represent graph: matrix ^ edgeSset #forward_list

Josh’s book [[data structures and other objects using c++]] has 5 pages devoted to three best-practice data structures representing a generic directed graph. (I think my [[discrete math] also covers the same topic.)

  1. boolean adjacency matrix — amat
  2. edge set — eset
  3. edge list — can be implemented using std::forward_list, marginally better than set… not discussed here
  • iterating all edges connected to a node — eset wins
  • sparse graph — amat wastes memory
  • test linkage between 2 nodes — amat wins
  • ease of implementation — amat wins

SCB-FM algo Q2a 2 slists

Q: Given two well-formed singly-linked lists (loop-free), that might be merged at some node, locate the merge point. If not merged, return null.

Note every node has exactly one child node, but may have two parent nodes.


Suppose list A has size P=55 and list B has size Q=32

–SolH: hashtable O(N+M) — optimal in time but not space

First populate hashtable with the short list’s 32 nodes. Then iterate the longer list and check each node’s address.

–SolR: reverse one list … how?

–SolA: array-based.  construct a simple 55-array of A node pointers, and another 32-array of B node pointers. Then compare the two final elements in the two arrays. If same then binary search in B. O(N+M) + O(log M)

–Sol2: 2-pointer algo O(1) space, usable on read-only lists 🙂

first scan to find the 2 end nodes and remember the sizes like 55 vs 32.

IFF the end nodes match, then 2nd scan:

skip first P-Q (23) nodes in the longer list and then increment 2 iterators in lock steps. Compare the 2 iterators at each step.

merge 2 sorted slists #2try

Q: Merge two sorted linked lists and return it as a new (ascending) list. The new list should be made by splicing together the nodes of the first two lists.

A test of implementation skill. A test of clear communication and clear thinking. My solution below is O(1) space and O(N) time without allocating any heap memory.

I would pick the list having the lowest value as the Yellow jersey (leading). The other list is the green. At any time there exist two head nodes for two slists.

I always keep a brown pointer to the last of the leading pack in the yellow’s list. The leading pack are the “retired” nodes.

Swap – In the event of a ‘swap’, the brown’s current child (current yellow jersey list head) gets detached and becomes the new green jersey, and the old green jersey moves into the leading pack. Then we increment the ptr — this old green jersey’s child becomes the yellow jersey.

Every time after we increment the pointer, we compare its child node vs the green jersey.

I won’t bother with std::list::splice() and will use python or my own c++ linked list.

generate no-loop paths: graph node A→B

Q1: given 2 nodes in a graph containing N (eg 121) nodes, potentially with cycles, generate all simple paths between the pair. A simple path has no cycle. (In other words, for a simple path, length + 1 ==  # unique nodes)

I think there are classic math algorithms for it, because this is part of basic graph theory. Here are some applications of this type of algorithms —

  • Q1b (special case of Q1): given 2 nodes in a C-by-R rectangular grid, where every node is connected to (up to) four neighbors, generate all cycle-free paths.
    • Below, I solved this problem in python
  • Q2 (simple application of Q1 algo): generate all simple paths between ALL node pair in a graph. The shortest simple path has length=0. Longest simple path can potentially visit every node exactly once.
  • A: first generate all 121-Choose-2 node pairs. For each pair, solve Q1. Lastly generate the 121 trivial paths of length=0.
    • repetitive 😦
  • Q2b (special case of Q1): In a C-by-R rectangular grid, where every node is connected to (up to) four neighbors, generate all cycle-free paths between all pairs.
  • Q2c (easy one based on Q2): given a binary tree containing no cycles, generate all paths.

— A1b: my DFT implementation (probably not 100% correct) , where each “trail” either fails or becomes a path.

