在两个市场上销售 N 件商品的最大利润

给定两个数组， A[]和B[]，每个数组的长度为N ，其中A[i]和B[i]分别是第i^个项目在A市场和B市场销售时的价格。任务是最大化销售所有N件商品的配置文件，但有一个问题：如果您去了市场B，那么您将无法返回。例如，如果您在 A 市场出售前 k 件商品，而您必须在 B 市场出售其余商品。

例子：

Input: A[] = {2, 3, 2}, B[] = {10, 3, 40} 
Output: 53 
Sell all the items in market B in order to 
maximize the profit i.e. (10 + 3 + 40) = 53.

Input: A[] = {7, 5, 3, 4}, B[] = {2, 3, 1, 3} 
Output: 19 

方法：

• 创建一个前缀总和数组preA[] ，其中preA[i]将存储A[0…i]商品在市场A中出售时的利润。
• 创建一个后缀和数组suffB[] ，其中suffB[i]将存储B[i…n-1]在市场B 中出售时的利润。
• 现在问题简化为找到一个索引i使得(preA[i] + suffB[i + 1])是最大值。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to calculate max profit
int maxProfit(int profitA[], int profitB[], int n)
{
 
    // Prefix sum array for profitA[]
    int preSum[n];
    preSum[0] = profitA[0];
    for (int i = 1; i < n; i++) {
        preSum[i] = preSum[i - 1] + profitA[i];
    }
 
    // Suffix sum array for profitB[]
    int suffSum[n];
    suffSum[n - 1] = profitB[n - 1];
    for (int i = n - 2; i >= 0; i--) {
        suffSum[i] = suffSum[i + 1] + profitB[i];
    }
 
    // If all the items are sold in market A
    int res = preSum[n - 1];
 
    // Find the maximum profit when the first i
    // items are sold in market A and the
    // rest of the items are sold in market
    // B for all possible values of i
    for (int i = 1; i < n - 1; i++) {
        res = max(res, preSum[i] + suffSum[i + 1]);
    }
 
    // If all the items are sold in market B
    res = max(res, suffSum[0]);
 
    return res;
}
 
// Driver code
int main()
{
    int profitA[] = { 2, 3, 2 };
    int profitB[] = { 10, 30, 40 };
    int n = sizeof(profitA) / sizeof(int);
 
    // Function to calculate max profit
    cout << maxProfit(profitA, profitB, n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
    // Function to calculate max profit
    static int maxProfit(int profitA[], int profitB[], int n)
    {
     
        // Prefix sum array for profitA[]
        int preSum[] = new int[n];
        preSum[0] = profitA[0];
        for (int i = 1; i < n; i++)
        {
            preSum[i] = preSum[i - 1] + profitA[i];
        }
     
        // Suffix sum array for profitB[]
        int suffSum[] = new int[n];
        suffSum[n - 1] = profitB[n - 1];
        for (int i = n - 2; i >= 0; i--)
        {
            suffSum[i] = suffSum[i + 1] + profitB[i];
        }
     
        // If all the items are sold in market A
        int res = preSum[n - 1];
     
        // Find the maximum profit when the first i
        // items are sold in market A and the
        // rest of the items are sold in market
        // B for all possible values of i
        for (int i = 1; i < n - 1; i++)
        {
            res = Math.max(res, preSum[i] + suffSum[i + 1]);
        }
     
        // If all the items are sold in market B
        res = Math.max(res, suffSum[0]);
     
        return res;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int profitA[] = { 2, 3, 2 };
        int profitB[] = { 10, 30, 40 };
        int n = profitA.length;
     
        // Function to calculate max profit
        System.out.println(maxProfit(profitA, profitB, n));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
 
# Function to calculate max profit
def maxProfit(profitA, profitB, n) :
 
    # Prefix sum array for profitA[]
    preSum = [0] * n;
    preSum[0] = profitA[0];
     
    for i in range(1, n) :
        preSum[i] = preSum[i - 1] + profitA[i];
 
    # Suffix sum array for profitB[]
    suffSum = [0] * n;
    suffSum[n - 1] = profitB[n - 1];
     
    for i in range(n - 2, -1, -1) :
        suffSum[i] = suffSum[i + 1] + profitB[i];
 
    # If all the items are sold in market A
    res = preSum[n - 1];
 
    # Find the maximum profit when the first i
    # items are sold in market A and the
    # rest of the items are sold in market
    # B for all possible values of i
    for i in range(1 , n - 1) :
        res = max(res, preSum[i] + suffSum[i + 1]);
 
    # If all the items are sold in market B
    res = max(res, suffSum[0]);
 
    return res;
 
# Driver code
if __name__ == "__main__" :
 
    profitA = [ 2, 3, 2 ];
    profitB = [ 10, 30, 40 ];
    n = len(profitA);
 
