//Constant prefix notation solver using bruh
//Could make it infix or postfix later
#include<string>
#include<vector>
#include<iostream>

std::vector<long double> bruhBuff;

long double operator ""bruh(long double a){
    bruhBuff.push_back(a);
    return a;
}

long double operator ""bruh(const char op){
    if(bruhBuff.size() < 2) throw "Bruh weak";
    
    long double b = bruhBuff.back();
    bruhBuff.pop_back();
    long double a = bruhBuff.back();
    bruhBuff.pop_back();
    
    
    switch(op){
        case (int)('+'):
            return a + b;
        case (int)('-'):
            return a - b;
        case (int)('*'):
            return a * b;
        case (int)('/'):
            return a / b;
    }
    return 69l;
}
 
int main(){
    1.0bruh;
    2.0bruh;
    std::cout << '+'bruh << std::endl;
    
    return 0;
}

#include <iostream>
using namespace std;
int main()
{
    int arr[] = {5, 1, 4, 20, 10, 2, 13, 11, 6, 21};
    int greed[] = {0, 0, 0, 0};
    int k = 0;
    int i;
    int set_index;
    while (k < 4)
    {
        i = 0;
        while (i < 10)
        {
            if (arr[i] > greed[k])
            {
                greed[k] = arr[i];
                set_index = i;
            }
            i++;
        }
        arr[set_index] = 0;
        k++;
    }
    cout << greed[0] << " " << greed[1] << " " << greed[2] << " " << greed[3] << endl;
}

int main()

#include <iostream>
using namespace std;

int main() {

cout << "Hello World!\n";
// Prints out "Hello World"
return 0;

}

/*
Good morning! Here's your coding interview problem for today.

This problem was asked by Stripe.

Given an array of integers, find the first missing positive integer in linear time and constant space. In other words, find the lowest positive integer that does not exist in the array. The array can contain duplicates and negative numbers as well.

For example, the input [3, 4, -1, 1] should give 2. The input [1, 2, 0] should give 3.

You can modify the input array in-place.
*/
#include <iostream>
using namespace std;

int calcMissing(int* input, int size)
{
	int sum = 0;
	int n = 1; //add one to account for missing value
	for(int i = 0; i < size; i++)
	{
	if(input[i] > 0)
	{
	sum += input[i];
	n++;
	}
	}

	//If no numbers higher than 0, answer is 1
	if(sum == 0)
	return 1;

	return (n*(n+1)/2) - sum; //Formula is expectedSum - actualSum
	/* expectedSum = n*(n+1)/2, the formula for sum(1, n) */
}

int main()
{

	cout << calcMissing(new int[4]{3, 4, -1, 1}, 4) << endl;
	cout << calcMissing(new int[3]{1, 2, 0}, 3) << endl;

	//No positive numbers
	cout << calcMissing(new int[1]{0}, 1) << endl;
}

#include <iostream>
using namespace std;

int main()
{
    cout << "Hello, World!" << endl;
    return 0;
}