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.

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

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

Analyse: Assume global[i][j] represents the maximum profit at day i with j transactions, local [i][j] is the maximum profit at day i  with j transactions at the last transaction occurred on day i. Then we have: global[i][j] = max(local[i][j], global[i - 1][j]) (the left one is doing the transaction at day i while the right one is not doing the transaction at day i), local[i][j] = max(local[i - 1][j] + diff, global[i - 1][j - 1] + max(0, diff)). (we have to do the j-th transaction on day i, so if we already did j transactions at day i - 1 and did the last transaction on the (i - 1)th day, or we did j - 1 transactions by the (i - 1)th day, then we choose the larger one. 

Runtime: 8ms.

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        if(prices.size() <= 1 || k == 0) return 0;
                
        int result = 0;
        if(k >= prices.size()){//cannot do k transactions, then do all operations which can earn money
            for(int i = 1; i < prices.size(); i++){
                if(prices[i] > prices[i - 1])
                    result += prices[i] - prices[i - 1];
            }
            return result;
        }
       
        const int n = k + 1;
        int l[n] = {0}, g[n] = {0};
        
        for(int i = 0; i < prices.size() - 1; i++){
            int diff = prices[i + 1] - prices[i];
            for(int j = k; j >= 1; j--){
                l[j] = max(g[j - 1] + max(0, diff), l[j] + diff);
                g[j] = max(l[j], g[j]);
            }
        }
        return g[k];
    }
};