/*  Harry is a postman. He's got a post office with a size of n*m(a matrix / 2D array). Each slot at the 2D array 
 *  represents the number of letters in that spot. Harry can only go right and down. He starts at (0, 0), and ends at (n-1, m-1). 
 *  n represents the height, and m the length. Return the maximum amount of letters he can pick up. He can only pick up letters if he is on that spot.
 *  Examples
 *  harry(2, 2, [[5, 2], [5, 2]]) ➞ 12
 *  (5+5+2)
 *  harry(3, 5, [[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15]]) ➞ 72
 *  (1+6+11+12+13+14+15)
 *  harry(0, 0, [[]]) ➞ -1
 *  Notes
 *  Like you saw in example 3, if the matrix is empty, return -1.
 *  If the amount of letters is the same down and right, go right.
 *  You can't go down if you're on the bottom row, or right if you're on the most right column.
 *  
 * 
 *  After solving this problem, we realized that one of the tests the writer made would always cause a failure. The unit tests did not match the instructions.
 */


#include <QCoreApplication>
int harry(int n, int m, std::vector<std::vector<int> > vv);
int main(int argc, char *argv[])
{
    QCoreApplication a(argc, argv);
    int answer = harry(5,5,std::vector<std::vector<int> >({{5,6,2,5,1},{7,2,4,1,2},{0,7,5,2,14},{9,5,12,5,9},{19,5,2,6,2}}));
    return a.exec();
}
int harry(int n, int m, std::vector<std::vector<int> > vv){
  int lett = 0;
  if (vv.empty()) return -1;
  for (int i = 0; i < n; i++){
    for (int j = 0; j < m; j++){
      if (i == vv.size()-1 && j == vv[i].size()-1) return lett;
      if (i == vv.size()-1) lett+=vv[i][j+1];
      else if (i == vv[i].size()-1) lett+=vv[i+1][j];
      else {if (vv[i+1][j] > vv[i][j+1]) lett+=vv[i+1][j]; else lett+=vv[i][j+1];}
    }
  }
  return lett;
}