Algorithms

I am currently taking a exercise is the shaker sort algorithm.Instead of separating two for loops, I have included a for-j loop inside a for-i loop. Please check my code.

public static void shakerSort(int array[], int n) 
{
    boolean swapped = true;
    while (swapped) 
    {
         swapped = false;
         for (int i = 0; i < n - 1; i++) 
         {
             if (array[i] > array[i + 1]) 
             {
                    int temp = array[i];
                    array[i] = array[i + 1];
                    array[i + 1] = temp;
                    swapped = true;
             }
                
             if (i == n - 2) 
             {
                 if (!swappped)
                 {
                    break;
                 }
                 else
                 {
                     swapped = false;
                     for (int j = n - 1; j > 0; j--) 
                     {
                          if (array[j - 1] > array[j]) 
                          {
                               int temp = array[j];
                               array[j] = array[j - 1];
                               array[j - 1] = temp;
                               swapped = true;
                          }
                     }
                     if (!swapped) 
                     {
                       break;
                     }
                 }     
             }
        }
    }

Output:

   -5
   -3
    5 
    9 
   11 
   12
   15
   18 

I hope you can rate my code .Thank you very much 

public int sumSqrtCeil(int n, String[] edges){
        ...

        List<Set<Integer>> edgeSetList = "converted first edges element to set of Integer"

        // iterate rest of edges elements -A
        for(int i=1;i<edges.length;i++){

            Set<Integer> nodeDuo = "converted this edges element to set of Integer"

            boolean found = false;
            //search for nodeDuo in set elements of edgeSetList

            //iterate the nodeDuo 2 element set -B
            for(Integer e : nodeDuo){
                //iterate the edgeSetList containing a linked set collection -C
                for (Set<Integer> edgeSet : edgeSetList) {
                    found = edgeSet.contains(e);
                    if (found) {
                        break;
                    }
                }
                if(found) break;
            }

            if(!found){
    
                //if nodeDuo has no element common to any element of edgeSetList then add a new set

                indexOfEdgeSetList++;
                edgeSetList.add(nodeDuo);
            }
            else{
                //else add nodeDuo to the existing set element of edgeSetList in which any of nodeDuo elements were found.
                edgeSetList.get(indexOfEdgeSetList).addAll(nodeDuo);
            }
        }


        Set<Integer> linkedNodes =         //identify a set of all nodes which are linked to a different node(s)

        //Add 1 for each unit nodes (the nodes which are no linked to any other nodes) -D
        for(int i=1;i<=n;i++){
            if(!linkedNodes.contains(i)) output+=1;
        }

        // Add ceil of square root of 'node count' of each linked set in linked nodes set collection -E
        output += edgeSetList.stream().mapToInt(integers -> (int) Math.ceil(Math.sqrt(integers.size()))).sum();
        return output;
    }

Is there any fast algorithm for this:
K people sit at a table. Each person holds a number (hi) which means: some of his neighbour has a cap of height (hi). I want to count the number of combinations for setting people up at the table according to their height of their cups.

Example:

in:
6
2 6 4 5 3 5
out:
2
(We have two combinations which are 1 2 6 4 5 3 and 6 2 1 4 5 3)
Explanation:
We know that first person points on the second person and last person points on the penultimate person. The neighbour of the second person on the left is two, so his neighbour with height 4 should be on the right. Next we know that left neighbour of fifth person is 4, so his neighbour with height 3 should be on the right. We have two combinations to place 1, 6 cups (1,6 and 6, 1):

Visualisation:
2 6 4 5 3 5
X 2 X X 5 X
X 2 X 4 5 3

Possibilities: 1 2 6 4 5 3 or 6 2 1 4 5 3

Any suggestions or help would be greatly appreciated.