1200A - Hotelier - CodeForces Solution


brute force data structures implementation *800

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Python Code:

n = int(input())
s = input()
ans = [0] * 10
for i in s:
    if i == 'L':
        for j in range(10):
            if not ans[j]:
                ans[j] = 1
                break
    elif i == 'R':
        for l in range(10):
            if not ans[9 - l]:
                ans[9 - l] = 1
                break
    else:
        ans[int(i)] = 0
print(*ans, sep = '')

C++ Code:

#include<stdio.h>
#include<unordered_map>
#include<iostream>
#include<string>
#include<stack>
#include<set>
#include<string.h>
#include<algorithm>
#include<deque>
#include<vector>
using namespace std;

typedef unsigned int uint;

//char s[1000000];
int ans[10];

int main() {

    
    int n;
    cin >> n;
    //scanf("%s", s);
    for (int i = 0; i < n; i++) {
        char th;
        cin >> th;
        if (th == 'L') {
            int j = 0;
            while (ans[j]) j++;
            ans[j] = 1;
        }
        if (th == 'R') {
            int j = 9;
            while (ans[j]) j--;
            ans[j] = 1;
        }
        else {
            ans[th - '0'] = 0;
        }
    }
    
    for (int j = 0; j < 10; j++) {
        cout << ans[j];
    }

    return 0;
}


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