Sorting a sequence by swapping adjacent elements using minimum swaps

This is a classical algorithm problem. The minimum number of swaps is equal to the number of inversions in the array. If we have index i and index j such that a_i > a_j and i < j then this is called an inversion. Let’s prove this statement! I will need a few lemmas on the way:

Lemma 1: If there is no inversion of two adjacent elements then the array is sorted.
Proof: Let’s assume that no two adjacent elements form an inversion. This means that a_i <= a_i+1 for all i in the interval [0, n-1]. As <= is transitive this will mean that the array is sorted.

Lemma 2: A single swap of two adjacent elements will reduce the total number of inversions in the array by at most 1.
Proof: when we swap two adjacent elements a_i and a_i+1 their relative position with respect to all the other elements in the array will remain unchanged. That is for all elements that were after a_i+1, they will still be after a_i+1 and for all elements before a_i, they will still be before the a_i. This also means that if a_i or a_i+1 formed an inversion with an element a_j then, they will still form an inversion with it after the swap. Therefor if we swap a_i and a_i+1 we will affect only inversions that these two elements used to form. As two elements may participate in no more than one inversion we have also proved the lemma.

Lemma 3: We need to perform at least NI swaps of adjacent elements in order to sort the array where NI is the number of inversions in the array
Proof: In a sorted array there are no inversions. Also according to lemma 2, a single swap can reduce the number of inversions by at most one. Thus we need to perform at least as many swaps as is the number of inversions.

Lemma 4: We can always sort the array performing NI swaps of adjacent elements, where just like above NI is the number of inversions in the array.
Proof: If we assume that in our array there is no inversion of two adjacent elements, then according to lemma 1, the array will be sorted and we are done.
Otherwise there is at least one pair of adjacent elements that form an inversion. We can swap them and thus reduce the total number of inversions by exactly once. We can continue performing this operation exactly NI times.

Now I have proven my statement from the beginning of the answer.

The only question left is how to count the number of inversions in a given array. You can do that using a slight modification of merge sort where you accumulate the inversions in the merge phase. You can have a look at this answer for details on how to implement that. The overall complexity of the algorithm is O(n*log(n)).