#include <iostream>
using namespace std;
/*
 Description: uses switch case statements to determine whether it is hot or not outside.
 Also uses toupper() function which forces user input char to be uppercase in order to work for the switch statement
*/

int main() {
  char choice;
  cout << "S = Summer, F = Fall, W = Winter, P = Spring" << endl;
  cout << "Enter a character to represent a season: ";asdasdasdasd
  cin >> choice;

  enum Season {SUMMER='S', FALL='F', WINTER='W', SPRING='P'};

  switch(toupper(choice))  // This switch statement compares a character entered with values stored inside of an enum
  {
  case SUMMER:
    cout << "It's very hot outside." << endl;
    break;
  case FALL:
    cout << "It's great weather outside." << endl;
    break;
  case WINTER:
    cout << "It's fairly cold outside." << endl;
    break;
  case SPRING:
    cout << "It's rather warm outside." << endl;
    break;
  default:
    cout << "Wrong choice" << endl;
    break;
  }
  return 0;
}

#include <iostream>
using namespace std;

/* Function: get_coeff
Parameters: double& coeff, int pos passed from bb_4ac
Return: type is void so no return, but does ask for user to input data that establishes what a b and c are.
*/
void get_coeff(double& coeff, int pos) {
  char position;
  if(pos == 1) {
    position = 'a';
  } else if(pos == 2) {   //a simple system to determine what coefficient the program is asking for.
    position = 'b';
  } else {
    position = 'c';
  }
  cout << "Enter the co-efficient " << position << ":"; //prompt to input coeff
  coeff = 5; //input coeff
}

/* Function: bb_4ac
Parameters: no parameters passed from main, but 3 params established in function, double a, b, c.
Return: b * b - 4 * a * c
*/
double bb_4ac() {
  double a, b, c; //coefficients of a quadratic equation
  get_coeff(a, 1); // call function 1st time
  get_coeff(b, 2); // call function 2nd time
  get_coeff(c, 3); // call function 3rd time
  return b * b - 4 * a * c; //return b * b - 4 * a * c
}

int main() {
  cout << "Function to calculate the discriminant of the equation. . . " << endl;
  double determinate = bb_4ac(); //assign double determinate to bb_4ac function
  cout << "The discriminant for given values is: " << determinate << endl;  //output the determinate!
}

#define NUM_BITS 8

#include <iostream>


struct Number{
    int num : NUM_BITS;
    Number(){}
    Number(const int& bruh){
        num = bruh;
    }
    operator int() const { return num; }
    Number& operator=(const int& bruh){
        num = bruh;
        return (*this);
    }
};

using namespace std;

bool isNegative(const int& num){
    //This gets the bitwise and of num and 10000000000000000000000000000000
    //This implicit casts to bool, which means (num & (1 << 31)) != 0
    return (num & (1 << 31));
}

void printBinaryNumber(const int& num, const int numBits){
    for(int i = numBits; i > 0; --i){
        //8..1
        int bitMask = 1 << (i-1);
        if(num & bitMask){ //Test the bit
            cout << '1';
        }else{
            cout << '0';
        }
    }
}

void printCarryBits(const int& a, const int& b, const int numBits){
    int answer = 0;
    
    bool carry = false;
    for(int i = 0; i < numBits; ++i){
        //8..1
        int bitMask = 1 << i;
        bool aBit = a & bitMask;
        bool bBit = b & bitMask;
        if(aBit && bBit || aBit && carry || bBit && carry){ //Carry bit is true next
            if(carry)
                answer |= bitMask;
            carry = true;
        }else{
            if(carry)
                answer |= bitMask;
            carry = false;
        }
    }
    printBinaryNumber(answer, 8);
}


void printBorrowBits(const int& a, const int& b, const int numBits){
    int answer = 0;
    
    bool carry = false;
    for(int i = 0; i < numBits; ++i){
        //8..1
        int bitMask = 1 << i;
        bool aBit = a & bitMask;
        bool bBit = b & bitMask;
        if((!(aBit ^ carry)) && bBit){ //Carry bit is true next
            if(carry)
                answer |= bitMask;
            carry = true;
        }else{
            if(carry)
                answer |= bitMask;
            carry = false;
        }
    }
    printBinaryNumber(answer, 8);
}

