买卖股票后的最大利润

给定一个包含股票价格和交易费用的正整数数组，任务是找到最大利润以及获取最大利润的天数之差。
例子：

Input: arr[] = {6, 1, 7, 2, 8, 4}
      transactionFee = 2
Output: 8 1

Input: arr[] = {7, 1, 5, 3, 6, 4}
       transactionFee = 1
Output: 5 1

说明：考虑第一个示例：arr [] = {6，1，7，2，2，8，4}，transactionFee = 2

1. 如果我们在同一天进行买卖，我们将不会获得任何利润，这就是买卖之间的差额必须至少为1的原因。
2. 以1天的差额计算，如果我们买入1卢比的股票并以1天的差额卖出7卢比，这意味着在第2天购买，然后在第二天卖出，则在支付了2卢比的交易费用后，即7- 1-2 = 4，我们将获得4卢比的利润，就像我们在第4天购买并在第5天将其卖出的价格与第1天的差额一样，那么我们将获得4卢比的利润。因此，总利润为8卢比。
3. 相差2天，我们将不会获得任何利润。
4. 与3天的差额，如果我们买入1卢比的股票并以8天的差额卖出8卢比，这意味着在第2天进行购买，并在3天后出售，那么在支付了2卢比的交易费后，便获得最大利润，即8- 1-2 = 5我们将获得5卢比的利润。
5. 与4天的差额，如果我们买入1卢比的股票并卖出4卢比，而4天的差额则意味着在第2天购买，然后在4天后卖出，则在支付了2卢比的交易费用后，即4-1 -2 = 1，我们将获得1卢比的利润。
6. 相差5天，我们将不会获得任何利润。

方法：

1. 遍历整个数组，每天有不同的时间。
2. 通过减去包括交易费在内的每一天的价格来检查利润。
3. 跟踪最大利润，并存储获得最大利润的diff_days。
4. 重复上述步骤，直到循环终止。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
int max_profit(int a[],int b[],int n,int fee)
{
int i, j, profit;
 
int l, r, diff_day = 1, sum = 0;
 
//b[0] will contain the maximum profit
    b[0]=0;                
//b[1] will contain the day
//on which we are getting the maximum profit
    b[1]=diff_day;
for(i=1;i=i;j--)
        {
        //here finding the max profit
            profit=(a[r]-a[l])-fee;
     
        //if we get less then or equal to zero
        // it means we are not getting the profit
            if(profit>0)    
                {
                sum=sum+profit;
                }
            l++;
     
            r++;
            }
//check if sum is greater then maximum then store the new maximum
    if(b[0] < sum)
{
    b[0] = sum;
     
    b[1] = diff_day;
 
    }
diff_day++;
}
 
return 0;
}
 
// Driver code
int main()
{
    int arr[] = { 6, 1, 7, 2, 8, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int b[2];
    int tranFee = 2;
 
    max_profit(arr, b, n, tranFee);
 
    cout << b[0] << ", " << b[1] << endl;
 
    return 0;
}

Java

// Java implementation of above approach
import java.util.*;
 
class solution
{
 
static int max_profit(int a[],int b[],int n,int fee)
{
int i, j, profit;
 
int l, r, diff_day = 1, sum = 0;
 
//b[0] will contain the maximum profit
    b[0]=0;                
//b[1] will contain the day
//on which we are getting the maximum profit
    b[1]=diff_day;
for(i=1;i=i;j--)
        {
        //here finding the max profit
            profit=(a[r]-a[l])-fee;
     
        //if we get less then or equal to zero
        // it means we are not getting the profit
            if(profit>0)    
                {
                sum=sum+profit;
                }
            l++;
     
            r++;
            }
//check if sum is greater then maximum then store the new maximum
    if(b[0] < sum)
{
    b[0] = sum;
     
    b[1] = diff_day;
 
    }
diff_day++;
}
 
return 0;
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = { 6, 1, 7, 2, 8, 4 };
    int n = arr.length;
    int[] b = new int[2];
    int tranFee = 2;
 
    max_profit(arr, b, n, tranFee);
 
    System.out.println(b[0]+", "+b[1]);
 
}
}
 
//This code is contributed by Surendra_Gangwar

Python3

# Python3 implementation of above approach
def max_profit(a, b, n, fee):
 
    i, j, profit = 1, n - 1, 0
     
    l, r, diff_day = 0, 0, 1
     
    # b[0] will contain the maximum profit
    b[0] = 0   
     
    # b[1] will contain the day on which
    # we are getting the maximum profit
    b[1] = diff_day
 
    for i in range(1, n):
        l = 0
        r = diff_day
        Sum = 0
     
        for j in range(n - 1, i - 1, -1):
             
