Hello everyone,
problems about swapping adjacent elements are quite frequent in CP, but they can be tedious. In this tutorial we will see some easy ideas and use them to solve some problems of increasing difficulty.
Target: rating $$$[1400, 2100]$$$ on CF
Prerequisites: greedy, basic DP, Fenwick tree (or segment tree)
Counting inversions
Let's start from a simple problem.
You are given a permutation $$$a$$$ of length $$$n$$$. In one move, you can swap two elements in adjacent positions. What's the minimum number of moves required to sort the array?
Claim
The result $$$k$$$ is equal to the number of inversions, i.e. the pairs $$$(i, j)$$$ ($$$1 \leq i < j \leq n$$$) such that $$$a_i > a_j$$$.
Proof 1
Let $$$f(x)$$$ be the number of inversions after $$$x$$$ moves.
In one move, if you swap the values on positions $$$i, i + 1$$$, $$$f(x)$$$ either increases by $$$1$$$ or decreases by $$$1$$$. This is because the only pair $$$(a_i, a_j)$$$ whose relative order changed is $$$(a_i, a_{i+1})$$$. Since the sorted array has $$$0$$$ inversions, you need at least $$$k$$$ moves to sort the array.
For example, if you have the permutation $$$[2, 3, 7, 8, 6, 9, 1, 4, 5]$$$ ($$$16$$$ inversions) and you swap two adjacent elements such that $$$a_i > a_{i+1}$$$ (getting, for example, $$$[2, 3, 7, 6, 8, 9, 1, 4, 5]$$$), the resulting array has $$$15$$$ inversions, and if you swap two adjacent elements such that $$$a_i < a_{i+1}$$$ (getting, for example, $$$[3, 2, 7, 8, 6, 9, 1, 4, 5]$$$), the resulting array has $$$17$$$ inversions.
On the other hand, if the array is not sorted you can always find an $$$i$$$ such that $$$a_i > a_{i+1}$$$, so you can sort the array in $$$k$$$ moves.
Proof 2
For each $$$x$$$, let $$$f(x)$$$ be the number of inversions if you consider only the elements from $$$1$$$ to $$$x$$$ in the permutation.
First, let's put $$$x$$$ at the end of the permutation: this requires $$$x - pos(x)$$$ moves. That's optimal (the actual proof is similar to Proof 1; in an intuitive way, if you put the last element to the end of the array, it doesn't interfere anymore with the other swaps).
For example, if you have the permutation $$$[2, 3, 7, 8, 6, 9, 1, 4, 5]$$$ and you move the $$$9$$$ to the end, you get $$$[2, 3, 7, 8, 6, 1, 4, 5, 9]$$$ and now you need to sort $$$[2, 3, 7, 8, 6, 1, 4, 5]$$$. Hence, $$$f(x) = f(x-1) + x - pos(x)$$$. For each $$$x$$$, $$$x - pos(x)$$$ is actually the number of pairs $$$(i, j)$$$ ($$$1 \leq i < j \leq x$$$) such that $$$x = a_i > a_j$$$. So $$$f(x)$$$ is equal to the number of inversions.
Calculating inversions
You can use a Fenwick tree (or segment tree).
For each $$$j$$$, calculate the number of $$$i < j$$$ such that $$$a_i > a_j$$$.
The Fenwick tree should contain the frequency of each value in $$$[1, n]$$$ in the prefix $$$[1, j - 1]$$$ of the array.
So, for each $$$j$$$, the queries look like
- $$$\text{res} := \text{res} + \text{range_sum}(a_j + 1, n)$$$
- add $$$1$$$ in the position $$$a_j$$$ of the Fenwick tree
Observations / slight variations of the problem
By using a Fenwick tree, you are actually calculating the number of inversions for each prefix of the array.
You can calculate the number of swaps required to sort an array with distinct elements (not necessarily a permutation) by compressing the values of the array.
You can also calculate the number of swaps required to get an array $$$b$$$ starting from $$$a$$$, by renaming the values. For example,
$$$a = [2, 3, 7, 8, 6, 9, 1, 4, 5], b = [9, 8, 5, 2, 1, 4, 7, 3, 6]$$$ is equivalent to $$$a = [4, 8, 7, 2, 9, 1, 5, 6, 3], b = [1, 2, 3, 4, 5, 6, 7, 8, 9]$$$
$$$a^{-1}$$$ (a permutation such that $$$a^{-1}[a[x]] = x$$$) has the same number of inversions as $$$a$$$. For example, $$$[2, 3, 7, 8, 6, 9, 1, 4, 5]$$$ and $$$[7, 1, 2, 8, 9, 5, 3, 4, 6]$$$ have both $$$16$$$ inversions. Sketch of a proof: note that, when you swap two elements in adjacent positions in $$$a$$$, you are swapping two adjacent values in $$$a^{-1}$$$, and the number of inversions in $$$a^{-1}$$$ also increases by $$$1$$$ or decreases by $$$1$$$ (like in Proof 1).
