# Best Time to Buy and Sell Stock with Transaction Fee

Oct 28, 2017

Your are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer fee representing a transaction fee.

You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)

Return the maximum profit you can make. Example 1: Input: prices = [1, 3, 2, 8, 4, 9], fee = 2 Output: 8 Explanation: The maximum profit can be achieved by: Buying at prices[0] = 1 Selling at prices[3] = 8 Buying at prices[4] = 4 Selling at prices[5] = 9 The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8. Note: 0 < prices.length <= 50000. 0 < prices[i] < 50000. 0 <= fee < 50000.

int maxProfit(vector<int>& prices, int fee) {
if (prices.size() == 0) return 0;
vector<int> profit(prices.size(),0);
profit[0] = 0;
for(int i = 1;i < profit.size();i++) {
for(int j = i-1;j>=0;j--) {
profit[i] = max(profit[i],profit[j]);
int t = profit[j] + prices[i] - prices[j] - fee;
profit[i] = max(profit[i],t);
}
}
return *profit.rbegin();
}


int maxProfit(vector<int>& prices, int fee) {
int s0 = 0,s1 = INT_MIN;
for(auto p:prices) {
int t = s0;
s0 = max(s0,s1+p);      //sell
}
return s0;
}


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