HackerEarth Final Destination problem solution YASH PAL, 31 July 2024 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;} coding problems