An extension of Best Time to Buy and Sell Stock III. The idea is still to use dynamic programming (see here for detailed introduction). However, in this problem, some trick (the quickProfit function below) is required to pass the TLE.
The code is rewritten below.
1 class Solution { 2 public: 3 int maxProfit(int k, vector<int>& prices) { 4 int n = prices.size(); 5 if (k >= n / 2) return quickProfit(prices); 6 vector<vector<int> > dp(k + 1, vector<int>(n, 0)); 7 for (int i = 1; i <= k; i++) { 8 int temp = dp[i - 1][0] - prices[0]; 9 for (int j = 1; j < n; j++) { 10 dp[i][j] = max(dp[i][j - 1], prices[j] + temp); 11 temp = max(temp, dp[i - 1][j] - prices[j]); 12 } 13 } 14 return dp[k][n - 1]; 15 } 16 private: 17 int quickProfit(vector<int>& prices) { 18 int n = prices.size(), profit = 0; 19 for (int i = 1; i < n; i++) 20 if (prices[i] > prices[i - 1]) 21 profit += prices[i] - prices[i - 1]; 22 return profit; 23 } 24 };