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

• Last Updated : 21 May, 2021

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#includeusing 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")<

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 Norths = 0 # Count of Southe = 0 # Count of Eastw = 0 # Count of Westfor 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+= 1if(n == s and w == e):    print("YES")else:    print("NO")

C#

 // C# implementation of above approachusing 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



Javascript


Output:
YES

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