  1. from NodeA start a breadcrumb/trail. We can’t revisit any node already visited on current breadcrumb,
    1. if this is a matrix, then instead of a hashtable, we can also use a shadow matrix, but the breadcrumb is much smaller than a shadow matrix
  2. if we reach a node surrounded by nodes on the same breadcrumb, then the trail fails at a dead-end
  3. else we will reach NodeB 🙂 Print the breadcrumb

By construction, we won’t see duplicate paths 🙂 is the implementation

–BFT? I don’t think BFT can print each unique path

shortest path:2nodes]binary matrix #BFT

Q: given 2 cells in a binary matrix (1=black, 0=white=blocked), check the pair are connected and if yes return the shortest path. There exists a path of length 1 between any 2 cells IFF both are side by side or stack atop.

count paths between 2 bTree nodes #PimcoQ9 Ashish is arguably harder than this problem, but this problem allows moving in four directions.

binary-matrix island count #DeepakM technique is more applicable. A BFT path should work.

  • every reachable node is painted Green (like 2)
  • we give up after our queue is empty is the implementation, briefly tested.

count paths between 2 bTree nodes #PimcoQ9 Ashish

The DP idea — compare matrix-path-counter and EditDistance

  • showcasing efficient queue in python.
  • showcasing using (x,y) coordinates as dictionary key
  • showcasing find max value in a dictionary

–Requirement: ( is similar)

See Q9.pdf in the email to Ashish. Here are some key points:

Consider a maze mapped to a matrix with an upper left corner at coordinates (row, column) = (0, 0). Any movement must be in increasing row or column direction. You must determine the number of distinct paths through the maze. You will always start at position (0, 0), the top left, and end up at (max(row), max(column)), the bottom right.
1 1 0 1
1 1 1 1
As an example, consider the above matrix where 1 indicates an open cell and 0 indicates blocked. You can only travel through open cells, so no path can go through the cell at (0, 2). There are two distinct paths to the goal. As a 2nd example, matrix below has 10 paths:
1 1 1 1
1 1 1 1
1 1 1 1 is my solution.

==== Analysis ====
I see a binary tree, where each node is a cell. Cell (0,0) has down/left node (1,0) + right node (0,1). I feel this is similar to a more generic problem “count paths between 2 nodes in a binary tree”


😦 I implemented something like a DFT but it was too slow with some test cases

😦 can overflow stack

🙂 DFT can print each unique path. I think BFT can’t.

🙂 DFT is easier to implement than BFT

–DynamicProgramming BFT solution from origin, as I described to Bill Pinsky:

This solution is more versatile than the one from Ashish. It can handle any directed graph.

Mark each node as “computed” once we have computed a score denoting how many paths-from-root there are. Save the score in a shadow matrix.

Origin node has score 1. Blocked cell has score 0.

Start BFT from origin. The two immediate neighbors are set to 1 (or 0 if blocked). Every node can be computed by adding up above-score + left-score. (This algo is simpler than the EditDistance algo.)

Performance Keynote — My implementation was extremely slow (albeit correct) until I added an “if already computed, then continue” early in the loop

Q: why is score[1,1] accessed 4 times?
A: node[1,1] is added to the queue twice. Each dequeue would need one check.
A: Also score[1,2] need to access score[1,1] as a parent. Ditto score[2,1]

–Ashish Singh gave me a much simpler solution, as shown in my github code.

NxN matrix: graph@N nodes #IV

Simon Ma of CVA team showed me this simple technique. is my first usage of it.

  • I only needed half of all matrix cells (excluding the diagonal cells) because relationships are bilateral.
  • Otherwise, if graph edges are directed, then we need all (N-1)(N-1) cells since A->B is not same as B->A.

My discrete math textbook shows this is a simplified form of representation and can’t handle self-link or parallel edge. The vertex-edge matrix is more robust.


binary-matrix island count #DeepakM

Q: has my solution. I don’t want to spend the time passing all leetcode tests! Diminishing return is conceptually identical, though using a different criteria for “connected” — diagonal neighbors are “connected”

Hi Deepak

I realize for “connected” problems, there’s definitely a graph underneath, so graph traversal is required. I guess BFT or DST will both find all the nodes connected to “me”.

Given a 1000 x 1000 all-black matrix, I think DFT recursion will go into exactly 1,000,000 levels and overflow stack space.

