#include <iostream>
#include <string>	//Should already be in iostream
#include <cstdlib>

//A word score adds up the character values. a-z gets mapped to 1-26 for the values of the characters.

//wordScore [wordValue]
//Pipe in the input into stdin, or type the words yourself.

//Lowercase words only
int characterValue(const char b){
	return ((b >= 'a') && (b <= 'z'))? ((b - 'a') + 1) : 0;
}

int main(int argc, char** argv){
    //The first argument specifies if you are trying to look for a certain word score
	int wordValue = (argc > 1)? std::atoi(argv[1]) : 0;

	std::string line;
	while(std::getline(std::cin, line)){
		int sum = 0;
		for(const char c : line){
			sum += characterValue(c);
		}
		if(wordValue){	//If wordValue is 0 or the sum is the correct value
			if(wordValue == sum){
				std::cout << line << std::endl;
			}
		} else {
			std::cout << sum << "\t" << line << std::endl;
		}
	}
	return 0;
}

int main()

/*
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;
}

/*
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() {
  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;
}

using namespace std; 
  
class Hash 
{ 
    int BUCKET;    // No. of buckets 
  
    // Pointer to an array containing buckets 
    list<int> *table; 
public: 
    Hash(int V);  // Constructor 
  
    // inserts a key into hash table 
    void insertItem(int x); 
  
    // deletes a key from hash table 
    void deleteItem(int key); 
  
    // hash function to map values to key 
    int hashFunction(int x) { 
        return (x % BUCKET); 
    } 
  
    void displayHash(); 
}; 
  
Hash::Hash(int b) 
{ 
    this->BUCKET = b; 
    table = new list<int>[BUCKET]; 
} 
  
void Hash::insertItem(int key) 
{ 
    int index = hashFunction(key); 
    table[index].push_back(key);  
} 
  
void Hash::deleteItem(int key) 
{ 
  // get the hash index of key 
  int index = hashFunction(key); 
  
  // find the key in (inex)th list 
  list <int> :: iterator i; 
  for (i = table[index].begin(); 
           i != table[index].end(); i++) { 
    if (*i == key) 
      break; 
  } 
  
  // if key is found in hash table, remove it 
  if (i != table[index].end()) 
    table[index].erase(i); 
} 
  
// function to display hash table 
void Hash::displayHash() { 
  for (int i = 0; i < BUCKET; i++) { 
    cout << i; 
    for (auto x : table[i]) 
      cout << " --> " << x; 
    cout << endl; 
  } 
} 
  
// Driver program  
int main() 
{ 
  // array that contains keys to be mapped 
  int a[] = {15, 11, 27, 8, 12}; 
  int n = sizeof(a)/sizeof(a[0]); 
  
  // insert the keys into the hash table 
  Hash h(7);   // 7 is count of buckets in 
               // hash table 
  for (int i = 0; i < n; i++)  
    h.insertItem(a[i]);   
  
  // delete 12 from hash table 
  h.deleteItem(12); 
  
  // display the Hash table 
  h.displayHash(); 
  
  return 0; 
}