# Check if it is possible to return to the starting position after moving in the given directions

Given a string S having N directions in which a person travels. The task is to check if he/she will be able to return to the same place where he/she started. On Day i(1 <= i <= N), he will travel a positive distance in the following direction:

North if the i-th letter of str is N

West if the i-th letter of str is W

South if the i-th letter of str is S

East if the i-th letter of str is E

If he can return back to the place where he starts after nth day, print “YES” else print “NO”.**Examples:**

Input:str = “NNNWEWESSS”Output:YES

On the 1st, 2nd, and 3rd day he goes to north and on the 4th day he goes west, then eventually

returns where he was standing on the 3rd day on the 5th day, then on the 6th day he again goes to

west.On the 7th day he again return exactly to where he was standing on the 5th day.And on the

10th day he returns home safely.Input:str = “NW”Output:NO

**Approach:** There has to be a same number of N as there are a number of S and also the same number of E as there is W. So, count each type of directions given and just check if they are equal or not.

Below is the implementation of the above approach:

## C++

`// C++ implementation of above approach` `#include<bits/stdc++.h>` `using` `namespace` `std;` `int` `main()` ` ` `{` ` ` `string st = ` `"NNNWEWESSS"` `;` ` ` `int` `len = st.length();` ` ` `int` `n = 0 ; ` `// Count of North` ` ` `int` `s = 0 ; ` `// Count of South` ` ` `int` `e = 0 ; ` `// Count of East` ` ` `int` `w = 0 ; ` `// Count of West` ` ` `for` `(` `int` `i = 0; i < len ; i++ )` ` ` `{` ` ` `if` `(st[i]==` `'N'` `)` ` ` `n += 1;` ` ` `if` `(st[i] == ` `'S'` `)` ` ` `s += 1;` ` ` `if` `(st[i] == ` `'W'` `)` ` ` `w+= 1 ;` ` ` `if` `(st[i] == ` `'E'` `)` ` ` `e+= 1 ;` ` ` `}` ` ` ` ` `if` `(n == s && w == e)` ` ` `cout<<(` `"YES"` `)<<endl;` ` ` `else` ` ` `cout<<(` `"NO"` `)<<endl;` ` ` ` ` `}` ` ` `// This code is contributed by` ` ` `// Sahil_Shelangia` |

## Java

`// Java implementation of above approach` `public` `class` `GFG {` ` ` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `String st = ` `"NNNWEWESSS"` `;` ` ` `int` `len = st.length();` ` ` ` ` `int` `n = ` `0` `; ` `// Count of North` ` ` `int` `s = ` `0` `; ` `// Count of South` ` ` `int` `e = ` `0` `; ` `// Count of East` ` ` `int` `w = ` `0` `; ` `// Count of West` ` ` ` ` `for` `(` `int` `i = ` `0` `; i < len ; i++ )` ` ` `{` ` ` `if` `(st.charAt(i)==` `'N'` `)` ` ` `n+= ` `1` `;` ` ` `if` `(st.charAt(i) == ` `'S'` `)` ` ` `s+= ` `1` `;` ` ` `if` `(st.charAt(i) == ` `'W'` `)` ` ` `w+= ` `1` `;` ` ` `if` `(st.charAt(i) == ` `'E'` `)` ` ` `e+= ` `1` `;` ` ` `}` ` ` `if` `(n == s && w == e)` ` ` `System.out.println(` `"YES"` `);` ` ` `else` ` ` `System.out.println(` `"NO"` `) ;` ` ` ` ` `}` ` ` `// This Code is contributed by ANKITRAI1` `}` |

## Python

`# Python implementation of above approach` `st ` `=` `"NNNWEWESSS"` `length ` `=` `len` `(st)` `n ` `=` `0` `# Count of North` `s ` `=` `0` `# Count of South` `e ` `=` `0` `# Count of East` `w ` `=` `0` `# Count of West` `for` `i ` `in` `range` `(length):` ` ` `if` `(st[i]` `=` `=` `"N"` `):` ` ` `n` `+` `=` `1` ` ` `if` `(st[i]` `=` `=` `"S"` `):` ` ` `s` `+` `=` `1` ` ` `if` `(st[i]` `=` `=` `"W"` `):` ` ` `w` `+` `=` `1` ` ` `if` `(st[i]` `=` `=` `"E"` `):` ` ` `e` `+` `=` `1` `if` `(n ` `=` `=` `s ` `and` `w ` `=` `=` `e):` ` ` `print` `(` `"YES"` `)` `else` `:` ` ` `print` `(` `"NO"` `)` |