            # here finding the max profit
            profit = (a[r] - a[l]) - fee
         
            # if we get less then or equal to zero
            # it means we are not getting the profit
            if(profit > 0):
                Sum = Sum + profit
                     
            l += 1
            r += 1
                 
        # check if Sum is greater then maximum
        # then store the new maximum
        if(b[0] < Sum):
            b[0] = Sum
            b[1] = diff_day
     
    diff_day += 1
     
    return 0
     
# Driver code
arr = [6, 1, 7, 2, 8, 4]
n = len(arr)
b = [0 for i in range(2)]
tranFee = 2
 
max_profit(arr, b, n, tranFee)
 
print(b[0], ",", b[1])
 
# This code is contributed by
# Mohit kumar 29

C#

// C# implementation of above approach
using System;
 
class GFG
{
     
static int max_profit(int []a, int []b,
                      int n, int fee)
{
int i, j, profit;
 
int l, r, diff_day = 1, sum = 0;
 
// b[0] will contain the
// maximum profit
b[0] = 0;
 
// b[1] will contain the day on which
// we are getting the maximum profit
b[1] = diff_day;
for(i = 1; i < n; i++)
{
    l = 0; r = diff_day; sum = 0;
 
    for(j = n - 1; j >= i; j--)
        {
            // here finding the max profit
            profit = (a[r] - a[l]) - fee;
     
            // if we get less then or equal
            // to zero it means we are not
            // getting the profit
            if(profit > 0)
            {
                sum = sum + profit;
            }
            l++;
     
            r++;
        }
         
    // check if sum is greater then maximum
    // then store the new maximum
    if(b[0] < sum)
    {
        b[0] = sum;
         
        b[1] = diff_day;
     
    }
    diff_day++;
}
 
return 0;
}
 
// Driver code
static public void Main ()
{
    int []arr = { 6, 1, 7, 2, 8, 4 };
    int n = arr.Length;
    int[] b = new int[2];
    int tranFee = 2;
     
    max_profit(arr, b, n, tranFee);
     
    Console.WriteLine(b[0] + ", " + b[1]);
}
}
 
// This code is contributed by Sachin

PHP

= $i; $j--)
        {
            // here finding the max profit
            $profit = ($a[$r] - $a[$l]) - $fee;
     
            // if we get less then or equal to zero
            // it means we are not getting the profit
            if($profit > 0)    
            {
                $sum = $sum + $profit;
            }
            $l++;
     
            $r++;
        }
         
        // check if sum is greater then maximum
        // then store the new maximum
        if($b[0] < $sum)
        {
            $b[0] = $sum;
             
            $b[1] = $diff_day;
        }
        $diff_day++;
    }
 
}
 
// Driver code
$arr = array(6, 1, 7, 2, 8, 4 );
$n = sizeof($arr);
$b = array();
$tranFee = 2;
 
max_profit($arr, $b, $n, $tranFee);
echo($b[0]);
echo(", ");
echo($b[1]);
 
// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

 
// Javascript implementation of above approach
 
    function max_profit(a , b , n , fee) {
        var i, j, profit;
 
        var l, r, diff_day = 1, sum = 0;
 
        // b[0] will contain the maximum profit
        b[0] = 0;
        // b[1] will contain the day
        // on which we are getting the maximum profit
        b[1] = diff_day;
        for (i = 1; i < n; i++) {
            l = 0;
            r = diff_day;
            sum = 0;
 
            for (j = n - 1; j >= i; j--) {
                // here finding the max profit
                profit = (a[r] - a[l]) - fee;
 
                // if we get less then or equal to zero
                // it means we are not getting the profit
                if (profit > 0) {
                    sum = sum + profit;
                }
                l++;
 
                r++;
            }
            // check if sum is greater then maximum
            // then store the new maximum
            if (b[0] < sum) {
                b[0] = sum;
 
                b[1] = diff_day;
 
            }
            diff_day++;
        }
 
        return 0;
    }
 
    // Driver code
     
        var arr = [ 6, 1, 7, 2, 8, 4 ];
        var n = arr.length;
        var b = Array(2).fill(0);
        var tranFee = 2;
 
        max_profit(arr, b, n, tranFee);
 
        document.write(b[0] + ", " + b[1]);
 
 
// This code contributed by Rajput-Ji
 

输出：

8, 1

时间复杂度： O(N ² )