Shubhi

#include <bits/stdc++.h>
using namespace std;

int main(){
      int n; cin>>n;
      vector<int> a(n);
      for(int i=0;i<n;i++) cin>>a[i];

      // General format for most of 2 pointer prob
      for(int L=0,R=0;L<n;L++){
            // R is the index which can't be in our range
            R = L; // we should do this naively but for O(N) comp we will do
            R = max(R,L);
            while(we_can_include_R_in_current_range){
                  include it
                  R++;
            }
            // Now L....R-1 is the max range satisfying
            // condition for this fixed L

            // take max or count something as per ques.....

            // Now we will shift L and find the best R for L+1
            // But our main goal is to retain the info for L+1...R-1
            // which we have computed right now

            if(we_have_L_included_in_our_range(means R>L)) {
                  remove L from the current range means
                  basically remove from the aggregate info in which we are interested
            }
      }
}