检查在给定方向移动后是否可以返回到起始位置

给定一个人行进的具有 N 个方向的字符串S。任务是检查他/她是否能够回到他/她开始的同一个地方。在第 i(1 <= i <= N) 天，他将向以下方向移动正距离：

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

如果他能在第 n 天后回到他开始的地方，打印“YES”，否则打印“NO”。
例子：

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

方法： N 的数量必须与 S 的数量相同，E 的数量必须与 W 的数量相同。因此，计算给定的每种类型的方向，然后检查它们是否相等。
下面是上述方法的实现：

C++

// C++ implementation of above approach
#include
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")<

---

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

Javascript

 
// 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") ;
   

输出：

YES

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