Rahul and Ankit are the only two waiters in the Royal Restaurant. Today, the restaurant received N orders. The amount of tips may differ when handled by different waiters and given as arrays A[] and B[] such that if Rahul takes the ith Order, he would be tipped A[i] rupees, and if Ankit takes this order, the tip would be B[i] rupees. In order to maximize the total tip value, they decided to distribute the order among themselves. One order will be handled by one person only. Also, due to time constraints, Rahul cannot take more than X orders and Ankit cannot take more than Y orders. It is guaranteed that X + Y is greater than or equal to N, which means that all the orders can be handled by either Rahul or Ankit. The task is to find out the maximum possible amount of total tip money after processing all the orders.
Examples:
Input: N = 5, X = 3, Y = 3, A[] = {1, 2, 3, 4, 5}, B[] = {5, 4, 3, 2, 1}
Output: 21
Explanation:
Step 1: 5 is included from Ankit’s array
Step 2: 4 is included from Ankit’s array
Step 3: As both of them has same value 3 then choose any one of them
Step 4: 4 is included from Rahul’s array
Step 5: 5 is included from Rahul’s array
Therefore, the maximum possible amount of total tip money sums up to 21.Input: N = 7, X = 3, Y = 4, A[] = {8, 7, 15, 19, 16, 16, 18}, B[] = {1, 7, 15, 11, 12, 31, 9}
Output: 110
Naive Approach: The naive approach to solve this article is discussed in the previous article.
Efficient Approach: The idea is to use the greedy technique to solve the problem. Sort the N orders in decreasing order on the basis of the difference between Rahul’s tip and Ankit’s tip. Then, traverse all the orders and take the maximum tip of the ith order if it is possible. Follow the steps below to solve the problem:
- Initialize ans as 0 to store the maximum possible tip.
- Create a vector of pair of integers, V of size N such that the first and second element of V[i] stores the tip of the ith order of Rahul and Ankit respectively.
- Sort the vector, V in decreasing order on the basis of the difference between the tips.
- Traverse the vector, V using the variable i
- If the value of Y is 0 OR the condition V[i].first ≥ V[i].second holds true, then add the value of V[i].first to ans and decrement X by 1.
- Otherwise, add the value of V[i].second to ans and decrement Y by 1.
- After completing the above steps, print the value of the ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Utility Compare function // for sorting the arraybool compare(pair<int, int> p1, pair<int, int> p2){ return abs(p1.first - p1.second) > abs(p2.first - p2.second);}// Function to find the maximum possible amount of total// tip money after processing all the ordersvoid maxTip(int A[], int B[], int X, int Y, int N){ // Store all the N orders vector<pair<int, int> > tips(N); // Traverse all the orders and add them // to the vector, V for (int i = 0; i < N; i++) { tips[i].first = A[i]; tips[i].second = B[i]; } // Sort the vector in decreasing // order of absolute // difference of the tips value sort(tips.begin(), tips.end(), compare); // Store the required result int ans = 0; // Traverse all the orders for (int i = 0; i < N; i++) { // Check if Y is 0 or Rahul's tip value // is greater than or equal to that of Ankit if (Y == 0 || (X > 0 && tips[i].first >= tips[i].second)) { // Update the overall result and // decrement X by 1 ans += tips[i].first; X--; } else { // Update the overall result and // decrement Y by 1 ans += tips[i].second; Y--; } } // Print the result cout << ans;}// Driver Codeint main(){ // Given input int N = 5, X = 3, Y = 3; int A[] = { 1, 2, 3, 4, 5 }; int B[] = { 5, 4, 3, 2, 1 }; // Function Call maxTip(A, B, X, Y, N); return 0;} |
Java
