n = int(input())
seq = input()
rBegin = rEnd = lBegin = lEnd = 0
R = L = False
for i in range(n):
if seq[i] == 'R':
if R == False:
R = True
rBegin = i+1
rEnd = i+1
elif seq[i] == 'L':
if L == False:
L = True
lBegin = i+1
lEnd = i+1
if R==True and L==True:
print(rBegin, lBegin-1)
elif R==True:
print(rBegin, rEnd+1)
else:
print(lEnd, lBegin-1)
#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <vector>
#include <map>
#define test_case int t;cin>>t;while(t--)
#define ll long long
#define endl "\n"
#define all(v) v.begin(),v.end()
using namespace std;
void fast()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
}
string base_converter(ll x,ll y)
{
char buffer [33];
itoa (x,buffer,y);
string s=buffer;
return s;
}
string findSum(string str1, string str2)
{
// Before proceeding further, make sure length
// of str2 is larger.
if (str1.length() > str2.length())
swap(str1, str2);
// Take an empty string for storing result
string str = "";
// Calculate length of both string
int n1 = str1.length(), n2 = str2.length();
// Reverse both of strings
reverse(str1.begin(), str1.end());
reverse(str2.begin(), str2.end());
int carry = 0;
for (int i=0; i<n1; i++)
{
// Do school mathematics, compute sum of
// current digits and carry
int sum = ((str1[i]-'0')+(str2[i]-'0')+carry);
str.push_back(sum%10 + '0');
// Calculate carry for next step
carry = sum/10;
}
// Add remaining digits of larger number
for (int i=n1; i<n2; i++)
{
int sum = ((str2[i]-'0')+carry);
str.push_back(sum%10 + '0');
carry = sum/10;
}
// Add remaining carry
if (carry)
str.push_back(carry+'0');
// reverse resultant string
reverse(str.begin(), str.end());
return str;
}
long long gcd(long long a, long long b) {
return a ? gcd(b % a, a) : b;
}
long long lcm(long long a, long long b) {
return a / gcd(a, b) * b;
}
long long fact(long long n) {
long long f = 1;
for (int i = 2; i <= n; i++)
{
f *= i;
}
return f;
}
long long ncr(long long n, long long r) {
return fact(n) / (fact(r) * fact(n - r));
}
long long npr(long long n, long long r) {
return fact(n) / fact(n - r);
}
bool prime (long long x)
{
long long i=2;
bool ch=true;
if (x==1)
ch=false;
while (x>=i*i)
{
if (x%i==0)
{
ch=false;
break;
}
i++;
}
if (ch==false)
{
return false;
}
else
return true;
}
int main()
{
int f=0 , l = 0;
int size ; cin >> size ;
string str ; cin >> str ;
for(int i=0 ; i<size ;i++) {
if (str[i] != '.') {
f = i;
break;
}
}
for(int i=size-1 ; i>=0 ;i--) {
if (str[i] != '.') {
l = i;
break;
}
}
if(str[f]==str[l] and str[l]=='R')
cout << f+1 << " " << l+1+1 ;
else if(str[f]==str[l] and str[l]=='L')
cout << l+1 << " " << f ;
else
{
for(int i=f ; i<size ; i++)
{
l = i ;
if(str[i]=='L')
break;
}
cout << ++f << " " << l ;
}
}
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946. Validate Stack Sequences | 921. Minimum Add to Make Parentheses Valid |
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528. Random Pick with Weight | 470. Implement Rand10() Using Rand7() |
866. Prime Palindrome | 1516A - Tit for Tat |
622. Design Circular Queue | 814. Binary Tree Pruning |
791. Custom Sort String | 787. Cheapest Flights Within K Stops |
779. K-th Symbol in Grammar | 701. Insert into a Binary Search Tree |
429. N-ary Tree Level Order Traversal | 739. Daily Temperatures |
647. Palindromic Substrings | 583. Delete Operation for Two Strings |
518. Coin Change 2 | 516. Longest Palindromic Subsequence |
468. Validate IP Address | 450. Delete Node in a BST |