A high-level (vague) idea is

  • Scan in type-writer fashion. Suppose there are 10×10 = 100 cells either black or write. I paint each cell brown [1] once visited. I also use a integer counter to keep track how many cells already visited.
  • During the scan. If a cell is already visited, I will ignore it and move on
  • Any time I find a black cell, I will start a BFT (I dislike DST) to find all connected black cells.  Once I find all the black cells connected to “me”, I resume the type-writer scan.
  • Once my counter hits 100, I have visited all cells and will exit.

[1] you paint it white, which is probably better.

I thought this problem was too hard and not worth study, but you convinced me that it may come up and I can at least give a vague solution  … better than no solution.

Compared to a binary tree, walking over a matrix is /operationally/ (not conceptually) harder because

  • need to check array bounds. Error-prone and time consuming in coding tests
  • The (invisible) links will not cover all cells. To avoid missing a cell, we still need a full matrix scan. The integration of tree walk + matrix scan is unintuitive, to put it mildly.
  • During the tree-walk, you don’t want to visit any cell too many times or get lost in an endless loop. A brute-force technique — shadow matrix to remember all visited cells.
  • … However, once you get over these nitty-gritty operational complexities, then tree-walk in matrix is not really harder. These algo questions can therefore frustrate and fail many candidates untrained on The technique.

bbg-Eq: fewest increments on odometer

There’s a special odometer with N (eg 5) digits. Each digit can increment (never decrement, to keep the problem simpler) independently and cyclically, like 8->9->0->1. You can think of them as 5 independent mini-odometers. The starting sequence is input (perhaps something like 07748). Ending sequence can be any other value.

Each increment on any mini-odometer has cost of $1. Find the cheapest way to go from start to end. One simple solution (from me) –increment each digit cyclically to the target value. So a “8” will increment to 9 -> 0 -> 1. Other mini-odometers are not affected by this mini-odometer.

However, 18021 is one of many illegal sequences — barriers.

Key insight — When N==2, this becomes a 2D matrix Tom-n-Jerry problem with barriers. We can simplify the problem to a 3×3 matrix i.e. 2-digit odometer from 00 to 22. Nine possible sequences.

We need to build a graph of nine nodes, which helps answer

  • Q: feasible? [2,2] would be unreachable if [1,2] and [2,1] are blocked.
  • Q: which path is shortest path

Once we build the graph, shortest path btw 2 graph nodes #binary matrix as illutration will solve it.

shortest path: linked-in distance #Flextrade

I haven’t searched online for this common question.

Q1: given a snapshot of the entire linked-in network and two personIDs A1 and B1, return the shortest distance between them. If they are directly connected, then return 1.
%%A: breadth-first search

Q2: what if you are given another pair of personID’s like A2 to B2, then A3 to B3 … Will your answer change?
%%A: I will pre-construct a jagged 2D int-array (more efficient than a matrix [1]). The outer array is indexed by personID. The element at position 2503 is for person #2503(say, Ed) and is an int array holding all of Ed’s first-degree friends’ personIDs.

This jagged array offers cache efficiency since I can load entire #2503 child-array in one memory access, even if Ed has 500 friends. Without this jagged array, I need to visit 500 pointers.

The jagged array alone can answer my question, so I don’t need the graph.

O() complexity is identical to graph traversal.

Q3: what if the snapshot is replaced by a live network? I feel we have to assume at the time of query we treat the network as momentarily frozen.
%%A: I would update my jagged array in real time

[1] jagged array can achieve 99% memory footprint saving. If the most connected person has 9999 friends then the matrix needs 9999 columns, but a typical person only has a few hundred friends.

generate paths in ternary matrix #Promethean TomJerry


Similar to binary-matrix cluster count #DeepakM #DST+fullScan but the connection is modeled differently.

