HackerEarth Meeting the origin problem solution YASH PAL, 31 July 2024 In this HackerEarth Meeting the origin problem solution You are given a string path S which consists of only four characters ‘L’, ‘R’, ‘U’, and ‘D’. You are initially at origin (0,0). You have to follow the directions as provided in path S in order. For ‘L’ you have to move 1 unit left, for ‘R’ move 1 unit right, for ‘U’ move 1 unit up and for ‘D’ move 1 unit down. You can do these operations any number of times: Delete any single move and remove it from the string path . Convert any single move ‘L’ to ‘R’ or vice-versa. Convert any single move ‘U’ to ‘D’ or vice-versa. Print the minimum operations required, after which going through the new path, you will meet the origin again. HackerEarth Meeting the origin problem solution. #include<bits/stdc++.h>using namespace std;#define F first#define S second#define pb push_back#define lb lower_bound#define ub upper_bound#define vi vector<int>#define all(x) x.begin(),x.end()#define fix fixed<<setprecision(10)#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)#define repb(i,b,a) for(int i=int(b);i>=int(a);i--)#define FastIO ios_base::sync_with_stdio(0),cin.tie(0)typedef double db;typedef long long ll;const int N=2e5+5;const int mod=1e9+7;void solve(){ string s; cin>>s; int l=count(all(s),'L'),r=count(all(s),'R'),u=count(all(s),'U'),d=count(all(s),'D'); int ans=(l+r)%2+(u+d)%2+abs(l-r)/2+abs(u-d)/2; cout<<ans<<'n';}signed main(){ FastIO; int t; cin>>t; while(t--) solve(); return 0;} Second solution t = int(input())while t > 0: t -= 1 s = input() print((abs(s.count('L') - s.count('R')) + 1) // 2 + (abs(s.count('U') - s.count('D')) + 1) // 2) coding problems