## C#

`// C# implementation of above approach` `using` `System;` `class` `GFG {` ` ` ` ` `// Main Method` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` ` ` `string` `st = ` `"NNNWEWESSS"` `;` ` ` `int` `len = st.Length;` ` ` ` ` `int` `n = 0 ; ` `// Count of North` ` ` `int` `s = 0 ; ` `// Count of South` ` ` `int` `e = 0 ; ` `// Count of East` ` ` `int` `w = 0 ; ` `// Count of West` ` ` ` ` `for` `(` `int` `i = 0; i < len ; i++ )` ` ` `{` ` ` `if` `(st[i]==` `'N'` `)` ` ` `n += 1 ;` ` ` `if` `(st[i] == ` `'S'` `)` ` ` `s += 1 ;` ` ` `if` `(st[i] == ` `'W'` `)` ` ` `w += 1 ;` ` ` `if` `(st[i] == ` `'E'` `)` ` ` `e += 1 ;` ` ` `}` ` ` ` ` `if` `(n == s && w == e)` ` ` `Console.WriteLine(` `"YES"` `);` ` ` `else` ` ` `Console.WriteLine(` `"NO"` `) ;` ` ` ` ` `}` `}` `// This code is contributed by Subhadeep` |

## PHP

`<?php` `// PHP implementation of above approach` `$st` `= ` `"NNNWEWESSS"` `;` `$len` `= ` `strlen` `(` `$st` `);` `$n` `= 0; ` `// Count of North` `$s` `= 0; ` `// Count of South` `$e` `= 0; ` `// Count of East` `$w` `= 0; ` `// Count of West` `for` `(` `$i` `= 0; ` `$i` `< ` `$len` `; ` `$i` `++ )` `{` ` ` `if` `(` `$st` `[` `$i` `] == ` `'N'` `)` ` ` `$n` `+= 1;` ` ` `if` `(` `$st` `[` `$i` `] == ` `'S'` `)` ` ` `$s` `+= 1;` ` ` `if` `(` `$st` `[` `$i` `] == ` `'W'` `)` ` ` `$w` `+= 1 ;` ` ` `if` `(` `$st` `[` `$i` `] == ` `'E'` `)` ` ` `$e` `+= 1;` `}` `if` `(` `$n` `== ` `$s` `&& ` `$w` `== ` `$e` `)` ` ` `echo` `"YES\n"` `;` `else` ` ` `echo` `"NO\n"` `;` `// This code is contributed by` `// Rajput-Ji` `?>` |

## Javascript

`<script>` `// JavaScript implementation of the approach` `// driver code` ` ` `let st = ` `"NNNWEWESSS"` `;` ` ` `let len = st.length;` ` ` ` ` `let n = 0 ; ` `// Count of North` ` ` `let s = 0 ; ` `// Count of South` ` ` `let e = 0 ; ` `// Count of East` ` ` `let w = 0 ; ` `// Count of West` ` ` ` ` `for` `(let i = 0; i < len ; i++ )` ` ` `{` ` ` `if` `(st[i]==` `'N'` `)` ` ` `n += 1 ;` ` ` `if` `(st[i] == ` `'S'` `)` ` ` `s += 1 ;` ` ` `if` `(st[i] == ` `'W'` `)` ` ` `w += 1 ;` ` ` `if` `(st[i] == ` `'E'` `)` ` ` `e += 1 ;` ` ` `}` ` ` ` ` `if` `(n == s && w == e)` ` ` `document.write(` `"YES"` `);` ` ` `else` ` ` `document.write(` `"NO"` `) ;` ` ` `</script>` |

**Output:**

YES