Feat/graph questions

book/D-interview-questions-solutions.asc

@@ -655,3 +655,194 @@ We could also have used a Map and keep track of the indexes, but that's not nece
 
 - Time: `O(n)`. We visit each letter once.
 - Space: `O(W)`, where `W` is the max length of non-repeating characters. The maximum size of the Set gives the space complexity. In the worst-case scenario, all letters are unique (`W = n`), so our space complexity would be `O(n)`. In the avg. case where there are one or more duplicates, it uses less space than `n`, because `W < n`.
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+:leveloffset: +1
+
+=== Solutions for Graph Questions
+(((Interview Questions Solutions, Graph)))
+
+:leveloffset: -1
+
+
+[#graph-q-course-schedule]
+include::content/part03/graph-search.asc[tag=graph-q-course-schedule]
+
+Basically, we have to detect if the graph has a cycle or not.
+There are multiple ways to detect cycles on a graph using BFS and DFS.
+
+One of the most straightforward ways to do it is using DFS one each course (node) and traverse their prerequisites (neighbors). If we start in a node, and then we see that node again, we found a cycle! (maybe)
+
+A critical part of solving this exercise is coming up with good test cases. Let's examine these two:
+
+[graphviz, course-schedule-examples, png]
+....
+digraph G {
+  subgraph cluster_1 {
+    a0 -> a1 -> a2
+    a0 -> a2 [color=gray]
+    label = "Example A"
+  }
+
+  subgraph cluster_2 {
+    b0 -> b1 -> b2 -> b3
+    b3 -> b1 [color=red]
+    label = "Example B";
+  }
+}
+....
+
+Let's say we are using a regular DFS, where we visit the nodes and keep track of visited nodes. If we test the example A, we can get to the course 2 (a2) in two ways. So, we can't blindly assume that "seen" nodes are because of a cycle. To solve this issue, we can keep track of the parent.
+
+For example B, if we start in course 0 (b0), we can find a cycle. However, the cycle does not involve course 0 (parent). When we visit course 1 (b1) and mark it as the parent, we will see that reach to course 1 (b1) again. Then, we found a cycle!
+
+[source, javascript]
+----
+include::interview-questions/course-schedule.js[tags=brute1]
+----
+
+We built the graph on the fly as an adjacency list (Map + Arrays).
+Then we visited each node, checking if there it has cycles. If none has cyles, then we return true.
+
+The cycle check uses DFS. We keep track of seen nodes and also who the parent is. If we get to the parent more than once, we have a cycle like examples A and B.
+
+What's the time complexity?
+
+We visite every node/vertex: `O(|V|)` and then for every node, we visite all it's edges, so we have `O(|V|*|E|)`.
+
+Can we do better?
+
+There's no need to visit nodes more than once. Instead of having a local `seen` variable for each node, we can move it outside the loop. However, it won't be a boolean anymore (seen or not seen). We could see nodes more than once, without being in a cycle (example A). One idea is to have 3 states: `unvisited` (0), `visiting` (1) and `visited` (2). Let's devise the algorithm:
+
+*Algorithm*:
+
+* Build a graph as an adjacency list (map + arrays).
+* Fill in every prerequisite as an edge on the graph.
+* Visit every node and if there's a cycle, return false.
+** When we start visiting a node, we mark it as 1 (visiting)
+** Visit all its adjacent nodes
+** Mark current node as 2 (visited) when we finish visiting neighbors.
+** If we see a node in visiting state more than once, it's a cycle!
+** If we see a node in a visited state, skip it.
+
+*Implementation*:
+
+[source, javascript]
+----
+include::interview-questions/course-schedule.js[tags=description;solution]
+----
+
+In the first line, we initialize the map with the course index and an empty array.
+This time the `seen` array is outside the recursion.
+
+*Complexity Analysis*:
+
+- Time: `O(|V| + |E|)`. We go through each node and edge only once.
+- Space: `O(|V| + |E|)`. The size of the adjacency list.
+
+
+
+
+//
+[#graph-q-critical-connections-in-a-network]
+include::content/part03/graph-search.asc[tag=graph-q-critical-connections-in-a-network]
+
+On idea to find if a path is critical is to remove it. If we visit the graph and see that some nodes are not reachable, then, oops, It was critical!
+
+We can code precisely that. We can remove one link at a time and check if all other nodes are reachable. It's not very efficient, but it's a start.
+
+[source, javascript]
+----
+include::interview-questions/critical-connections-in-a-network.js[tags=criticalConnectionsBrute1]
+----
+
+We are using a function `areAllNodesReachable`, which implements a BFS for visiting the graph, but DFS would have worked too. The runtime is `O(|E| + |V|)`, where `E` is the number of edges and `V` the number of nodes/servers. In `criticalConnectionsBrute1`, We are looping through all `connections` (`E`) to remove one connection at a time and then checking if all servers are still reachable with `areAllNodesReachable`.
+
+The time complexity is `O(|E|^2 * |V|)`. Can we do it on one pass? Sure we can!
+
+*Tarjan's Strongly Connected Components Algorithms*
+
+A connection is critical only if it's not part of the cycle.
+
+In other words, a critical path is like a bridge that connects islands; if you remove it you won't cross from one island to the other.
+
+Connections that are part of the cycle (blue) have redundancy. If you eliminate one, you can still reach other nodes. Check out the examples below.
+
+[graphviz, critical-connections-sol-examples, png]
+....
+graph G {
+  subgraph cluster_0 {
+    a0 -- a1 [color=blue]
+    a1 -- a2 [color=blue]
+    a2 -- a0 [color=blue]
+    a1 -- a3 [color=blue]
+    a3 -- a2 [color=blue]
+    label = "Example A";
+  }
+
+  subgraph cluster_3 {
+    b0 -- b1 [color=blue]
+    b1 -- b2 [color=blue]
+    b2 -- b0 [color=blue]
+    b1 -- b3 [color=red]
+    b3 -- b2 [color=transparent] // removed
+    label = "Example B";
+  }
+
+  subgraph cluster_1 {
+    c0 -- c1 -- c2 -- c3 [color=red]
+    label = "Example C";
+  }
+}
+....
+
+The red connections are critical; if we remove any, some servers won't be reachable.
+
+We can solve this problem in one pass using DFS. But for that, we keep track of the nodes that are part of a loop (strongly connected components). To do that, we use the time of visit (or depth in the recursion) each node.
+
+For example C, if we start on `c0`, it belongs to group 0, then we move c1, c2, and c3, increasing the depth counter. Each one will be on its own group since there's no loop.
+
+For example B, we can start at `b0`, and then we move to `b1` and `b2`. However, `b2` circles back to `b0`, which is on group 0. We can update the group of `b1` and `b2` to be 0 since they are all connected in a loop.
+
+For an *undirected graph*, If we found a node on our dfs, that we have previously visited, we found a loop! We can mark all of them with the lowest group number. We know we have a critical path when it's a connection that links two different groups. For example A, they all will belong to group 0, since they are all in a loop. For Example B, we will have `b0`, `b1`, and `b2` on the same group while `b3` will be on a different group.
+
+*Algorithm*:
+
+* Build the graph as an adjacency list (map + array)
+* Run dfs on any node. E.g. `0`.
+** Keep track of the nodes that you have seen using `group` array. But instead of marking them as seen or not. Let's mark it with the `depth`.
+** Visit all the adjacent nodes that are NOT the parent.
+** If we see a node that we have visited yet, do a dfs on it and increase the depth.
+** If the adjacent node has a lower grouping number, update the current node with it.
+** If the adjacent node has a higher grouping number, then we found a critical path.
+
+*Implementation*:
+
+[source, javascript]
+----
+include::interview-questions/critical-connections-in-a-network.js[tags=description;solution]
+----
+
+This algorithm only works with DFS.
+
+*Complexity Analysis*:
+
+- Time: `O(|E| + |V|)`. We visit each node and edge only once.
+- Space: `O(|E| + |V|)`. The graph has all the edges and nodes. Additionally, we use the `group` variable with a size of `|V|`.
+
+
+
+
+
+
+
+//
+
+
+
+

