**Best Time to Buy and Sell Stock IV**

2015.4.17 05:27

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).

**Solution:**

The "Best Time to Buy and Sell Stock" series has been appearing in several version. Apparently this one is about dynamic programming.

At first I was a bit confused about the definition of "transaction" in the problem desription.

It was one "buy" **or** one "sell" that is defined as a "transaction", not **one buy and one sell**.

Thus here comes the typical two-demensional DP. Please see the code below, notice that the key idea lies in the local optimal and global optimal.

I haven't understood the algorithm well enough, though. Yet to think about it.

One thing is for sure: when k is greater than n, this problem is simplified to a greedy version.

**Accepted code:**

1 // Yet to grasp the mechanism here... DP wasn't easy. 2 #include <algorithm> 3 #include <vector> 4 using namespace std; 5 6 class Solution { 7 public: 8 int maxProfit(int k, vector<int> &prices) { 9 int n = prices.size(); 10 11 if (k >= n) { 12 return easyMaxProfit(prices); 13 } 14 15 int i, j; 16 vector<int> local, global; 17 18 local.resize(n + 1, 0); 19 global.resize(n + 1, 0); 20 int diff; 21 for (i = 1; i < n; ++i) { 22 diff = prices[i] - prices[i - 1]; 23 for (j = k; j >= 1; --j) { 24 local[j] = max(global[j - 1] + max(diff, 0), local[j] + diff); 25 global[j] = max(global[j], local[j]); 26 } 27 } 28 return global[k]; 29 } 30 private: 31 int easyMaxProfit(vector<int> &prices) { 32 int res; 33 int i, len = prices.size(); 34 35 res = 0; 36 for(i = 0; i < len - 1; ++i){ 37 if(prices[i] < prices[i + 1]){ 38 res += prices[i + 1] - prices[i]; 39 } 40 } 41 42 return res; 43 } 44 };