Given two arrays A1 and A2 of N numbers. There are two people A and B who select numbers out of N. If A selects i-th number, then he will be paid A1[i] amount of money and if B selects i-th number then he will be paid A2[i] amount of money but A cannot select more than X numbers and B cannot select more than Y numbers. The task is to select N numbers in such a way that the amount of total money is maximized at the end.
Note: X + Y >= N
Examples:
Input: N = 5, X = 3, Y = 3
A1[] = {1, 2, 3, 4, 5},
A2= {5, 4, 3, 2, 1}
Output: 21
B will take the first 3 orders and A
will take the last two orders.
Input: N = 2, X = 1, Y = 1
A1[] = {10, 10}, A2= {20, 20}
Output: 30
Approach: Let us create a new array C such that C[i] = A2[i] – A1[i]. Now we will sort the array C in decreasing order. Note that the condition X + Y >= N guarantees that we will be able to assign the number to any one of the persons. Assume that for some i, A1[i] > A2[i] and you assigned an order to B, loss encountered due to this assignment is C[i]. Similarly, for some i, A2[i] > A1[i] and you assigned a number to A, loss encountered due to this assignment is C[i]. As we want to minimize the loss encountered, it is better to process the numbers having high possible losses, because we can try to reduce the loss in the starting part. There is no point of selecting a number with high loss after assigning a number with less loss. Hence we initially assign all numbers to A initially, then subtract the loss from them greedily. Once the assigned order number is under X, then we store the maximal of that.
Below is the implementation of the above approach:
C++
// C++ program to maximize profit #include <bits/stdc++.h> using namespace std; // Function that maximizes the sum int maximize(int A1[], int A2[], int n, int x, int y) { // Array to store the loss int c[n]; // Initial sum int sum = 0; // Generate the array C for (int i = 0; i < n; i++) { c[i] = A2[i] - A1[i]; sum += A1[i]; } // Sort the array elements // in descending order sort(c, c + n, greater<int>()); // Variable to store the answer int maxi = -1; // Iterate in the array, C for (int i = 0; i < n; i++) { // Subtract the loss sum += c[i]; // Check if X orders are going // to be used if (i + 1 >= (n - x)) maxi = max(sum, maxi); } return maxi; } // Driver Code int main() { int A1[] = { 1, 2, 3, 4, 5 }; int A2[] = { 5, 4, 3, 2, 1 }; int n = 5; int x = 3, y = 3; cout << maximize(A1, A2, n, x, y); return 0; } |
Java
// Java program to maximize profit import java.util.*; class GFG { // Function that maximizes the sum static int maximize(int A1[], int A2[], int n, int x, int y) { // Array to store the loss int[] c = new int[n]; // Initial sum int sum = 0; // Generate the array C for (int i = 0; i < n; i++) { c[i] = A2[i] - A1[i]; sum += A1[i]; } // Sort the array elements // in descending order int temp; for(int i = 0; i < n - 1; i++) { if(c[i] < c[i+1]) { temp = c[i]; c[i] = c[i + 1]; c[i + 1] = temp; } } // Variable to store the answer int maxi = -1; // Iterate in the array, C for (int i = 0; i < n; i++) { // Subtract the loss sum += c[i]; // Check if X orders are going // to be used if (i + 1 >= (n - x)) maxi = Math.max(sum, maxi); } return maxi; } // Driver Code public static void main(String args[]) { int A1[] = { 1, 2, 3, 4, 5 }; int A2[] = { 5, 4, 3, 2, 1 }; int n = 5; int x = 3, y = 3; System.out.println(maximize(A1, A2, n, x, y)); } } // This code is contributed by // Surendra_Gangwar |
Python3
# Python3 program to maximize profit # Function that maximizes the Sum def maximize(A1, A2, n, x, y): # Array to store the loss c = [0 for i in range(n)] # Initial Sum Sum = 0 # Generate the array C for i in range(n): c[i] = A2[i] - A1[i] Sum += A1[i] # Sort the array elements # in descending order c.sort() c = c[::-1] # Variable to store the answer maxi = -1 # Iterate in the array, C for i in range(n): # Subtract the loss Sum += c[i] # Check if X orders are going # to be used if (i + 1 >= (n - x)): maxi = max(Sum, maxi) return maxi # Driver Code A1 = [ 1, 2, 3, 4, 5 ] A2 = [ 5, 4, 3, 2, 1 ] n = 5x, y = 3, 3 print(maximize(A1, A2, n, x, y)) # This code is contributed # by Mohit Kumar |
C#
// C# program to maximize profit using System; class GFG { // Function that maximizes the sum static int maximize(int [] A1, int [] A2, int n, int x, int y) { // Array to store the loss int [] c = new int[n]; // Initial sum int sum = 0; // Generate the array C for (int i = 0; i < n; i++) { c[i] = A2[i] - A1[i]; sum += A1[i]; } // Sort the array elements // in descending order int temp; for(int i = 0; i < n - 1; i++) { if(c[i] < c[i+1]) { temp = c[i]; c[i] = c[i + 1]; c[i + 1] = temp; } } // Variable to store the answer int maxi = -1; // Iterate in the array, C for (int i = 0; i < n; i++) { // Subtract the loss sum += c[i]; // Check if X orders are going // to be used if (i + 1 >= (n - x)) maxi = Math.Max(sum, maxi); } return maxi; } // Driver Code public static void Main() { int [] A1 = { 1, 2, 3, 4, 5 }; int [] A2 = { 5, 4, 3, 2, 1 }; int n = 5; int x = 3, y = 3; Console.WriteLine(maximize(A1, A2, n, x, y)); } } // This code is contributed by ihritik |
PHP
<?php // PHP program to maximize profit // Function that maximizes the sum function maximize($A1, $A2, $n, $x, $y) { # Array to store the loss $c = array(); # Initial sum $sum = 0; // Generate the array C for ($i = 0; $i < $n; $i++) { $c[$i] = $A2[$i] - $A1[$i]; $sum += $A1[$i]; } // Sort the array elements // in descending order rsort($c); // Variable to store the answer $maxi = -1; // Iterate in the array, C for ($i = 0; $i < $n; $i++) { // Subtract the loss $sum += $c[$i]; // Check if X orders are going // to be used if ($i + 1 >= ($n - $x)) $maxi = max($sum, $maxi); } return $maxi; } # Driver Code $A1 = array( 1, 2, 3, 4, 5 ); $A2 = array( 5, 4, 3, 2, 1 ); $n = 5; $x = 3; $y = 3; echo maximize($A1, $A2, $n, $x, $y); // This code is contributed by Ryuga ?> |
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