**Bill Division**

### Problem Statement :

Two friends Anna and Brian, are deciding how to split the bill at a dinner. Each will only pay for the items they consume. Brian gets the check and calculates Anna's portion. You must determine if his calculation is correct. For example, assume the bill has the following prices: bill = [2, 4, 6]. Anna declines to eat item k= bill[2] which costs 6. If Brian calculates the bill correctly, Anna will pay (2+4)/2 = 3. If he includes the cost of bill[2], he will calculate (2 + 4 +6)/2 =6. In the second case, he should refund 3 to Anna. Function Description Complete the bonAppetit function in the editor below. It should print Bon Appetit if the bill is fairly split. Otherwise, it should print the integer amount of money that Brian owes Anna. bonAppetit has the following parameter(s): bill: an array of integers representing the cost of each item ordered k: an integer representing the zero-based index of the item Anna doesn't eat b: the amount of money that Anna contributed to the bill Input Format The first line contains two space-separated integers n and k, the number of items ordered and the 0-based index of the item that Anna did not eat. The second line contains n space-separated integers bill[i] where 0 <= i < n. The third line contains an integer, b, the amount of money that Brian charged Anna for her share of the bill. Constraints 2 <= n <= 10^5 0 <= k < n 0 <= bill[i] <= 10^4 0 <= b <= bill[0] + bill[1] +.......+ bill[n-1] The amount of money due Anna will always be an integer Output Format If Brian did not overcharge Anna, print Bon Appetit on a new line; otherwise, print the difference (i.e., b charged - b actual ) that Brian must refund to Anna. This will always be an integer.

### Solution :

` ````
Solution in C :
python 3 :
n, k = map(int, input().split())
a = list(map(int, input().split()))
b = int(input())
x = sum(a) - a[k]
if 2 * b == x: print("Bon Appetit")
else: print(a[k] // 2)
Java :
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int k = in.nextInt();
int sum = 0;
for(int i = 0; i < n; i++){
if(i != k){
sum += in.nextInt();
}else{
in.nextInt();
}
}
int res = in.nextInt();
int t = res - sum / 2;
System.out.println(t == 0 ? ("Bon Appetit") : t);
}
}
C ++ :
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int n, k, sum=0;
cin >> n >> k;
for (int i=0;i<n;i++) {
int a;
cin >> a;
if (i!=k) sum+=a;
}
int l;
cin >> l;
if (sum/2==l) cout << "Bon Appetit" << endl;
else cout << l-sum/2 << endl;
}
C :
#include<stdio.h>
int main()
{
int n,k;
scanf("%d %d",&n,&k);
int i,a[n];
for(i=0;i<n;i++)
scanf("%d",&a[i]);
int sum = 0;
for(i=0;i<n;i++)
sum += a[i];
int paid;
scanf("%d",&paid);
int toBePaid = sum-a[k];
if((toBePaid)/2==paid)
printf("Bon Appetit\n");
else
printf("%d\n",paid-(toBePaid)/2);
return 0;
}
```

## View More Similar Problems

## Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

View Solution →## Cycle Detection

A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer

View Solution →## Find Merge Point of Two Lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share

View Solution →## Inserting a Node Into a Sorted Doubly Linked List

Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function

View Solution →## Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

View Solution →## Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

View Solution →