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Final cell position in the matrix
  • Difficulty Level : Easy
  • Last Updated : 21 May, 2021

Given an array of list of commands U(Up), D(Down), L(Left) and R(Right) and initial cell position (x, y) in a matrix. Find the final cell position of the object in the matrix after following the given commands. It is assumed that the final required cell position exists in the matrix. 
Examples
 

Input : command[] = "DDLRULL"
        x = 3, y = 4
Output : (1, 5)

Input : command[] = "LLRUUUDRRDDDULRLLUDUUR"
        x = 6, y = 5
Output : (6, 3)

Source: Flipkart Interview (SDE-1 On Campus).
 

Approach: Following are the steps:
 

  1. Count cup, cdown, cleft and cright for U(Up), D(Down), L(Left) and R(Right) movements respectively.
  2. Calculate final_x = x + (cright – cleft) and final_y = y + (cdown – cup).

The final cell position is (final_x, final_y)
 

C++




// C++ implementation to find the final cell position
// in the given matrix
#include <bits/stdc++.h>
 
using namespace std;
 
// function to find the final cell position
// in the given matrix
void finalPos(char command[], int n,
              int x, int y)
{
    // to count up, down, left and cright
    // movements
    int cup, cdown, cleft, cright;
 
    // to store the final coordinate position
    int final_x, final_y;
 
    cup = cdown = cleft = cright = 0;
 
    // traverse the command array
    for (int i = 0; i < n; i++) {
        if (command[i] == 'U')
            cup++;
        else if (command[i] == 'D')
            cdown++;
        else if (command[i] == 'L')
            cleft++;
        else if (command[i] == 'R')
            cright++;
    }
 
    // calculate final values
    final_x = x + (cright - cleft);
    final_y = y + (cdown - cup);
 
    cout << "Final Position: "
         << "("
         << final_x << ", " << final_y << ")";
}
 
// Driver program to test above
int main()
{
    char command[] = "DDLRULL";
    int n = (sizeof(command) / sizeof(char)) - 1;
    int x = 3, y = 4;
    finalPos(command, n, x, y);
    return 0;
}

Java




// Java implementation to find
// the final cell position
// in the given matrix
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
// function to find the
// final cell position
// in the given matrix
static void finalPos(String command, int n,
                              int x, int y)
{
    // to count up, down, left
    // and cright movements
    int cup, cdown, cleft, cright;
 
    // to store the final
    // coordinate position
    int final_x, final_y;
 
    cup = cdown = cleft = cright = 0;
 
    // traverse the command array
    for (int i = 0; i < n; i++)
    {
        if (command.charAt(i) == 'U')
            cup++;
        else if (command.charAt(i) == 'D')
            cdown++;
        else if (command.charAt(i)== 'L')
            cleft++;
        else if (command.charAt(i) == 'R')
            cright++;
    }
 
    // calculate final values
    final_x = x + (cright - cleft);
    final_y = y + (cdown - cup);
 
    System.out.println("Final Position: " +
                       "(" + final_x + ", " +
                              final_y + ")");
}
 
// Driver Code
public static void main(String []args)
{
    String command = "DDLRULL";
    int n = command.length();
    int x = 3, y = 4;
    finalPos(command, n, x, y);
}
}
 
// This code is contributed
// by Subhadeep

C#




// C# implementation to find
// the final cell position
// in the given matrix
class GFG
{
// function to find the
// final cell position
// in the given matrix
static void finalPos(string command, int n,
                              int x, int y)
{
    // to count up, down, left
    // and cright movements
    int cup, cdown, cleft, cright;
 
    // to store the final
    // coordinate position
    int final_x, final_y;
 
    cup = cdown = cleft = cright = 0;
 
    // traverse the command array
    for (int i = 0; i < n; i++)
    {
        if (command[i] == 'U')
            cup++;
        else if (command[i] == 'D')
            cdown++;
        else if (command[i]== 'L')
            cleft++;
        else if (command[i] == 'R')
            cright++;
    }
 
    // calculate final values
    final_x = x + (cright - cleft);
    final_y = y + (cdown - cup);
 
    System.Console.WriteLine("Final Position: " +
                             "(" + final_x + ", " +
                                    final_y + ")");
}
 
// Driver Code
public static void Main()
{
    string command = "DDLRULL";
    int n = command.Length;
    int x = 3, y = 4;
    finalPos(command, n, x, y);
}
}
 
// This code is contributed
// by mits

PHP




<?php
// PHP implementation to find the final
// cell position in the given matrix
 
// function to find the final cell
// position in the given matrix
function finalPos($command, $n, $x, $y)
{
    // to count up, down, left and
    // cright movements
    $cup; $cdown; $cleft; $cright;
 
    // to store the final coordinate
    // position
    $final_x; $final_y;
 
    $cup = $cdown = $cleft = $cright = 0;
 
    // traverse the command array
    for ($i = 0; $i < $n; $i++)
    {
        if ($command[$i] == 'U')
            $cup++;
        else if ($command[$i] == 'D')
            $cdown++;
        else if ($command[$i] == 'L')
            $cleft++;
        else if ($command[$i] == 'R')
            $cright++;
    }
 
    // calculate final values
    $final_x = $x + ($cright - $cleft);
    $final_y = $y + ($cdown - $cup);
 
    echo "Final Position: " . "(" .
                  $final_x . ", " .
                  $final_y . ")";
}
 
// Driver Code
$command = "DDLRULL";
$n = strlen($command);
$x = 3; $y = 4;
finalPos($command, $n, $x, $y);
 
// This code is contributed
// by Akanksha Rai

Javascript




<script>
 
// JavaScript implementation to find
// the final cell position
// in the given matrix
 
// function to find the
// final cell position
// in the given matrix
function finalPos(command, n, x, y)
{
    // to count up, down, left
    // and cright movements
    let cup, cdown, cleft, cright;
   
    // to store the final
    // coordinate position
    let final_x, final_y;
   
    cup = cdown = cleft = cright = 0;
   
    // traverse the command array
    for (let i = 0; i < n; i++)
    {
        if (command[i] == 'U')
            cup++;
        else if (command[i] == 'D')
            cdown++;
        else if (command[i]== 'L')
            cleft++;
        else if (command[i] == 'R')
            cright++;
    }
   
    // calculate final values
    final_x = x + (cright - cleft);
    final_y = y + (cdown - cup);
   
    document.write("Final Position: " +
                             "(" + final_x + ", " +
                                    final_y + ")");
}
  
 
// driver code
 
    let command = "DDLRULL";
    let n = command.length;
    let x = 3, y = 4;
    finalPos(command, n, x, y);
   
</script>

Output:  

Final Position: (1, 5)

Time Complexity: O(n), where n  is the number of commands.
 

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