Graph - Data Structures in Javascript 

Can you please provide a code snippet so we could discuss it in more details? Sounds like a very interesting topic :)

@@QuestionableCoding it's a bit messy but here is my buildPath function: function buildPath(listOfEdges, source, destination) { let path = []; let currentNode = destination while (currentNode !== source) { for (let i = 0; i < listOfEdges.length; i++) { // find first instance of current node as an edge destination if (listOfEdges[i][1] === currentNode) { // add edge to route path.unshift(listOfEdges[i]); // re-assign current node currentNode = listOfEdges[i][0]; // remove rest of edge list and jump out of loop listOfEdges.pop(listOfEdges.length - i) break; } } } // flatten array and remove duplicates return [...new Set([].concat(...path))]; } to initially build the list of edges in the main bfs function i did: listOfEdges.push([current, neigbour[i]]) at each iteration.

That will be an improvement but it won't affect our time complexity because the worst case is still there - we can expect up to N elements in our queue since our first element can be a neighbor of every other element. But still, for the sake of not doing redundant work, I would use your improvement and more to that, reorganise the code a bit not to have the first check for 'visited'. We'll need to mark our first node as visited before starting the while loop. All together it will be a more logical and optimised solution. Also there is a significant and tricky mistake with calling 'shift()'. In Javascript it will be O(N) because you need to reindex the entire array. So it is not acceptable and shouldn't be used. I'll fix that in the upcoming video about the BFS.

I think I didn't mention traversing the graph. I found a place where I was stating that the lookup operation is faster for the matrix. And by that I mean finding that one node has an edge with another which is O(1) for the matrix and O(N) for the adjacency list implementation.