Given two arrays, 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. For example, if you sell the first k items in market A and 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 item B[i…n-1] is sold in market B.
- Now the problem is reduced to finding an index i such that (preA[i] + suffB[i + 1]) is the maximum.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to calculate max profitint 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 codeint 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
<script>// Javascript implementation of the approach// Function to calculate max profitfunction 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 codelet profitA = [2, 3, 2];let profitB = [10, 30, 40];let n = profitA.length;// Function to calculate max profitdocument.write(maxProfit(profitA, profitB, n));</script> |
Output:
80
Time Complexity: O(n)
Auxiliary Space: O(n)
Alternate Implementation:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;int maxProfit(vector<int> a, vector<int> 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 Program11111111111111111111111int main(){ vector<int> a = { 2, 3, 2 }; vector<int> b = { 10, 30, 40 }; cout << maxProfit(a, b, 4); return 0;}// This code is contributed by pankajsharmagfg. |
Java
// Java implementation of the approachclass GFG { static int maxProfit(int[] a, int[] 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 < Math.min(i, a.length); 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.length; j++) sumB += b[j]; // Replace the value of Max Profit with a // bigger value among maxP and sumA+sumB maxP = Math.max(maxP, sumA + sumB); } // Return the value of Max Profit return maxP; } // Driver Program public static void main(String args[]) { int[] a = { 2, 3, 2 }; int[] b = { 10, 30, 40 }; System.out.println(maxProfit(a, b, 4)); }}// This code is contributed by Lovely Jain |
Python3
# 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 |
C#
// Include namespace systemusing System;// C# implementation of the approachpublic class GFG{ public static int maxProfit(int[] a, int[] b, int n) { // Max profit will be saved here var 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 var sumA = 0; for (int j = 0; j < Math.Min(i,a.Length); j++) { sumA += a[j]; } // the sum of the profit after the sale // for products i to n in market B var sumB = 0; for (int j = i; j < b.Length; j++) { sumB += b[j]; } // Replace the value of Max Profit with a // bigger value among maxP and sumA+sumB maxP = Math.Max(maxP,sumA + sumB); } // Return the value of Max Profit return maxP; } // Driver Program public static void Main(String[] args) { int[] a = {2, 3, 2}; int[] b = {10, 30, 40}; Console.WriteLine(GFG.maxProfit(a, b, 4)); }}// This code is contributed by sourabhdalal0001. |
Javascript
<script>// JavaScript implementation of the approachfunction maxProfit(a, b, n){ // Max profit will be saved here let maxP = -1; // loop to check all possible combinations of sales for (let i = 0; i < n + 1; i++) { // the sum of the profit after the sale // for products 0 to i in market A let sumA = 0; for (let j = 0; j < Math.min(i, a.length); j++) sumA += a[j]; // the sum of the profit after the sale // for products i to n in market B let sumB = 0; for (let j = i; j < b.length; j++) sumB += b[j]; // Replace the value of Max Profit with a // bigger value among maxP and sumA+sumB maxP = Math.max(maxP, sumA + sumB); } // Return the value of Max Profit return maxP;}// Driver Programlet a = [ 2, 3, 2 ];let b = [ 10, 30, 40 ];document.write(maxProfit(a, b, 4));// This code is contributed by shinjanpatra</script> |
Output:
80
Time Complexity : O(N)
Auxiliary Space : O(1)
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