    # Function to calculate max profit
    print(maxProfit(profitA, profitB, n));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to calculate max profit
    static int maxProfit(int []profitA,
                        int []profitB, int n)
    {
     
        // Prefix sum array for profitA[]
        int []preSum = new int[n];
        preSum[0] = profitA[0];
        for (int i = 1; i < n; i++)
        {
            preSum[i] = preSum[i - 1] + profitA[i];
        }
     
        // Suffix sum array for profitB[]
        int []suffSum = new int[n];
        suffSum[n - 1] = profitB[n - 1];
        for (int i = n - 2; i >= 0; i--)
        {
            suffSum[i] = suffSum[i + 1] + profitB[i];
        }
     
        // If all the items are sold in market A
        int res = preSum[n - 1];
     
        // Find the maximum profit when the first i
        // items are sold in market A and the
        // rest of the items are sold in market
        // B for all possible values of i
        for (int i = 1; i < n - 1; i++)
        {
            res = Math.Max(res, preSum[i] +
                            suffSum[i + 1]);
        }
     
        // If all the items are sold in market B
        res = Math.Max(res, suffSum[0]);
     
        return res;
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        int []profitA = { 2, 3, 2 };
        int []profitB = { 10, 30, 40 };
        int n = profitA.Length;
     
        // Function to calculate max profit
        Console.WriteLine(maxProfit(profitA, profitB, n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

// Javascript implementation of the approach
// Function to calculate max profit
function maxProfit(profitA, profitB, n) {
 
    // Prefix sum array for profitA[]
    let preSum = new Array(n);
 
    preSum[0] = profitA[0];
 
    for (let i = 1; i < n; i++) {
        preSum[i] = preSum[i - 1] + profitA[i];
    }
 
    // Suffix sum array for profitB[]
    let suffSum = new Array(n);
    suffSum[n - 1] = profitB[n - 1];
 
    for (let i = n - 2; i >= 0; i--) {
        suffSum[i] = suffSum[i + 1] + profitB[i];
    }
 
    // If all the items are sold in market A
    let res = preSum[n - 1];
 
    // Find the maximum profit when the first i
    // items are sold in market A and the
    // rest of the items are sold in market
    // B for all possible values of i
    for (let i = 1; i < n - 1; i++) {
        res = Math.max(res, preSum[i] + suffSum[i + 1]);
    }
 
    // If all the items are sold in market B
    res = Math.max(res, suffSum[0]);
 
    return res;
}
 
// Driver code
 
let profitA = [2, 3, 2];
let profitB = [10, 30, 40];
let n = profitA.length;
 
// Function to calculate max profit
document.write(maxProfit(profitA, profitB, n));

输出：

80

Python的替代实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
int maxProfit(vector a, vector b, int n)
{
 
    // Max profit will be saved here
    int maxP = -1;
 
    // loop to check all possible combinations of sales
    for (int i = 0; i < n + 1; i++) {
 
        // the sum of the profit after the sale
        // for products 0 to i in market A
        int sumA = 0;
 
        for (int j = 0; j < min(i, (int)a.size()); j++)
            sumA += a[j];
 
        // the sum of the profit after the sale
        // for products i to n in market B
        int sumB = 0;
        for (int j = i; j < b.size(); j++)
            sumB += b[j];
 
        // Replace the value of Max Profit with a
        // bigger value among maxP and sumA+sumB
        maxP = max(maxP, sumA + sumB);
    }
 
    //  Return the value of Max Profit
    return maxP;
}
 
// Driver Program11111111111111111111111
int main()
{
    vector a = { 2, 3, 2 };
    vector b = { 10, 30, 40 };
    cout << maxProfit(a, b, 4);
    return 0;
}
 
// This code is contributed by pankajsharmagfg.

蟒蛇3

# Python3 implementation of the approach
def maxProfit (a, b, n):
 
    # Max profit will be saved here
    maxP = -1
 
    # loop to check all possible combinations of sales
    for i in range(0, n+1):
 
        # the sum of the profit after the sale
        # for products 0 to i in market A
        sumA = sum(a[:i])
 
        # the sum of the profit after the sale
        # for products i to n in market B
        sumB = sum(b[i:])
 
        # Replace the value of Max Profit with a
        # bigger value among maxP and sumA+sumB
        maxP = max(maxP, sumA+sumB)
 
    #  Return the value of Max Profit
    return maxP
  
# Driver Program
if __name__ == "__main__" : 
    a = [2, 3, 2]
    b = [10, 30, 40]
    print(maxProfit(a, b, 4))
      
# This code is contributed by aman_malhotra

输出：

80

时间复杂度： O(N)
辅助空间： O(N)