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;}`