# 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

Live Demo

#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, ]