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AtCoder Beginner Contest 237(D - LR insertion)

Problem solving criteria:stack (https://www.cplusplus.com/reference/stack/stack/)

Think about the first input: LRRLR
That means 0->(L)->1->(R)->2->(R)->3->(L)->4->(R)->5
so we just need to save first R right number in an array(vector) & other number saves in a stack.
0->(L)->1  general array(vector)       stack(0)   top of stack=0
1->(R)->2  general array(vector)=1     stack(0)   top of stack=0
2->(R)->3  general array(vector)=1,2   stack(0)   top of stack=0
3->(L)->4  general array(vector)=1,2   stack(0,4) top of stack=4
4->(R)->5  general array(vector)=1,2,4 stack(0,4) top of stack=4
so answer will be (general array->last number(in that case 5)->all stack's element)
Answer:1 2 4 5 3 0

///Bismillahir Rahmanir Rahim
///Author: Tanvir Ahmmad
///CSE,Islamic University,Bangladesh

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<vector>
#include<iterator>
#include <functional> ///sort(arr,arr+n,greater<int>()) for decrement of array
/*every external angle sum=360 degree
  angle find using polygon hand(n) ((n-2)*180)/n*/
///Floor[Log(b)  N] + 1 = the number of digits when any number is represented in base b
using namespace std;
typedef long long ll;


int main()
{
    vector<int>vec;
    stack<int>stk;
    int n;
    string str;
    cin>>n>>str;
    for(int i=0;i<str.size();i++)
    {
        if(str[i]=='R') vec.push_back(i);
        else stk.push(i);
    }
    vec.push_back(str.size());
    while(!stk.empty())
    {
        vec.push_back(stk.top());
        stk.pop();
    }
    for(int i=0;i<vec.size();i++) cout<<vec[i]<<" ";
    cout<<endl;
    return 0;
}

///Alhamdulillah 
Problem Link:https://atcoder.jp/contests/abc237/tasks/abc237_d




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