HackerEarth Final Destination problem solution

In this HackerEarth Final Destination problem solution, Bob and Khatu are stuck in a matrix. The command center sent them a string that decodes to their final destination. Since Bob and Khatu are not good at problem-solving help them to figure out their final destination. They are initially at (0, 0). The string contains L, R, U, D denoting left, right, up, and down. In each command, they will traverse 1 unit distance in the respective direction. For example, if they are at (2, 0) and the command is they will go to (1, 0).

HackerEarth Final Destination problem solution.

#include <bits/stdc++.h>
using namespace std;
int main()
{
  string S;
  int X=0, Y=0;
  cin>>S;
  for(int i=0;i<S.size(); i++)
  {
    if(S[i]=='L')
      X--;
    else if(S[i]=='R')
      X++;
    else if(S[i]=='U')
      Y++;
    else if(S[i]=='D')
      Y--;
    else
      assert(0);
  }
  cout<<X<<" "<<Y<<endl;
  return 0;
}

Second solution

#include<bits/stdc++.h>

using namespace std;

#define vi vector < int >
#define pii pair < int , int >
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define foreach(it,v) for( __typeof((v).begin())it = (v).begin() ; it != (v).end() ; it++ )
#define ll long long
#define llu unsigned long long
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define dbg(x) { cout<< #x << ": " << (x) << endl; }
#define dbg2(x,y) { cout<< #x << ": " << (x) << " , " << #y << ": " << (y) << endl; }
#define all(x) x.begin(),x.end()
#define mset(x,v) memset(x, v, sizeof(x))
#define sz(x) (int)x.size()

int main()
{
    string s;
    cin >> s;
    int x = 0 , y = 0;
    int n = sz(s) , i;
    assert(1 <= n && n <= 100000);
    for(i=0;i<n;i++)
    {
        if(s[i] == 'L')
        {
            x--;
        }
        else if(s[i] == 'R')
        {
            x++;
        }
        else if(s[i] == 'U')
        {
            y++;
        }
        else if(s[i] == 'D')
        {
            y--;
        }
        else
        {
            assert(0);
        }
    }
    cout << x << " " << y << endl;
    return 0;
}