/*package whatever //do not write package name here */import java.io.*;import java.util.*;class Pair{ int x, y; Pair(int a, int b){ x = a; y = b; }}class GFG { public static void main (String[] args) { // Given input int N = 5, X = 3, Y = 3; int A[] = { 1, 2, 3, 4, 5 }; int B[] = { 5, 4, 3, 2, 1 }; // Function Call int ans = maxTip(A, B, N, X, Y); System.out.println(ans); } static int maxTip(int[] a, int[] b, int n, int x, int y) { // list to store pairs ArrayList<Pair> tips = new ArrayList<>(); // adding tip values to list for(int i = 0; i < n; i++){ tips.add(new Pair(a[i], b[i])); } // sorting list by absolute difference Collections.sort(tips, (a1, b1)-> { return Math.abs(b1.x - b1.y) - Math.abs(a1.x - a1.y) ; }); // System.out.println(tips); int ans = 0; // traversing the list for(int i = 0; i < n; i++){ Pair p = tips.get(i); // if first should be considered if(y == 0 || (p.x >= p.y && x > 0)){ ans += p.x; x--; } else if(y > 0) { ans += p.y; y--; } } return ans; }} |
Python3
from functools import cmp_to_key# Utility Compare function# for sorting the arraydef compare(p1, p2): return abs(p2[0] - p2[1]) - abs(p1[0] - p1[1])# Function to find the maximum possible amount of total# tip money after processing all the ordersdef maxTip(A, B, X, Y, N): # Store all the N orders tips = [] # Traverse all the orders and add them # to the list, tips for i in range(N): tips.append((A[i], B[i])) # Sort the list in decreasing # order of absolute # difference of the tips value tips.sort(key=lambda x: -abs(x[0] - x[1])) # Store the required result ans = 0 # Traverse all the orders for i in range(N): # Check if Y is 0 or Rahul's tip value # is greater than or equal to that of Ankit if Y == 0 or (X > 0 and tips[i][0] >= tips[i][1]): # Update the overall result and # decrement X by 1 ans += tips[i][0] X -= 1 else: # Update the overall result and # decrement Y by 1 ans += tips[i][1] Y -= 1 # Print the result print(ans)# Driver Code# Given inputN, X, Y = 5, 3, 3A = [1, 2, 3, 4, 5]B = [5, 4, 3, 2, 1]# Function CallmaxTip(A, B, X, Y, N)# This code is contributed by phasing17. |
C#
using System;using System.Collections.Generic;using System.Linq;class Pair{ public int x, y; public Pair(int a, int b) { x = a; y = b; }}class GFG{ static void Main(string[] args) { // Given input int N = 5, X = 3, Y = 3; int[] A = { 1, 2, 3, 4, 5 }; int[] B = { 5, 4, 3, 2, 1 }; // Function Call int ans = maxTip(A, B, N, X, Y); Console.WriteLine(ans); } static int maxTip(int[] a, int[] b, int n, int x, int y) { // list to store pairs List<Pair> tips = new List<Pair>(); // adding tip values to list for (int i = 0; i < n; i++) { tips.Add(new Pair(a[i], b[i])); } // sorting list by absolute difference tips.Sort((a1, b1) => Math.Abs(b1.x - b1.y) - Math.Abs(a1.x - a1.y)); int ans = 0; // traversing the list for (int i = 0; i < n; i++) { Pair p = tips[i]; // if first should be considered if (y == 0 || (p.x >= p.y && x > 0)) { ans += p.x; x--; } else if (y > 0) { ans += p.y; y--; } } return ans; }} |
Javascript
<script>// JavaScript program for the above approach// Utility Compare function// for sorting the arrayfunction compare(p1, p2){ return Math.abs(p2[0] - p2[1]) - Math.abs(p1[0] - p1[1]);}// Function to find the maximum possible amount of total// tip money after processing all the ordersfunction maxTip(A, B, X, Y, N){ // Store all the N orders let tips = new Array(N); // Traverse all the orders and add them // to the vector, V for (let i = 0; i < N; i++) { tips[i] = new Array(); tips[i][0] = A[i]; tips[i][1] = B[i]; } // Sort the vector in decreasing // order of absolute // difference of the tips value tips.sort(compare); // Store the required result let ans = 0; // Traverse all the orders for (let i = 0; i < N; i++) { // Check if Y is 0 or Rahul's tip value // is greater than or equal to that of Ankit if (Y == 0 || (X > 0 && tips[i][0] >= tips[i][1])) { // Update the overall result and // decrement X by 1 ans += tips[i][0]; X--; } else { // Update the overall result and // decrement Y by 1 ans += tips[i][1]; Y--; } } // Print the result document.write(ans,"</br>");}// Driver Code// Given inputlet N = 7, X = 3, Y = 4;let A = [8, 7, 15, 19, 16, 16, 18];let B = [1, 7, 15, 11, 12, 31, 9];// Function CallmaxTip(A, B, X, Y, N);// This code is contributed by shinjanpatra.</script> |
21
Time Complexity: O(N * log(N))
Auxiliary Space: O(N)
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