book/content/part03/graph-search.asc

@@ -7,9 +7,9 @@ endif::[]
 
 Graph search allows you to visit search elements.
 
-WARNING:  Graph search is very similar to <<Tree Search & Traversal>>. So, if you read that sections some of the concepts here will be familiar to you.
+WARNING:  Graph search is very similar to <<Tree Search & Traversal>>. So, if you read that section, some of the concepts here will be familiar to you.
 
-There are two ways to navigate the graph, one is using Depth-First Search (DFS) and the other one is Breadth-First Search (BFS). Let's see the difference using the following graph.
+There are two ways to navigate the graph, one is using Depth-First Search (DFS), and the other one is Breadth-First Search (BFS). Let's see the difference using the following graph.
 
 image::directed-graph.png[directed graph]
 
@@ -44,10 +44,10 @@ image::directed-graph.png[directed graph]
 
 ==== Depth-First Search for Graphs
 
-With Depth-First Search (DFS) we go deep before going wide.
+With Depth-First Search (DFS), we go deep before going wide.
 
 Let's say that we use DFS on the graph shown above, starting with node `0`.
-A DFS, will probably visit 5, then visit `1` and continue going down `3` and `2`. As you can see, we need to keep track of visited nodes, since in graphs we can have cycles like `1-3-2`.
+A DFS will probably visit 5, then visit `1` and continue going down `3` and `2`. As you can see, we need to keep track of visited nodes, since in graphs, we can have cycles like `1-3-2`.
 Finally, we back up to the remaining node `0` children: node `4`.
 
 So, DFS would visit the graph: `[0, 5, 1, 3, 2, 4]`.
@@ -56,13 +56,13 @@ So, DFS would visit the graph: `[0, 5, 1, 3, 2, 4]`.
 
 ==== Breadth-First Search for Graphs
 
-With Breadth-First Search (BFS) we go wide before going deep.
+With Breadth-First Search (BFS), we go wide before going deep.
 
