//Leif Messinger
//Compile with C++ 20
#include <iostream>
#include <ranges>
#include <vector>
#include <functional>
#include <cctype> //toupper
#include <cxxabi.h>

template <typename T>
void printType(){
    std::cout << abi::__cxa_demangle(typeid(T).name(), NULL, NULL, NULL) << std::endl;
}

template <typename T>
class Slicer{
    public:
        T begin_;
        T end_;
        T trueEnd;
        Slicer(T begin, T end): begin_(begin), end_(begin), trueEnd(end){}
        template<typename U>
        Slicer(U&& vec) : begin_(vec.begin()), end_(vec.begin()), trueEnd(vec.end()){}
        Slicer& finish(){
            begin_ = end_;
            end_ = trueEnd;
            return (*this);
        }
        Slicer& to(long int index){
            begin_ = end_;
            if(index > 0){
                end_ = (begin_ + index);
            }else{
                index *= -1;
                end_ = (trueEnd - index);
            }
            return (*this);
        }
        Slicer& operator[](long int index){
            return to(index);
        }
        T begin(){
            return this->begin_;
        }
        T end(){
            return this->end_;
        }
        Slicer& operator()(std::function<void(decltype(*begin_))> func) {
            for(decltype(*begin_) thing : (*this)){
                func(thing);
            }
            return (*this);
        }
};
static_assert(std::ranges::range< Slicer<std::vector<int>::const_iterator> >);

int main(){
    std::string vec = "abcdefghijklmnopqrstuvwxyz";
    Slicer<std::string::const_iterator> bruh(vec);
    
    //printType<decltype(bruh)>();

    bruh.to(3)([](char yeet){
        std::cout << yeet;
    })
    .to(-1)([](char yeet){
        std::cout << char(std::toupper(yeet));
    }).finish()([](char yeet){
        std::cout << yeet << yeet << yeet << yeet << yeet;
    });
    
    std::cout << std::endl << std::endl;
    
    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8};
    Slicer<int*> arrSlicer(arr, arr + (sizeof(arr)/sizeof(int)));
    
    std::cout << "[";
    arrSlicer.to(-1)([](int yeet){
        std::cout << yeet << ", ";
    }).finish()([](int yeet){
        std::cout << yeet << "]" << std::endl;
    });

    return 0;
}

#include <iostream>
#include "PlaylistNode.h"
using namespace std;
void PrintMenu(string title);

int main() {
   string plTitle;
   cout << "Enter playlist's title:" << endl;
   getline(cin, plTitle);
   PrintMenu(plTitle);
   return 0;
}

void PrintMenu(string title) {
   Playlist list;
   string id;
   string sname;
   string aname;
   int length;
   int oldPos;
   int newPos;
   char choice;
   
   while(true) {
      cout << endl << title << " PLAYLIST MENU" << endl;
      cout << "a - Add song" << endl;
      cout << "d - Remove song" << endl;
      cout << "c - Change position of song" << endl;
      cout << "s - Output songs by specific artist" << endl;
      cout << "t - Output total time of playlist (in seconds)" << endl;
      cout << "o - Output full playlist" << endl;
      cout << "q - Quit" << endl << endl;
      
      cout << "Choose an option:" << endl;
      cin >> choice;
      cin.ignore();
      
      if (choice == 'q') {
         exit(1);
      }
      else if (choice == 'a') {
         cout << "\nADD SONG" << endl;
         cout << "Enter song's unique ID: ";
         cin >> id;
         cin.ignore();
         cout << "Enter song's name: ";
         getline(cin,sname);
         cout << "Enter artist's name: ";
         getline(cin,aname);
         cout << "Enter song's length (in seconds): ";
         cin >> length;
         list.AddSong(id, sname, aname, length);
      }
      else if (choice == 'd') {
         cout << "\nREMOVE SONG" << endl;
         cout << "Enter song's unique ID: ";
         cin >> id;
         list.RemoveSong(id);
      }
      else if (choice == 'c') {
         cout << "\nCHANGE POSITION OF SONG" << endl;
         cout << "Enter song's current position: ";
         cin >> oldPos;
         cout << "Enter new position for song: ";
         cin >> newPos;
         list.ChangePosition(oldPos, newPos);
      }
      else if (choice == 's') {
         cout << "\nOUTPUT SONGS BY SPECIFIC ARTIST" << endl;
         cout << "Enter artist's name: ";
         getline(cin, aname);
         list.SongsByArtist(aname);
      }
      else if (choice == 't') {
         cout << "\nOUTPUT TOTAL TIME OF PLAYLIST (IN SECONDS)" << endl;
         cout << "Total time: " << list.TotalTime() << " seconds" << endl;
      }
      else if (choice == 'o') {
         cout << endl << title << " - OUTPUT FULL PLAYLIST" << endl;
         list.PrintList();
      }
      else {
         cout << "Invalid menu choice! Please try again." << endl;
      }
   }
}

#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;
}

/*
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;
}

// Iterative C++ program to
// implement Stein's Algorithm
//#include <bits/stdc++.h>
#include <bitset>
using namespace std;

// Function to implement
// Stein's Algorithm
int gcd(int a, int b)
{
    /* GCD(0, b) == b; GCD(a, 0) == a,
    GCD(0, 0) == 0 */
    if (a == 0)
        return b;
    if (b == 0)
        return a;

    /*Finding K, where K is the
    greatest power of 2
    that divides both a and b. */
    int k;
    for (k = 0; ((a | b) & 1) == 0; ++k)
    {
        a >>= 1;
        b >>= 1;
    }

    /* Dividing a by 2 until a becomes odd */
    while ((a & 1) == 0)
        a >>= 1;

    /* From here on, 'a' is always odd. */
    do
    {
        /* If b is even, remove all factor of 2 in b */
        while ((b & 1) == 0)
            b >>= 1;

        /* Now a and b are both odd.
        Swap if necessary so a <= b,
        then set b = b - a (which is even).*/
        if (a > b)
            swap(a, b); // Swap u and v.

        b = (b - a);
    } while (b != 0);

    /* restore common factors of 2 */
    return a << k;
}

// Driver code
int main()
{
    int a = 12, b = 780;
    printf("Gcd of given numbers is %d\n", gcd(a, b));
    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;
}