void doProblem(const int& a, const int& b, const char& sign, const int& result, const int& numBits){
    if(sign == '+'){
        cout << ' '; printCarryBits(a, b, numBits); cout << endl;
    }else{
        cout << ' '; printBorrowBits(a, b, numBits); cout << endl;
    }
    cout << ' '; printBinaryNumber(a, numBits); cout << endl;
    cout << sign; printBinaryNumber(b, numBits); cout << endl;
    cout << "----------" << endl;
    cout << ""; printBinaryNumber(result, numBits + 1); cout << " = " << result;
    cout << endl;
}

int main(){
    
    Number a = 0b110;
    Number b = 0b011;
    cout<< a << endl << b << endl;
    
    doProblem(a, b, '+', a + b, NUM_BITS);
    doProblem(a, b, '-', a - b, NUM_BITS);
    doProblem(-a, b, '+', -a + b, NUM_BITS);
    doProblem(a, b, '-', -a - b, NUM_BITS);
    

    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;
}

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

This problem was asked by LinkedIn.

A wall consists of several rows of bricks of various integer lengths and uniform height. Your goal is to find a vertical line going from the top to the bottom of the wall that cuts through the fewest number of bricks. If the line goes through the edge between two bricks, this does not count as a cut.

For example, suppose the input is as follows, where values in each row represent the lengths of bricks in that row:

[[3, 5, 1, 1],
	[2, 3, 3, 2],
	[5, 5],
	[4, 4, 2],
	[1, 3, 3, 3],
	[1, 1, 6, 1, 1]]

The best we can we do here is to draw a line after the eighth brick, which will only require cutting through the bricks in the third and fifth row.

Given an input consisting of brick lengths for each row such as the one above, return the fewest number of bricks that must be cut to create a vertical line.

AUTHORS NOTE:
Makes following assumptions:
- Each row is same length
- Data is in file called "data.dat" and formatted in space-separated rows
- The cuts at the beginning and end of the wall are not solutions

This requires the following file named data.dat that is a space separated file, or similar formatted file:
----START FILE----
3 5 1 1
2 3 3 2
5 5
4 4 2
1 3 3 3
1 1 6 1 1
----END FILE----
*/

#include <algorithm>
#include <iostream>
#include <fstream>
#include <map>
#include <sstream>
#include <string>
#include <vector>
using namespace std;

int main()
{
	vector<vector<int>> wall;

	ifstream in;
	in.open("data.dat");
	if(!in.good())
	{
	cout << "ERROR: File failed to open properly.\n";
	}

	/* Get input from space separated file */
	string line;
	while(!in.eof())
	{
	getline(in, line);

	int i;
	vector<int> currv;
	stringstream strs(line);
	while(strs >> i)
	currv.push_back(i);
	wall.push_back(currv);
	}

	

	/* Convert each value from "length of brick" to "position at end of brick" */
	for(int y = 0; y < wall.size(); y++)
	{
	wall.at(y).pop_back(); //Delet last val
	for(int x = 1; x < wall.at(y).size(); x++) //Skip the first bc data doesn't need change
	wall.at(y).at(x) += wall.at(y).at(x-1);
	}

	/* Check output. COMMENT OUT */
	// for(auto row : wall)
	// {
	// for(int pos : row)
	// cout << pos << " ";
	// cout << endl;
	// }

	/* Determine which ending position is most common, and cut there */
	//Exclude final position, which will be the size of the wall

	int mode = -1;
	int amt = -1;
	vector<int> tried;
	for(auto row : wall)
	{
	for(int pos : row) //For each pos in the wall
	{
	//Guard. If pos is contained in the list, skip pos
	if(find(tried.begin(), tried.end(), pos) != tried.end())
	continue;
	tried.push_back(pos);

	/* Cycle through each row to see if it contains the pos */
	int curramt = 0;
	for(auto currrow : wall)
	{
	if( find( currrow.begin(), currrow.end(), pos ) != currrow.end() )
	curramt++;
	}
	//cout << pos << " " << curramt << endl; 

	if(curramt > amt)
	{
	amt = curramt;
	mode = pos;
	}
	}
	}

	cout << "Please cut at position " << mode << endl;
	cout << "This will cut through " << (wall.size() - amt) << " bricks." << 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;
}