–Original requirements: Input matrix has an object in each cell

  • 1 means wall or barrier
  • 0 means passage
  • 2 means cheese station

Jerry is [x][y]. Tom starts at [0][0] and must collect all cheese and deliver to Jerry. Find shortest path. Tom can only move one (U/D/L/R) step at a time. is my half-finished code. It builds a graph connecting all the cells (each a Node object). It’s able to check feasibility. Todo:

  • breadth-first walk from Tom to Jerry, without revisiting any node. Hopefully we can discover all possible paths (i.e. collecting all cheese)
  • determine the shortest.


tokens shared among friends #Promethean

See also NxN matrix for graph of N nodes

Requirement: There are N friends numbered from 1 to N. There are M pairs of links, where each (x , y ) pair is connected by a shared integer token described by tokenId. Any two friends, x and y , can be connected directly by multiple tokens, or indirectly (without directly shared token) because if friends x and y share token t and friends y and z also share token t , then x and z are also said to share token t.

Note if x/y shares t and u/v share t, then x and u may be unconnected!

Find the maximal product of x and y for any directly or indirectly connected (x , y ) pair such that x and y share the maximal number of tokens with each other. If x/y have 3 tokens connecting them, and u/v also have 3 tokens, then we compare x*y vs u*v. showcasing nested containers like map<int, list<set<int>>>. I believe STL is harder than java, python etc.

construct graph from list of connections #BGC java

Given an input file showing a list of {string, string} pairs, build a connection graph.

If you have a correct connection graph, then you can easily determine the connectedness (bool) of any 2 nodes. In a social-network, this bool flag indicates whether 2 individuals are completely unconnected or somehow connected.

I see this as a social-network. Any pair represents an edge connecting 2 nodes.  At any time there are a number of disconnected islands. The next pair could 1) merge 2 islands or 2) add a node to an existing island or 3) create a new island 4) do nothing, if the 2 nodes are already in some existing island

  • Any known node appears exactly once in the entire graph, in exactly one of the islands.
  • All nodes are contained in a lookup table or hashmap  {node -> island}
  • Each island can be a implemented as a hashset of nodes.

So here’s a proposed algo to process a new pair {A, B}. Look for A and B in the  graph. 3 scenarios + a dummy scenario:

  • (Scenario 3) If both A an B are new comers, then they form a new island.
  • if both A and B are already in the graph,
    • (Scenario 4) if they are in the same island, then exit. Nothing to do
    • (Scenario 1) else we can merge the 2 islands
  • (Scenario 2) If A is in island 3 but B is new comer, then B joins island 3

The merge operation is expensive. The big lookup table needs update but here’s an alternative:

  • At merge time, the smaller island would have all the nodes moved to the bigger island. When the island is empty, it gets a pointer “this.redirect” to the bigger island.
  • lookup table needs no update, avoiding locking a global object.
  • At query time, we look up the table to get the original island, then we follow its pointer (defaults to null) until the island is non-empty.
  • endless loop? would only be a programming error.

equals method of a class


You are spot on about linked list — If a class N has-a field of type N, then N is almost always, by definition, a node in a graph. That N field is probably a parent node. So allow me to put in some meaningful names — Each GraphNode has-a field named this.parent. Now the question becomes “how to override equals() in GraphNode and deal with the unbounded recursion”.

It’s an unusual technical requirement to make equals() to compare all ancestor nodes. However, It’s a reasonable business requirement to compare 2 GraphNodes by comparing all ancestors. Such a business requirement calls for a (static) utility method, NOT an instance method in A static utility method like compareAllAncestor(GraphNode, GraphNode) can be iterative and avoid recursion and stack overflow. Once this static method is in place, I might (grudgingly) create an instance method compare(GraphNode other) which simply returns compareAllAncestor(this, other), without unbounded recursion or stack overflow.

If 2 threads both perform this comparison, then I feel the method may need to lock the entire graph — expensive.

Even in a single-threaded environment, this comparison is expensive. (The recursive version would add an additional memory cost.) Potentially a performance issue. For most graph data structures in business applications, GraphNode should be Serializable and collections-friendly. Therefore hashCode() and equals() should be cheap.

For most graph data structures in business applications, each graph node usually represents a real world entity like a member in a MLM network. Now, if a graph node represents a real world entity, then it’s always, without exception, identifiable by an immutable and unique ID. Usually this ID is saved in database (could also be generated in application). Therefore, in most cases, equals() should compare ID only.