##### Canadian Computing Competition: 2021 Stage 1, Junior #4

Valentina wants books on a shelf to be arranged in a particular way. Every time she sees a shelf of books, she rearranges the books so that all the large books appear on the left, followed by all the medium-sized books, and then all the small books on the right. She does this by repeatedly choosing any two books and exchanging their locations. Exchanging the locations of two books is called a swap.

Help Valentina determine the fewest number of swaps needed to arrange a shelf of books as she wishes.

#### Input Specification

The input will consist of exactly one line containing at least one character. The following table shows how the available marks are distributed.

Subtask | Number of books | Book types |
---|---|---|

marks | at most characters | each character will be `L` or `S` |

marks | at most characters | each character will be `L` or `S` |

marks | at most characters | each character will be `L` , `M` or `S` |

marks | at most characters | each character will be `L` , `M` or `S` |

#### Output Specification

Output a single integer which is equal to the minimum possible number of swaps needed to
arrange the books so that all occurrences of `L`

come first, followed by all occurrences of `M`

,
and then all occurrences of `S`

.

#### Sample Input 1

`LMMMS`

#### Output for Sample Input 1

`0`

#### Explanation of Output for Sample Input 1

The books are already arranged according to Valentina's wishes.

#### Sample Input 2

`LLSLM`

#### Output for Sample Input 2

`2`

#### Explanation of Output for Sample Input 2

By swapping the `S`

and `M`

, we end up with `LLMLS`

. Then the `M`

can be swapped with the `L`

to
its right. This is one way to use two swaps to arrange the books as Valentina wants them to
be. It is not possible to use fewer swaps because both `S`

and `M`

need to be moved but using
one swap to exchange them does not leave the books arranged as Valentina wants them to
be.

## Comments

This is the same as this one from usaco 2007 except with higher constraints. A very interesting problem, I have to say!

I don't understand what is wrong with my algorithm

I start by finding the highest index L and the lowest index S and swapping them if the index of S is less. Then If there is a M, then I swap the highest index M with the lowest index S if the index of S is once again less. If all those have passed false then I swap the highest index L with the lowest index M is the index of M is less. I don't understand what is wrong with this and for all the quick tests I have done for myself, it seemed to word so I am really confused on what is the problem.

I got 7/15 points with this algorithm, getting the perfect marks on the first 3 batches.

You're getting TLE (

Time Limit Exceeded), try optimizing your algorithm and then check if you're getting WA or AC. How your algorithm works looks correct though.EDIT: You may find this helpful, if you hadn't already read it - https://dmoj.ca/tips