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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:
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 * 1041 <= prices[i] < 5 * 1040 <= fee < 5 * 104We design a function $dfs(i, j)$, which represents the maximum profit that can be obtained starting from day $i$ with state $j$. Here, $j$ can take the values $0$ and $1$, representing not holding and holding a stock, respectively. The answer is $dfs(0, 0)$.
The execution logic of the function $dfs(i, j)$ is as follows:
If $i \geq n$, there are no more stocks to trade, so we return $0$.
Otherwise, we can choose not to trade, in which case $dfs(i, j) = dfs(i + 1, j)$. We can also choose to trade stocks. If $j \gt 0$, it means that we currently hold a stock and can sell it. In this case, $dfs(i, j) = prices[i] + dfs(i + 1, 0) - fee$. If $j = 0$, it means that we currently do not hold a stock and can buy one. In this case, $dfs(i, j) = -prices[i] + dfs(i + 1, 1)$. We take the maximum value as the return value of the function $dfs(i, j)$.
The answer is $dfs(0, 0)$.
To avoid redundant calculations, we use memoization to record the return value of $dfs(i, j)$ in an array $f$. If $f[i][j]$ is not equal to $-1$, it means that we have already calculated it, so we can directly return $f[i][j]$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $prices$.
class Solution:
def maxProfit(self, prices: List[int], fee: int) -> int:
@cache
def dfs(i: int, j: int) -> int:
if i >= len(prices):
return 0
ans = dfs(i + 1, j)
if j:
ans = max(ans, prices[i] + dfs(i + 1, 0) - fee)
else:
ans = max(ans, -prices[i] + dfs(i + 1, 1))
return ans
return dfs(0, 0)
class Solution {
private Integer[][] f;
private int[] prices;
private int fee;
public int maxProfit(int[] prices, int fee) {
f = new Integer[prices.length][2];
this.prices = prices;
this.fee = fee;
return dfs(0, 0);
}
private int dfs(int i, int j) {
if (i >= prices.length) {
return 0;
}
if (f[i][j] != null) {
return f[i][j];
}
int ans = dfs(i + 1, j);
if (j > 0) {
ans = Math.max(ans, prices[i] + dfs(i + 1, 0) - fee);
} else {
ans = Math.max(ans, -prices[i] + dfs(i + 1, 1));
}
return f[i][j] = ans;
}
}
class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
int n = prices.size();
int f[n][2];
memset(f, -1, sizeof(f));
function<int(int, int)> dfs = [&](int i, int j) {
if (i >= prices.size()) {
return 0;
}
if (f[i][j] != -1) {
return f[i][j];
}
int ans = dfs(i + 1, j);
if (j) {
ans = max(ans, prices[i] + dfs(i + 1, 0) - fee);
} else {
ans = max(ans, -prices[i] + dfs(i + 1, 1));
}
return f[i][j] = ans;
};
return dfs(0, 0);
}
};
func maxProfit(prices []int, fee int) int {
n := len(prices)
f := make([][2]int, n)
for i := range f {
f[i] = [2]int{-1, -1}
}
var dfs func(i, j int) int
dfs = func(i, j int) int {
if i >= n {
return 0
}
if f[i][j] != -1 {
return f[i][j]
}
ans := dfs(i+1, j)
if j > 0 {
ans = max(ans, prices[i]+dfs(i+1, 0)-fee)
} else {
ans = max(ans, -prices[i]+dfs(i+1, 1))
}
f[i][j] = ans
return ans
}
return dfs(0, 0)
}
function maxProfit(prices: number[], fee: number): number {
const n = prices.length;
const f: number[][] = Array.from({ length: n }, () => [-1, -1]);
const dfs = (i: number, j: number): number => {
if (i >= n) {
return 0;
}
if (f[i][j] !== -1) {
return f[i][j];
}
let ans = dfs(i + 1, j);
if (j) {
ans = Math.max(ans, prices[i] + dfs(i + 1, 0) - fee);
} else {
ans = Math.max(ans, -prices[i] + dfs(i + 1, 1));
}
return (f[i][j] = ans);
};
return dfs(0, 0);
}
We define $f[i][j]$ as the maximum profit that can be obtained up to day $i$ with state $j$. Here, $j$ can take the values $0$ and $1$, representing not holding and holding a stock, respectively. We initialize $f[0][0] = 0$ and $f[0][1] = -prices[0]$.
