freeCodeCamp/curriculum/challenges/english/08-coding-interview-prep/data-structures/breadth-first-search.english.md

id, title, challengeType

id	title	challengeType
587d825c367417b2b2512c90	Breadth-First Search	1

Description

So far, we've learned different ways of creating representations of graphs. What now? One natural question to have is what are the distances between any two nodes in the graph? Enter graph traversal algorithms. Traversal algorithms are algorithms to traverse or visit nodes in a graph. One type of traversal algorithm is the breadth-first search algorithm. This algorithm starts at one node, first visits all its neighbors that are one edge away, then goes on to visiting each of their neighbors and so on until all nodes have been reached. Visually, this is what the algorithm is doing. To implement this algorithm, you'll need to input a graph structure and a node you want to start at. First, you'll want to be aware of the distances from the start node. This you'll want to start all your distances initially some large number, like Infinity. This gives a reference for the case where a node may not be reachable from your start node. Next, you'll want to go from the start node to its neighbors. These neighbors are one edge away and at this point you should add one unit of distance to the distances you're keeping track of. Last, an important data structure that will help implement the breadth-first search algorithm is the queue. This is an array where you can add elements to one end and remove elements from the other end. This is also known as a FIFO or First-In-First-Out data structure.

Instructions

Write a function bfs() that takes an adjacency matrix graph (a two-dimensional array) and a node label root as parameters. The node label will just be the integer value of the node between 0 and n - 1, where n is the total number of nodes in the graph. Your function will output a JavaScript object key-value pairs with the node and its distance from the root. If the node could not be reached, it should have a distance of Infinity.

Tests

tests:
  - text: 'The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>1</code> should return <code>{0: 1, 1: 0, 2: 1, 3: 2}</code>'
    testString: 'assert((function() { var graph = [[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]; var results = bfs(graph, 1); return isEquivalent(results, {0: 1, 1: 0, 2: 1, 3: 2})})(), ''The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>1</code> should return <code>{0: 1, 1: 0, 2: 1, 3: 2}</code>'');'
  - text: 'The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]</code> with a start node of <code>1</code> should return <code>{0: 1, 1: 0, 2: 1, 3: Infinity}</code>'
    testString: 'assert((function() { var graph = [[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]; var results = bfs(graph, 1); return isEquivalent(results, {0: 1, 1: 0, 2: 1, 3: Infinity})})(), ''The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]</code> with a start node of <code>1</code> should return <code>{0: 1, 1: 0, 2: 1, 3: Infinity}</code>'');'
  - text: 'The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>0</code> should return <code>{0: 0, 1: 1, 2: 2, 3: 3}</code>'
    testString: 'assert((function() { var graph = [[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]; var results = bfs(graph, 0); return isEquivalent(results, {0: 0, 1: 1, 2: 2, 3: 3})})(), ''The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>0</code> should return <code>{0: 0, 1: 1, 2: 2, 3: 3}</code>'');'
  - text: 'The input graph <code>[[0, 1], [1, 0]]</code> with a start node of <code>0</code> should return <code>{0: 0, 1: 1}</code>'
    testString: 'assert((function() { var graph = [[0, 1], [1, 0]]; var results = bfs(graph, 0); return isEquivalent(results, {0: 0, 1: 1})})(), ''The input graph <code>[[0, 1], [1, 0]]</code> with a start node of <code>0</code> should return <code>{0: 0, 1: 1}</code>'');'

Challenge Seed

function bfs(graph, root) {
  // Distance object returned
  var nodesLen = {};

  return nodesLen;
};

var exBFSGraph = [
  [0, 1, 0, 0],
  [1, 0, 1, 0],
  [0, 1, 0, 1],
  [0, 0, 1, 0]
];
console.log(bfs(exBFSGraph, 3));

After Test

// Source: http://adripofjavascript.com/blog/drips/object-equality-in-javascript.html
function isEquivalent(a, b) {
    // Create arrays of property names
    var aProps = Object.getOwnPropertyNames(a);
    var bProps = Object.getOwnPropertyNames(b);
    // If number of properties is different,
    // objects are not equivalent
    if (aProps.length != bProps.length) {
        return false;
    }
    for (var i = 0; i < aProps.length; i++) {
        var propName = aProps[i];
        // If values of same property are not equal,
        // objects are not equivalent
        if (a[propName] !== b[propName]) {
            return false;
        }
    }
    // If we made it this far, objects
    // are considered equivalent
    return true;
}

Solution

function bfs(graph, root) {
  // Distance object returned
  var nodesLen = {};
  // Set all distances to infinity
  for (var i = 0; i < graph.length; i++) {
    nodesLen[i] = Infinity;
  }
  nodesLen[root] = 0; // ...except root node
  var queue = [root]; // Keep track of nodes to visit
  var current; // Current node traversing
  // Keep on going until no more nodes to traverse
  while (queue.length !== 0) {
    current = queue.shift();
    // Get adjacent nodes from current node
    var curConnected = graph[current]; // Get layer of edges from current
    var neighborIdx = []; // List of nodes with edges
    var idx = curConnected.indexOf(1); // Get first edge connection
    while (idx !== -1) {
      neighborIdx.push(idx); // Add to list of neighbors
      idx = curConnected.indexOf(1, idx + 1); // Keep on searching
    }
    // Loop through neighbors and get lengths
    for (var j = 0; j < neighborIdx.length; j++) {
      // Increment distance for nodes traversed
      if (nodesLen[neighborIdx[j]] === Infinity) {
        nodesLen[neighborIdx[j]] = nodesLen[current] + 1;
        queue.push(neighborIdx[j]); // Add new neighbors to queue
      }
    }
  }
  return nodesLen;
}