Final Cell Position in the Matrix in C++

Suppose we have a set of commands as a string, the string will have four different letters for four directions. U for up, D for down, L for left and R for right. We also have initial cell position (x, y). Find the final cell position of the object in the matrix after following the given commands. We will assume that the final cell position is present in the matrix. Suppose the command string is like “DDLRULL”, initial position is (3, 4). The final position is (1, 5).

The approach is simple, count the number of up, down, left and right movements, then find final position (x’, y’) using the formula −

(x’, y’) = (x + count_right – count_left,y + (count_down – count_up))

Example

 Live Demo

#include<iostream>
using namespace std;
void getFinalPoint(string command, int x, int y) {
   int n = command.length();
   int count_up, count_down, count_left, count_right;
   int x_final, y_final;
   count_up = count_down = count_left = count_right = 0;
   for (int i = 0; i < n; i++) {
      if (command[i] == 'U')
         count_up++;
      else if (command[i] == 'D')
         count_down++;
      else if (command[i] == 'L')
         count_left++;
      else if (command[i] == 'R')
         count_right++;
   }
   x_final = x + (count_right - count_left);
   y_final = y + (count_down - count_up);  
   cout << "Final Position: " << "(" << x_final << ", " << y_final << ")";
}
int main() {
   string command = "DDLRULL";
   int x = 3, y = 4;
   getFinalPoint(command, x, y);
}

Output

Final Position: (1, 5)