 // TODO: BFS traversal
 Let's say that we use BFS on the graph shown above, starting with the same node `0`.
-A BFS, will visit 5 as well, then visit `1` and will not go down to it's children.
+A BFS will visit 5 as well, then visit `1` and not go down to its children.
 It will first finish all the children of node `0`, so it will visit node `4`.
-After all the children of node `0` are visited it continue with all the children of node `5`, `1` and `4`.
+After all the children of node `0` are visited, it will continue with all the children of node `5`, `1`, and `4`.
 
 In summary, BFS would visit the graph: `[0, 5, 1, 4, 3, 2]`
 
@@ -86,4 +86,131 @@ You might wonder what the difference between search algorithms in a tree and a g
 
 The difference between searching a tree and a graph is that the tree always has a starting point (root node). However, in a graph, you can start searching anywhere. There's no root.
 
-NOTE: Every tree is a graph, but not every graph is a tree.
+NOTE: Every tree is a graph, but not every graph is a tree. Only acyclic directed graphs (DAG) are trees.
+
+
+==== Practice Questions
+(((Interview Questions, graph)))
+
+
+
+
+// tag::graph-q-course-schedule[]
+===== Course Schedule
+
+*gr-1*) _Check if it's possible to take a number of courses while satisfying their prerequisites._
+
+// end::graph-q-course-schedule[]
+
+// _Seen in interviews at: Amazon, Facebook, Bytedance (TikTok)._
+
+
+*Starter code*:
+
+[source, javascript]
+----
+include::../../interview-questions/course-schedule.js[tags=description;placeholder]
+----
+
+
+*Examples*:
+
+[source, javascript]
+----
+canFinish(2, [[1, 0]]); // true
+// 2 courses: 0 and 1. One prerequisite: 0 -> 1
+// To take course 1 you need to take course 0.
+// Course 0 has no prerequisite, so you can take 0 and then 1.
+
+canFinish(2, [[1, 0], [0, 1]]); // false
+// 2 courses: 0 and 1. Two prerequisites: 0 -> 1 and 1 -> 0.
+// To take course 1, you need to take course 0.
+// To Course 0, you need course 1, so you can't any take them!
+
+canFinish(3, [[2, 0], [1, 0], [2, 1]]); // true
+// 3 courses: 0, 1, 2. Three prerequisites: 0 -> 2 and 0 -> 1 -> 2
+// To take course 2 you need course 0, course 0 has no prerequisite.
+// So you can take course 0 first, then course 1, and finally course 2.
+
+canFinish(4, [[1, 0], [2, 1], [3, 2], [1, 3]]); // false
+// 4 courses: 0, 1, 2, 3. Prerequisites: 0 -> 1 -> 2 -> 3 and 3 -> 1.
+// You can take course 0 first since it has no prerequisite.
+// For taking course 1, you need course 3. However, for taking course 3
+// you need 2 and 1. You can't finish then!
+----
+
+
+_Solution: <<graph-q-course-schedule>>_
+
+
+
+
+
+
+
+// tag::graph-q-critical-connections-in-a-network[]
+===== Critical Network Paths
+
+*gr-2*) _Given `n` servers and the connections between them, return the critical paths._
+
+// end::graph-q-critical-connections-in-a-network[]
+
+// _Seen in interviews at: Amazon, Google._
+
+Examples:
+
+[graphviz, critical-path-examples, png]
+....
+graph G {
+  subgraph cluster_1 {
+    a0 -- a1 -- a2 [color=firebrick1]
+    label = "Example A";
+  }
+
+  subgraph cluster_0 {
+    b0 -- b1 [color=blue]
+    b1 -- b2 [color=blue]
+    b2 -- b0 [color=blue]
+    b1 -- b3 [color=blue]
+    b3 -- b2 [color=blue]
+    label = "Example B";
+    b0, b1, b2, b3 [color=midnightblue]
+  }
+
+  subgraph cluster_3 {
+    c0 -- c1 [color=blue]
+    c1 -- c2 [color=blue]
+    c2 -- c0 [color=blue]
+    c1 -- c3 [color=firebrick1]
+    c3 -- c2 [color=transparent] // removed
+    label = "Example C";
+    c0, c1, c2 [color=midnightblue]
+    // c3 [color=red]
+  }
+}
+....
+
+[source, javascript]
+----
+// Example A
+criticalConnections(3, [[0, 1], [1, 2]]);// [[0, 1], [1, 2]]
+// if you remove any link, there will be stranded servers.
+
+// Example B
+criticalConnections(4, [[0, 1], [1, 2], [2, 0], [1, 3], [3, 2]]);// []
+// you can remove any connection and all servers will be reachable.
+
+// Example C
+criticalConnections(4, [[0, 1], [1, 2], [2, 0], [1, 3]]); // [[1, 3]]
+// if you remove [1, 3], then server 3 won't be reachable.
+// If you remove any other link. It will be fine.
+----
+
+Starter code:
+
+[source, javascript]
+----
+include::../../interview-questions/critical-connections-in-a-network.js[tags=description;placeholder]
+----
+
+_Solution: <<graph-q-critical-connections-in-a-network>>_