A person wants to go from origin to a particular location, he can move in only 4 directions(i.e East, West, North, South) but his friend gave him a long route, help a person to find minimum Moves so that he can reach to the destination.

Input – NESNWES

Output – E

E.g – SSSNEEEW

output – EESS

Input – NESNWES

Output – E

E.g – SSSNEEEW

output – EESS

Explanation : its very easy code :

1st way to think is with parenthesis matching there you have to match { , [ , ( and closing of it.

While here you have just four directions N,S,E,W.

instead of using Stack and increasing memory usage, we can take two variables say “north” and “east” which will take care of N,S and E,W.

for every occurrence of N or E, increment north or east by one.

and decrease for occurrences of S and W.

Finally in both north and east variable, there will be some value, just convert them in Direction and print it.

Thanks to Dhaval for suggesting this approach and Code

```
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
int NSValue(char exp){
if (exp == 'N' or exp == 'n') return 1;
else if(exp == 'S' or exp == 's') return -1;
}//NSValue
int EWValue(char exp){
if (exp == 'E' or exp == 'e') return 1;
else if(exp == 'W' or exp == 'w') return -1;
}//EWValue
int getDirection(char exp[]){
int east=0;
int north=0;
int i,j,temp,l;
l=strlen(exp);
for(i=0;i<l;i++){
if((exp[i]=='N' or exp[i]=='n') || (exp[i]=='S' or exp[i]=='s')){north+=NSValue(exp[i]);}
if((exp[i]=='E' or exp[i]=='e') || (exp[i]=='W' or exp[i]=='w')){east+=EWValue(exp[i]);}
}//for
if(east>0){for(i=0;i<east;i++){printf("%c",'E');}}//if
else if(east<0){east=-east;
for(i=0;i<east;i++){printf("%c",'S');}
}//if
if(north>0){for(i=0;i<north;i++){printf("%c",'N');}}//if
else if(north<0){north=-north;
for(i=0;i<north;i++){printf("%c",'S');}
}//elseif
}//getDirection
int main() {
//char exp[100] = "NESNWES";
char exp[100] = "SSSNEEEW";
getDirection(exp);
return 0;
}
```

You can check running code at http://ideone.com/e.js/U4yd4W