![Trending Articles on Technical and Non Technical topics](/images/trending_categories.jpeg)
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Number of Burgers with No Waste of Ingredients in C++
Suppose we have two integers tomatoSlices and cheeseSlices. These are the ingredients of different burgers −
- Jumbo Burger: 4 tomato slices and 1 cheese slice.
- Small Burger: 2 Tomato slices and 1 cheese slice.
We have to find [total_jumbo, total_small] so that the number of tomatoSlices that are left is equal to 0 and the number of cheeseSlices that are left is also 0. If it is not possible to make the remaining tomatoSlices and cheeseSlices equal to 0 return []. So if the input is tomatoSlices = 16 and chesseSlices = 7, then the output will be [1, 6]. So this indicates, to make one jumbo burger and 6 small burgers, we need 4*1 + 2*6 = 16 tomatoSlices and 1 + 6 = 7 cheeseSlices.
To solve this, we will follow these steps −
- make one array called ans
- if tomato is odd or cheese > tomato/2 or tomato > 4*cheese, then return ans
- x := (4 * cheese - tomato) / 2
- y := (tomato – (2*x)) / 4
- insert y then x into array ans
- return ans
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h> using namespace std; void print_vector(vector<auto> v){ cout << "["; for(int i = 0; i<v.size(); i++){ cout << v[i] << ", "; } cout << "]"<<endl; } class Solution { public: vector<int> numOfBurgers(int t, int c) { vector <int> ans; if(t % 2 != 0 || c > t/2 || t > c*4)return ans; int x = (4 * c - t) / 2; int y = ( t - (2 * x) )/ 4; ans.push_back(y); ans.push_back(x); return ans; } }; main(){ Solution ob; print_vector(ob.numOfBurgers(16,7)); }
Input
16 7
Output
[1, 6, ]
Advertisements