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

Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(1) as constant space is used.