In this HackerEarth Gift for Almas problem solution On his birthday, Almas was given a N x N(1 <= N <= 500) matrix of natural numbers up to 500 and instructions for it. The instruction consisted of symbols L and R, where if the symbol L is given you need to rotate the matrix 90 degrees to the left, and for the symbol R you need to rotate the matrix 90 degrees to the right. The instruction was only 3 characters in length so Almas could handle the twists with ease. Your task is to display the matrix that Almas had at the end of these turns.


HackerEarth Gift for Almas problem solution


HackerEarth Gift for Almas problem solution.

# include <bits/stdc++.h>

# include <ext/pb_ds/assoc_container.hpp>
# include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;
 
template<typename T> using ordered_set = tree <T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define _USE_MATH_DEFINES_
#define ll long long
#define ld long double
#define Accepted 0
#define pb push_back
#define mp make_pair
#define sz(x) (int)(x.size())
#define every(x) x.begin(),x.end()
#define F first
#define S second
#define lb lower_bound
#define ub upper_bound
#define For(i,x,y)  for (ll i = x; i <= y; i ++) 
#define FOr(i,x,y)  for (ll i = x; i >= y; i --)
#define debug(x) cerr << #x << " = " << x << endl
#define SpeedForce ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)

void setIn(string s) { freopen(s.c_str(),"r",stdin); }
void setOut(string s) { freopen(s.c_str(),"w",stdout); }
void setIO(string s = "") {
    if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}

const double eps = 0.000001;
const ld pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const ll MOD = 1e18 + 9;
const ll INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9; 
const int N = 2e5+5;                                         
const int M = 22;
const int pri = 997;
const int Magic = 2101;

const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
mt19937 gen(chrono::steady_clock::now().time_since_epoch().count());
 
int rnd (int l, int r) {
    return uniform_int_distribution<int> (l, r)(gen);
}

void rotate(vector<vector<int> > &a) {
    int n = a.size();
    vector<vector<int> > b = a;
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < n; ++j)
            b[j][n-i-1] = a[i][j];
            
    a.swap(b);
}

namespace ProblemA {
    
    void solve() {
        int n;
        cin >> n;
        vector < vector < int > > a(n, vector<int> (n));
        
        for (auto &ve : a)
            for (auto &x : ve)
                cin >> x;
                
        string s;
        cin >> s;
        int balance = 0;
        for(auto &it : s) {
            balance += it == 'R';
            balance -= it == 'L';
        }
        
        balance = (balance + 4) % 4;
        
        for (int it = 0; it < balance; ++it)
            rotate(a);
        
        for (int i = 0; i < n; ++i) {
            if(i) cout << '\n';
            for (int j = 0; j < n; ++j) {
                if(j) cout << ' ';
                cout << a[i][j];
            }
        }
    }
};

int32_t main () {
    SpeedForce;
    
    int TestCases = 1;
    
    for (int TestCase = 1; TestCase <= TestCases; ++TestCase) {
        
        ProblemA::solve();
    }
    
    return Accepted;
}

Second solution

n = int(input())
a = []
for i in range(n):
    a += [list(map(int, input().split()))]
rotate_l = sum(c == 'L' for c in input()) % 2 == 0
for j in (reversed(range(n)) if rotate_l else range(n)):
    print(*(a[i][j] for i in (range(n) if rotate_l else reversed(range(n)))))