When $i \geq 1$, if we do not hold a stock at the current day, then $f[i][0]$ can be obtained by transitioning from $f[i - 1][0]$ and $f[i - 1][1] + prices[i] - fee$, i.e., $f[i][0] = \max(f[i - 1][0], f[i - 1][1] + prices[i] - fee)$. If we hold a stock at the current day, then $f[i][1]$ can be obtained by transitioning from $f[i - 1][1]$ and $f[i - 1][0] - prices[i]$, i.e., $f[i][1] = \max(f[i - 1][1], f[i - 1][0] - prices[i])$. The final answer is $f[n - 1][0]$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $prices$.
We notice that the transition of the state $f[i][]$ only depends on $f[i - 1][]$ and $f[i - 2][]$. Therefore, we can use two variables $f_0$ and $f_1$ to replace the array $f$, reducing the space complexity to $O(1)$.
class Solution:
def maxProfit(self, prices: List[int], fee: int) -> int:
n = len(prices)
f = [[0] * 2 for _ in range(n)]
f[0][1] = -prices[0]
for i in range(1, n):
f[i][0] = max(f[i - 1][0], f[i - 1][1] + prices[i] - fee)
f[i][1] = max(f[i - 1][1], f[i - 1][0] - prices[i])
return f[n - 1][0]
class Solution {
public int maxProfit(int[] prices, int fee) {
int n = prices.length;
int[][] f = new int[n][2];
f[0][1] = -prices[0];
for (int i = 1; i < n; ++i) {
f[i][0] = Math.max(f[i - 1][0], f[i - 1][1] + prices[i] - fee);
f[i][1] = Math.max(f[i - 1][1], f[i - 1][0] - prices[i]);
}
return f[n - 1][0];
}
}
class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
int n = prices.size();
int f[n][2];
memset(f, 0, sizeof(f));
f[0][1] = -prices[0];
for (int i = 1; i < n; ++i) {
f[i][0] = max(f[i - 1][0], f[i - 1][1] + prices[i] - fee);
f[i][1] = max(f[i - 1][1], f[i - 1][0] - prices[i]);
}
return f[n - 1][0];
}
};
func maxProfit(prices []int, fee int) int {
n := len(prices)
f := make([][2]int, n)
f[0][1] = -prices[0]
for i := 1; i < n; i++ {
f[i][0] = max(f[i-1][0], f[i-1][1]+prices[i]-fee)
f[i][1] = max(f[i-1][1], f[i-1][0]-prices[i])
}
return f[n-1][0]
}
function maxProfit(prices: number[], fee: number): number {
const n = prices.length;
const f: number[][] = Array.from({ length: n }, () => [0, 0]);
f[0][1] = -prices[0];
for (let i = 1; i < n; ++i) {
f[i][0] = Math.max(f[i - 1][0], f[i - 1][1] + prices[i] - fee);
f[i][1] = Math.max(f[i - 1][1], f[i - 1][0] - prices[i]);
}
return f[n - 1][0];
}
class Solution:
def maxProfit(self, prices: List[int], fee: int) -> int:
f0, f1 = 0, -prices[0]
for x in prices[1:]:
f0, f1 = max(f0, f1 + x - fee), max(f1, f0 - x)
return f0
class Solution {
public int maxProfit(int[] prices, int fee) {
int f0 = 0, f1 = -prices[0];
for (int i = 1; i < prices.length; ++i) {
int g0 = Math.max(f0, f1 + prices[i] - fee);
f1 = Math.max(f1, f0 - prices[i]);
f0 = g0;
}
return f0;
}
}
class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
int f0 = 0, f1 = -prices[0];
for (int i = 1; i < prices.size(); ++i) {
int g0 = max(f0, f1 + prices[i] - fee);
f1 = max(f1, f0 - prices[i]);
f0 = g0;
}
return f0;
}
};
func maxProfit(prices []int, fee int) int {
f0, f1 := 0, -prices[0]
for _, x := range prices[1:] {
f0, f1 = max(f0, f1+x-fee), max(f1, f0-x)
}
return f0
}
function maxProfit(prices: number[], fee: number): number {
const n = prices.length;
let [f0, f1] = [0, -prices[0]];
for (const x of prices.slice(1)) {
[f0, f1] = [Math.max(f0, f1 + x - fee), Math.max(f1, f0 - x)];
}
return f0;
}
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