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QuestionTime Complexity of following method? Pin
Member 1269110623-Nov-21 4:09
Member 1269110623-Nov-21 4:09 
Here edges is a String array representing a set of space separated integer pair elements e.g. "e1 e2" where 0 < e1!=e2 <= n

n is an integer, let's say ranging 0 < n <= 100000

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;
    }


As per my understanding (updated) complexity should be-

O( O(A) * O(B) * O(C) + O(D) + O(E) )
O( O(edges.length) * O(2) * O(n/2) + O(n) + O(edges.length) )
O( O(edges.length) * O(n) + O(n) + O(edges.length) ) //constant removed
O( O(n*edges.length) + O(n) + O(edges.length) ) //evaluate
O( O(n^3) + O(n) + O(n^2) ) //evaluate
O( O(n^3) ) //taking biggest factor
O(n^3) - Final Time Complexity.

where,
O(edges.length) = O( n! / 2! * (n-2)! ) = O( ( n * (n-1)) / 2 ) = O( n^2 - n / 2) ==> O(n^2)

Please feel free to share also consider me a complete noob for Algorithm. Just started learning formally.

modified 24-Nov-21 1:01am.

AnswerRe: Time Complexity of following method? Pin
Greg Utas23-Nov-21 5:17
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GeneralRe: Time Complexity of following method? Pin
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AnswerRe: Time Complexity of following method? Pin
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AnswerRe: Time Complexity of following method? Pin
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QuestionWhich is the more preferred approach for novice coders, depth first search, or, breadth first search? Pin
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