#include<iostream>
using namespace std;

const int rows = 8;
const int cols = 8;
char chessboard[rows][cols];

void setBoard(char chessboard[][cols]);
void printBoard(char chessboard[][cols]);

void setBoard(char chessboard[][cols]) {
    for(int i = 0; i < rows; i++) {
        for(int j = 0; j < cols; j++) {
            if(i % 2 == 0 && j % 2 == 0) {
                chessboard[i][j] = 'x';
            } else {
                if(i % 2 != 0 && j % 2 == 1) {
                    chessboard[i][j] = 'x';
                } else {
                    chessboard[i][j] = '-';
                }
            }
        }
    }
    return;
}

void printBoard(char chessboard[][cols]) {
    for(int i = 0; i < rows; i++) {
        for(int j = 0; j < cols; j++) {
            cout << chessboard[i][j] << " ";
        }
        cout << endl;
    }
    return;
}

int main(int argc, char const *argv[])
{
    setBoard(chessboard);
    printBoard(chessboard);
    return 0;
}

/*
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>
#include <cmath>
#include <string.h>

using namespace std;

int main() {

    string tickerName;
    int numOfContracts;
    float currentOptionValue;

    cout << "Enter a stock ticker: ";
    getline(cin, tickerName);

    cout << "Enter the current number of " << tickerName << " contracts you are holding: ";
    cin >> numOfContracts;


    cout << "Enter the current price of the option: ";
    cin >> currentOptionValue;


    cout << "The value of your " << tickerName << " options are: $" << (currentOptionValue * 100.00) * (numOfContracts);

    cout << endl;

    return 0;
}

#include <iostream>
using namespace std;

int main() {
  const int ROW_SIZE = 2;
  const int COLUMN_SIZE = 5;              //establish all variables
  int matrix[ROW_SIZE][COLUMN_SIZE];
  int minVal;

  for (int i = 0; i < ROW_SIZE; ++i)    // for loop to ask user to enter data.
  {
    for (int h = 0; h < COLUMN_SIZE; ++h) {
      cout << "Enter data for row #" << i + 1 << " and column #" << h + 1 << ": ";
      cin >> matrix[i][h];
    }
  }
  cout << "You entered: " << endl;
  for (int i = 0; i < ROW_SIZE; ++i) //for statements to output the array neatly
  {
    for (int h = 0; h < COLUMN_SIZE; ++h) {
      cout << matrix[i][h] << "\t";
    }
    cout << endl;
  }

  cout << "Minimum for each row is: {";
  for (int i = 0; i < ROW_SIZE; i++) //for statements to find the minimum in each row
  {
    minVal = matrix[i][0];

    for (int h = 0; h < COLUMN_SIZE; h++) {
      if (matrix[i][h] < minVal) // if matrix[i][h] < minVal -> minVal = matrix[i][h];
      {
        minVal = matrix[i][h];
      }
    }
    cout << minVal << ", ";
  }
  cout << "}" << endl;

  cout << "Minimum for each column is: {";
  for (int i = 0; i < COLUMN_SIZE; i++) //for statements to find the minimum in each column
  {
    minVal = matrix[0][i];
    for (int h = 0; h < ROW_SIZE; h++) {
      if (matrix[h][i] < minVal) //replaces minVal with array index for that column that is lowest
      {
        minVal = matrix[h][i];
      }
    }
    cout << minVal << ", ";
  }
  cout << "}" << endl;

  return 0;
}

//From https://create.arduino.cc/projecthub/abhilashpatel121/easyfft-fast-fourier-transform-fft-for-arduino-9d2677
#include <cmath>
#include <iostream>
const unsigned char sine_data[] = {	//Quarter a sine wave
	0, 
	4, 9, 13, 18, 22, 27, 31, 35, 40, 44, 
	49, 53, 57, 62, 66, 70, 75, 79, 83, 87, 
	91, 96, 100, 104, 108, 112, 116, 120, 124, 127, 
	131, 135, 139, 143, 146, 150, 153, 157, 160, 164, 
	167, 171, 174, 177, 180, 183, 186, 189, 192, 195, //Paste this at top of program
	198, 201, 204, 206, 209, 211, 214, 216, 219, 221, 
	223, 225, 227, 229, 231, 233, 235, 236, 238, 240, 
	241, 243, 244, 245, 246, 247, 248, 249, 250, 251, 
	252, 253, 253, 254, 254, 254, 255, 255, 255, 255
};
float sine(int i){	//Inefficient sine
	int j=i;
	float out;
	while(j < 0) j = j + 360;
	while(j > 360) j = j - 360;
	if(j > -1 && j < 91) out = sine_data[j];
	else if(j > 90 && j < 181) out = sine_data[180 - j];
	else if(j > 180 && j < 271) out = -sine_data[j - 180];
	else if(j > 270 && j < 361) out = -sine_data[360 - j];
	return (out / 255);
}

float cosine(int i){	//Inefficient cosine
	int j = i;
	float out;
	while(j < 0) j = j + 360;
	while(j > 360) j = j - 360;
	if(j > -1 && j < 91) out = sine_data[90 - j];
	else if(j > 90 && j < 181) out = -sine_data[j - 90];
	else if(j > 180 && j < 271) out = -sine_data[270 - j];
	else if(j > 270 && j < 361) out = sine_data[j - 270];
	return (out / 255);
}

//Example data:

//-----------------------------FFT Function----------------------------------------------//
float* FFT(int in[],unsigned int N,float Frequency){	//Result is highest frequencies in order of loudness. Needs to be deleted.
	/*
	Code to perform FFT on arduino,
	setup:
	paste sine_data [91] at top of program [global variable], paste FFT function at end of program
	Term:
	1. in[] : Data array, 
	2. N : Number of sample (recommended sample size 2,4,8,16,32,64,128...)
	3. Frequency: sampling frequency required as input (Hz)

	If sample size is not in power of 2 it will be clipped to lower side of number. 
	i.e, for 150 number of samples, code will consider first 128 sample, remaining sample will be omitted.
	For Arduino nano, FFT of more than 128 sample not possible due to mamory limitation (64 recomended)
	For higher Number of sample may arise Mamory related issue,
	Code by ABHILASH
	Contact: [email protected] 
	Documentation:https://www.instructables.com/member/abhilash_patel/instructables/
	2/3/2021: change data type of N from float to int for >=256 samples
	*/

	unsigned int sampleRates[13]={1,2,4,8,16,32,64,128,256,512,1024,2048};
	int a = N;
	int o;
	for(int i=0;i<12;i++){		//Snapping N to a sample rate in sampleRates
		if(sampleRates[i]<=a){
			o = i;
		}
	}
		 
	int in_ps[sampleRates[o]] = {}; //input for sequencing
	float out_r[sampleRates[o]] = {}; //real part of transform
	float out_im[sampleRates[o]] = {}; //imaginory part of transform
	int x = 0; 
	int c1;
	int f;
	for(int b=0;b<o;b++){ // bit reversal
		c1 = sampleRates[b];
		f = sampleRates[o] / (c1 + c1);
		for(int j = 0;j < c1;j++){ 
			x = x + 1;
			in_ps[x]=in_ps[j]+f;
		}
	}

	
	for(int i=0;i<sampleRates[o];i++){ // update input array as per bit reverse order
		if(in_ps[i]<a){
			out_r[i]=in[in_ps[i]];
		}
		if(in_ps[i]>a){
			out_r[i]=in[in_ps[i]-a];
		} 
	}


	int i10,i11,n1;
	float e,c,s,tr,ti;

	for(int i=0;i<o;i++){ //fft
		i10 = sampleRates[i]; // overall values of sine/cosine :
		i11 = sampleRates[o] / sampleRates[i+1]; // loop with similar sine cosine:
		e = 360 / sampleRates[i+1];
		e = 0 - e;
		n1 = 0;

		for(int j=0;j<i10;j++){
			c=cosine(e*j);
			s=sine(e*j); 
			n1=j;

			for(int k=0;k<i11;k++){
				tr = c*out_r[i10 + n1]-s*out_im[i10 + n1];
				ti = s*out_r[i10 + n1]+c*out_im[i10 + n1];

				out_r[n1 + i10] = out_r[n1]-tr;
				out_r[n1] = out_r[n1]+tr;

				out_im[n1 + i10] = out_im[n1]-ti;
				out_im[n1] = out_im[n1]+ti; 

				n1 = n1+i10+i10;
			} 
		}
	}

	/*
	for(int i=0;i<sampleRates[o];i++)
	{
	std::cout << (out_r[i]);
	std::cout << ("\t"); // un comment to print RAW o/p 
	std::cout << (out_im[i]); std::cout << ("i"); 
	std::cout << std::endl;
	}
	*/


	//---> here onward out_r contains amplitude and our_in conntains frequency (Hz)
	for(int i=0;i<sampleRates[o-1];i++){ // getting amplitude from compex number
		out_r[i] = sqrt(out_r[i]*out_r[i]+out_im[i]*out_im[i]); // to increase the speed delete sqrt
		out_im[i] = i * Frequency / N;
		std::cout << (out_im[i]); std::cout << ("Hz");
		std::cout << ("\t");	// un comment to print freuency bin 
		std::cout << (out_r[i]);
		std::cout << std::endl;
	}




	x = 0; // peak detection
	for(int i=1;i<sampleRates[o-1]-1;i++){
		if(out_r[i]>out_r[i-1] && out_r[i]>out_r[i+1]){
			in_ps[x] = i; //in_ps array used for storage of peak number
			x = x + 1;
		} 
	}


	s = 0;
	c = 0;
	for(int i=0;i<x;i++){ // re arraange as per magnitude
		for(int j=c;j<x;j++){
			if(out_r[in_ps[i]]<out_r[in_ps[j]]){
				s=in_ps[i];
				in_ps[i]=in_ps[j];
				in_ps[j]=s;
			}
		}
		c=c+1;
	}
	float* f_peaks = new float[sampleRates[o]];
	for(int i=0;i<5;i++){ // updating f_peak array (global variable)with descending order
		f_peaks[i]=out_im[in_ps[i]];
	}
	return f_peaks;
}

//------------------------------------------------------------------------------------//
//main.cpp
int data[64]={
14, 30, 35, 34, 34, 40, 46, 45, 30, 4, -26, -48, -55, -49, -37,
-28, -24, -22, -13, 6, 32, 55, 65, 57, 38, 17, 1, -6, -11, -19, -34, 
-51, -61, -56, -35, -7, 18, 32, 35, 34, 35, 41, 46, 43, 26, -2, -31, -50,
-55, -47, -35, -27, -24, -21, -10, 11, 37, 58, 64, 55, 34, 13, -1, -7
};

int main(){
	const unsigned int SAMPLE_RATE = 48*1000;	//48khz
	auto result = FFT(data,64,SAMPLE_RATE);
	std::cout << result[0] << " " << result[1] << " " << result[2] << " " << result[3] << std::endl;
	delete[] result;
	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;
}