Question

714. Best Time to Buy and Sell Stock with Transaction Fee

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Description

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).
• The transaction fee is only charged once for each stock purchase and sale.

 

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

Solutions

Solution 1: Memoization

We 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);
}

Solution 2: Dynamic Programming

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];
}

Solution 3

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;
}