in Algorithms edited by
34 votes

An array contains four occurrences of $0$, five occurrences of $1$, and three occurrences of $2$ in any order. The array is to be sorted using swap operations (elements that are swapped need to be adjacent).

  1.  What is the minimum number of swaps needed to sort such an array in the worst case?
  2.  Give an ordering of  elements in the above array so that the minimum number of swaps needed to sort the array is maximum.
in Algorithms edited by


$\textbf{Just for knowledge – }$

If we remove the constraint of the $\textit{adjacent swap}$, then this problem can be solved in $O(n)$ with $O(1)$ extra space i.e, $\textit{inplace}$.

Here is the code:

class Solution {
    void Sort(vector<int>& a) {
        int i=0,j=a.size()-1,k=i;
            if(a[k]==0) swap(a[i++],a[k++]);
            else if(a[k]==1) k++;
            else if(a[k]==2) swap(a[j--],a[k]);



yes we can solve it with count sort or selection sort also,if they didn’t have mentioned about adjacent swapping right,in O(n)  swaps.


@jatinmittal199510 Just for extra information…. it wouldn't be stable.


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5 Answers

40 votes
Best answer
Since swaps are needed to be of adjacent elements only, the algorithm is actually Bubble sort.

In bubble sort, all smaller elements to right of an element are required to be swapped. So, if have ordering

$[2,2,2,1,1,1,1,1,0,0,0,0]$, then we need total $47$ swaps, and this will be the worst case.

So, it answers actually both parts.
edited by


we have to apply "Bubble sort" because as per question element that is swapped need to be adjacent.

array [ 2,2,2,1,1,1,1,1,0,0,0,0]   total 12 elements, as we know that bubble is stable in nature so repeated element order will not change.

pass 1> swap (9) ,pass 2 >sawp (9) ,pass 3>sawp (9)

pass 4> swap(4),pass 5>swap(4),pass 6>swap(4),pass 7>swap(4),pass 8>swap(4)

pass 9> sawp(0),pass10>swap(0),pass 11>swap(0)   

total swap = 47 


yes @ Prateek

Thanks :)

It can also be done with insertion sort.

pass 1 - swap(3), pass2- swap(3),pass3-swap(3), pass4-swap(3),pass5-swap(3), pass6-swap(8) ,pass7-swap(8), pass9-swap(8) ,pass10-swap(8)
insertion sort doesn't always give *adjacent swaps*

@ Prateek kumar

I could not understand how u have counted number of swaps.. Can u please explain in more detail
insertion sort also gives adjacent swaps???

The total no. of inversions present in the array = 47 , so can we say that no. of inversions = total no. of swaps required in case of bubble sort? I know for insertion sort its true.

insertion sort also has adjacent swaps always????

if not then please explain???
According to me adjacent swap means to put an element to its correct position you always have to swap that element with its adjacent element. In bubble sort in every iteration there is one element which gets sorted and to sort it in its correct position we swap it with it's adjacent element. Where in insertion sort we don't actually swap that element with everyone ,but we swap all greater elements to the right side and then we put original element on its right place.
if in the question they ask for no. of the pass then answer will be 9 or 11?
Why can't Insertion Sort be used here, bcoz it also swaps adjacent elements only, at least thats what I know? Can anybody give an example of insertion sort not using adjacent swaps.

 do you mean to say that in case of insertion sort ,we take an element(suppose we call it key) and compare with not only with its adjacent one rather with all those present to the left of it and which are greater than it ,they are shifted to right and not swapped and then we find the correct position of key we place it 

but in bubble sort ,adjacent swaps takes place ,am i correct or wrong?



@codingo1234 what I think is in Insertion sort when we traverse to the left of the key in the list, we swap each time we find an element greater than the key as we maintain a single pointer to traverse left and sorting happens in place. Shifting right would replace the key and key would be lost and to preserve it we will need extra space to store key which is not the way insertion sort works.

And that extra space would be constant ,so how does it violate inplace property of insertion sort,?

My bad! Space complexity would remain constant. Thanks a lot. Got it, In insertion sort key is not swapped but placed at the correct position, hence bubble sort. Thanks @codingo1234 

If its asked maximum numbers of swap in the best case and adjacent swap is not necessary then can we use selection sort with 9 Swaps?
In insertion sort, we $\textbf{shift}$ and then move the key at its appropriate position.
38 votes

Reference :

The number of swaps an algorithm performs is equal to the number of inversions present in the input array.

The question is somewhat directly taken from the exercise of coremen

I'll solve this later first question :P

(a) Minimum number of swaps in worst case

   This would occur when all the elements in the array would be present in descending order.

