188. Best Time to Buy and Sell Stock IV

Say you have an array for which the i-th 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).

Example 1:

Input: [2,4,1], k = 2
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Example 2:

Input: [3,2,6,5,0,3], k = 2
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
             Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

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

dp(i, j) ==> max profit of considering prices[0:j] and transaction of k times.

dp(i, j) will be max of :

• for every j, we assume we sell it at j，then we must have at least profit of prices[j] 。

• Then we should find where to buy in, which surely happens before j.

• so we should find the max of profit of 「dp(i-1, m) - prices[m] 」for every m<j. （meaning get profit of m prices and buy-in the m-th stock）