You took ptr1 and ptr2 but essentially what you are doing is equivalent to this.. ll books=0; for(int i=1;i<=n;++i){ ll count=0LL,sum=0LL; for(int j=i+1;j<=n;++j,++count){ if(sum+a[j]>t)
break; sum+=a[j]; } books=max(books,count); }

This has time complexity of O(n^2) so with large values of n you end up getting a TLE.. Instead you can try something like this.. You can have two pointers left and right; initially both are 1; keep incrementing right till sum+a[right]>t; at that point you can increment left and subtract a[left(before increment)] from sum and continue the process..