So, Number of inversions for 2=(9*3)=27

     Number of inversions for 1 = (4*5)=20

     0 would have no inversions.

So, total inversions=47

2 2 2 1 1 1 1 1 0 0 0 0

(b) The above configuration of the array will give the maximum number of swaps to sort the array.

Now coming to coremen exercise

We know insertion sort works by placing the unsorted element in it's correct position within sorted elements.

The running time of insertion sort depends on the number of inversions present in the array.

Assuming array has n distinct elements.

Case A : Array is sorted in ascending order

In this case, number of Inversions would be 0, so 0 swaps, however, a total of n-1 comparisons will still be made.

So, best case complexity comes out to be O(n).

Case B: Array is sorted in descending order

Total inversions come out to be 

(n-1)+(n-2)+(n-3)+.......1 = $\frac{n(n-1)}{2}$  =  O(n2).

This will be exactly equal to the number of swaps and comparisons insertion sort will perform in this case.

So, worst case complexity comes out to be O(n2).


u mean swap and inversion same thing?

But those two are not exacly same
No they both are not same. But a swap will happen only if there is an inversion pair in case of Insertion sort

"The number of  adjacent swaps an algorithm performs to sort an input in a non-decreasing order is equal to the number of inversions present in the input array."

This should be a more precise statement in my opinion.

@Nirmal Gaur-It's an idea of Insertion sort and it always sorts adjacent elements only.

The number of swaps an algorithm performs is equal to the number of inversions present in the input array.

The way it's written, it seems like a generalized statement so i was clearifing for those may find it little bit confusing. 


you made this one simple thanks.

In worst case the maximum number of swaps for the best running algorithm would be O(n) right ?


@Ayush Upadhyaya

It has been told that we have adjacent swaps so.In insertion sort we donot have adjacent swaps between 2 elements so .This is a question in bubble sort,whose time complexity also depends on the number of inversions.


@Doraemon-I think in insertion sort we have adjacent swaps.

no we don't have adjacent swaps.


fix 4

then check 3 with 4 and swap


then check 2 with 4 then 3

2<4 so shift 4 right.

2<3 so shift 3 right.

now place 2 in 1st position. this swap of 2 was not adjacent.


Similarly, shift 4,3,2 right and then place 1 at first position.

So this swapping of 1 is not adjacent.

Note :- when we were comparing an element to place in correct position it was stored in a temp variable.

@Ayush Upadhyaya 

We have adjacent shiftings not swapings.


What will be answer if elements to be swaped, need not be adjacent?
Then need not be use bubble sort, using merge sort we can do it.
If we use selection sort then we can do in 7 swaps worst case..right?

For any input  selection sort can do sorting in atmost n-1 swaps..
but TC of selection sort is $O\left ( n^{2} \right )$, but for merge sort $O\left ( n\log n \right )$
In merge sort comparison and copying is done..not swaping..

Here complexity is not an issue.
Can u tell me an example??

yes, if time is not issue, we can use any sorting, depending on how elements are inserting
if time is not an issue .

then selection sort is the best I think.because the number of swaps in selection sort is always theta(n).

@Ayush Upadhyaya, I think there is no swap in Insertion sort. There is only left or right shift depending on sorting. Question asks for adjacent swaps which take place in Bubble sort.

4 votes

$a)$ Bubble sort

Source :



$\downarrow$ After $9$ swaps


$\downarrow$ Similarly After $18$ swaps


$\downarrow$ After $4$ swaps


$\downarrow$ Similarly After $16$ swaps


$9+9+9+4+4+4+4+4=47$ Swaps



0 votes

elements that are swapped need to be adjacent

This screams bubble sort. The worst case would be when the input is sorted in descending order.

$\left \{ 2,2,2,1,1,1,1,1,0,0,0,0 \right \}$


No of swaps = Number of inversions.

For all three of the 2's, all others are inversions => $15+12$

For all five of the 1's => $20$ (count just with 0, because 2 already covered)

For all the 0's, inversions are already covered.

Total inversions = $20+15+12=47$

Hence, swaps required = $47$

–1 vote
Ans - 7 swaps


someone has used bubble sort to give answer as 47 swaps. Is it correct? and if yes how?
To be honest for this question particularly we have to use Quick sort three way partitioning algorithm.which i haven't learned. I used program to find the number of swaps.But if you want learn this  three-way partitioning. the best source i found is sir Robert Sedgewick course.

make sure you follow : "The array needs to be sorted using swap operations, elements which are swapped needs to be adjacent only".

Quick sort = adjacent elements sort only? 

It is CSE GATE 2000 Question Number 17
check it out.


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