Maximum profit by selling N items at two markets

Given two array A[] and B[] each of length N where A[i] and B[i] are the prices of the ith item when sold in market A and market B respectively. The task is to maximize the profile of selling all the N items but there is a catch if you went to market B then you can not return back. For example, if you sell the first k items in market A then you have to sell the rest of the items in market B.

Examples:

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

Approach:

  • Create a prefix sum array preA[] where preA[i] will store the profit when the items A[0…i] are sold in market A.
  • Create a suffix sum array suffB[] where suffB[i] will store the profit when the items B[i…n-1] are sold in market B.
  • Now the problem gets reduced to finding an index i such that (preA[i] + suffB[i + 1]) is maximum.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java














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// 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 



















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Python3














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# 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 



















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C#














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// 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 



















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Output:


80






          
          
          

          

    

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