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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 // 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

Output:

80

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