HackerEarth Meeting the origin problem solution

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)

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