#include <iostream>
using namespace std;

int main() {

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

}

#include <iostream>
using namespace std;

int main() {


    int arr[5];
    
    for(int i = 0; i < 5; i++) {
        arr[i] = i;
    }

    for(int i = 0; i < 5; i++) {
        cout << "Outputting array info at position " << i + 1 << ": " << arr[i] << endl;
    }

    for(int i=0;i<5;i++)
	{		
		for(int j=i+1;j<5;j++)
		{
			if(arr[i]>arr[j])
			{
				int temp=arr[i];
				arr[i]=arr[j];
				arr[j]=temp;
			}
		}
	}

    cout << endl;

    for(int i = 0; i < 5; i++) {
        cout << "Outputting sorted array info at position " << i + 1 << ": " << arr[i] << endl;
    }

    return 0;
}

#include <iostream>
int main(){
    const char* const hello = "Hello, world!";
    
    const char* bruh = hello;
    
    char* const yeet = hello;
    
    std::cout << bruh << std::endl;
    
    std::cout << yeet << std::endl;
    
    return 0;
}

/*
    Place your bets!
    
    Will the program:
    a.) Print "Hello, world!" twice?
    b.) Compile error on line 5 (bruh initialize line) because the pointer gets implicit cast to non-const?
    c.) Compile error on line 7 (yeet initialize line) because the char gets implicit cast to non-const?
    d.) Both b and c?
    e.) Compile error line 11 (print yeet) because the pointer is constant and can't be incremented
    f.) Print "Hello, world!" then print the pointer address in hexadecimal
    g.) Both b and e?
    h.) Both c and e?
    i.) B, c, and e?
    
*/

// The answer is in this base 64 string:
// 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

#include <iostream>
using namespace std;

main
{
	cout << "No tabbing. That's very sad :(\n";
	cout << "No in-editor highlighting either :(((\n";
	cout << "Descriptions might be niice too.";
}

#include <iostream>
#include <cstring>
int main(int argc, char** argv){
	//With decimal
	if(strstr(argv[1], ".") != nullptr){
		int i = 0;
		//Skip i to first non 0 digit
		while(argv[1][i] < '1' || argv[1][i] > '9') ++i;
		//If digit comes before decimal
		if((argv[1] + i) < strstr(argv[1], ".")){	//Good example of pointer arithmetic
			std::cout << strlen(argv[1] + i) - 1 << std::endl;	//Another good example
		}else{
			//If digit is after decimal
			std::cout << strlen(argv[1] + i) << std::endl;
		}
	}else{
	//Without decimal
		int m = 0;
		int i = 0;
		while(argv[1][i] < '1' || argv[1][i] > '9') ++i;	//In case of some number like 0045
		for(; argv[1][i] != '\0'; ++i){
			if(argv[1][i] >= '1' && argv[1][i] <= '9') m = i + 1;
		}
		std::cout << m << std::endl;
	}
	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;
}