Creative solving of searching redundant connection in graph

Question

I am trying to solve problem in leetcode: https://leetcode.com/problems/redundant-connection/description/ *finding redundant connection in undirected graph

And now I am writing a solution, inpired by Kruskal's algorithm: https://en.wikipedia.org/wiki/Disjoint-set_data_structure#/media/File:UnionFindKruskalDemo.gif

In short: I just add vertices in sets if I find a pair, when both vertices exist in set then it is an additional edge...

var findRedundantConnection = function(edges) {
  let additionals = [];
  let sets = [];
  for (let i = 0; i < edges.length; i++) {
    //find connections of each vertices in edge
    let check_0_indexes = [];
    let check_1_indexes = [];
    for (let j = 0; j < sets.length; j++) {
      if (sets[j]&&sets[j].includes(edges[i][0])) { //if vertices exist in sets
        check_0_indexes.push(j);
      } else if (sets[j]&&sets[j].includes(edges[i][1])) {
        check_1_indexes.push(j);
      }
    }
    //add to sets (but if not an additional, if additional then add to additionals) 
    check_0_indexes.map(el=>{
        if(sets[el].includes(edges[i][1])){ //check first vertice: if pair exist in set
            additionals.push(edges[i]);     
        } else {
            sets[el].push(edges[i][1]);
      }
    });  
    check_1_indexes.map(el=>{
        if(sets[el].includes(edges[i][0])){
            additionals.push(edges[i]);    //check second vertice: if pair exist in set
        } else {
            sets[el].push(edges[i][0]);
        }
    });  
    if(!check_0_indexes.length&&!check_1_indexes.length){ 
        //if both vertices not in sets
        sets.push([]); //create new set and add vertices
        sets[sets.length-1].push(edges[i][0]);
        sets[sets.length-1].push(edges[i][1]); 
    }   
  }
  return additionals.pop();
};

*JavaScript

But in this case: edges = [[3,7],[1,4],[2,8],[1,6],[7,9],[6,10],[1,7],[2,3],[8,9],[5,9]]
I do not know how to catch loop ( (8,9) and (7,9) ) Any ideas about it?

Answer 1

This step is incomplete:

sets[el].push(edges[i][1]);

For edge $(2,3)$, only $3$ is getting added to the existing set $\{2,8\}$. And, only $2$ is getting added to existing set $\{3,7,9, \dotsc\}$.

You should also add all the elements of the set $\{2,8\}$ to $\{3,7,9, \dotsc\}$ and vice-versa. Note that you have to merge the two sets $\{2,8\}$ to $\{3,7,9,\dotsc\}$ into one.