i - 1], dp[i - 1][3])
  • sell stock twice in first i days, written as dp[i][4]:
    • dp[i][4] = max(dp[i - 1][3] + prices[i - 1], dp[i - 1][4])

Initializaiton:

  • dp[0][0] = 0;
  • dp[0][1] = -prices[0]; // buy once
  • dp[0][2] = 0; // buy->sell
  • dp[0][3] = -prices[0]; // buy->sell->buy
  • dp[0][4] = 0; // buy->sell->buy->sell

Code Link to heading 

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