Algorithm
Problem Name: 714. Best Time to Buy and Sell Stock with Transaction Fee
You are given an array prices
where prices[i]
is the price of a given stock on the ith
day, and an integer fee
representing a transaction fee.
Find the maximum profit you can achieve. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction.
Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
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.
Example 2:
Input: prices = [1,3,7,5,10,3], fee = 3 Output: 6
Constraints:
1 <= prices.length <= 5 * 104
1 <= prices[i] < 5 * 104
0 <= fee < 5 * 104
Code Examples
#1 Code Example with C Programming
Code -
C Programming
int _max(a, b) {
if (a > b) return a;
return b;
}
int maxProfit(int* prices, int pricesSize, int fee) {
int i, p;
#if 1
buy = 0x80000000;
sell = 0;
for (i = 0; i < pricesSize; i ++) {
p = prices[i];
buy = _max(buy, sell - p);
sell = _max(sell, buy + p - fee);
}
#else
buy = - prices[0];
sell = 0;
for (i = 1; i < pricesSize; i ++) {
p = prices[i];
sell = _max(sell, buy + p - fee);
buy = _max(buy, sell - p);
}
#endif
return sell;
}
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#2 Code Example with Javascript Programming
Code -
Javascript Programming
const maxProfit = function(prices, fee) {
let cash = 0,
hold = -prices[0];
for (let i = 1; i < prices.length; i++) {
cash = Math.max(cash, hold + prices[i] - fee);
hold = Math.max(hold, cash - prices[i]);
}
return cash;
};
console.log(maxProfit([1, 3, 2, 8, 4, 9], 2));
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#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def maxProfit(self, prices, fee):
pre = [0, -float("inf")]
for p in prices:
p0, p1 = pre[1] + p - fee, pre[0] - p
if p0 > pre[0]: pre[0] = p0
if p1 > pre[1]: pre[1] = p1
return pre[0]
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