Dijkstra’s Algorithm Shortest Path

Dijkstra’s algorithm is the most popular algorithm to find the shortest paths from a certain vertex in a weighted graph.
In fact, the algorithm will find the shortest paths to every vertex from the start vertex.
The algorithm takes both greedy and dynamic programming approach that it will try to find local optimum paths first and use them to find the shortest paths.
Let’s take a look at how it works!

How does Dijkstra’s algorithm work?

Let’s say we need to find the shortest path from node A to node E.

dijkstra's algorithm shortest path

There are five nodes and each edge is weighted with the cost to traverse from a node to a node.
For example, it costs 5 to traverse from node A to B.

What would be the most efficient way to find out the shortest path?
In order to find the shortest path from node A to E, we need to find out what is the shortest path to nodes that are right before E – A, B, D.
Can you see where we are going?
In order to find out the shortest path to nodes B, D (I skipped A since it’s the start node), we need to find the shortest path to nodes right before – A, C.

Let’s take a look at each step.

dijkstra's algorithm shortest path

From node A, we can reach to nodes B, C, D, and E.
But let’s start with the nodes B and C first.
What is the shortest path from A to B? There is only one path and it costs 5.
What is the shortest path from A to C? There is only one path and it costs 1.
The nodes B and C are very simple but it’s very important because this will be stepping stone to find the shortest paths to further nodes.

dijkstra's algorithm shortest path

Now, based on the shortest paths to the node C, we can find out the shortest path from the node A to the node D.
There are two paths – A to D directly and A to C to D.
As we can see the shortest path from A to D is A -> C -> D with the total cost 3.

Finally, we have found all the shortest paths from A to B, C, and D which are the nodes right before E.
Based on the shortest paths we found so far, we just need to find which is the cheapest way to reach to E.

dijkstra's algorithm shortest path

There are a total of four possible routes from A to E.

  • A -> E : 10
  • A -> B -> E : 11
  • A -> D -> E : 9
  • A -> C -> D -> E : 8

The shortest path from A to E is A -> C -> D -> E with the cost 8.

The graph itself is pretty simple.
However, the steps we took to find the shortest path is like this.
It really finds the shortest paths to adjacent nodes of A then builds the path based on previous best paths each round.

  • Start from the start node A
  • Just like breadth-first search, find shortest paths to adjacent nodes which are B, C, D, and E.
  • Make sure you keep track of those records.
  • From nodes B, C, and D, find the shortest path to the node E.

Code Example

This code example has the implementation of Dijkstra’s algorithm which uses the graph in the above example.

#include <iostream>
#include <vector>
#include <limits>
#include <algorithm>

using namespace std;

struct Edge
    int destNodeId;
    int weight;
    Edge(int _destNodeId, int _weight) : 
        destNodeId(_destNodeId), weight(_weight)

struct Node
    // node id which is index of vector in this case
    int nodeId;
    // set of edges belong to this node
    vector<Edge> edges;

    // indicates if the node is processed - the shortest path found.    
    bool processed;
    Node(int _nodeId, vector<Edge> _edges) : 
        nodeId(_nodeId), edges(_edges), processed(false)

struct Graph
    // list of nodes in this graph
    vector<Node> nodes;
    // indicates the shortest paths up to this node so far.
    vector<int> shortestDistSoFar;
    // indicates parent node id for each node id, which is represented by index
    vector<int> parentNodes;

void initializeGraph(Graph &graph)
    // initialize node A
    vector<Edge> edges;
    edges.emplace_back(1, 5);
    edges.emplace_back(2, 1);
    edges.emplace_back(3, 4);
    edges.emplace_back(4, 10);
    graph.nodes.emplace_back(0, edges);
    // initialize node B
    edges.emplace_back(4, 6);
    graph.nodes.emplace_back(1, edges);
    // initialize node C
    edges.emplace_back(3, 2);
    graph.nodes.emplace_back(2, edges);
    // initialize node D
    edges.emplace_back(4, 5);
    graph.nodes.emplace_back(3, edges);
    // initialize node E
    graph.nodes.emplace_back(4, edges);
    graph.shortestDistSoFar.resize(5, numeric_limits<int>::max());
    graph.parentNodes.resize(5, -1);

vector<Node> dijkstra(Graph &graph, int startNodeId)
    int currNodeId = startNodeId;
    // distance from A to A is 0
    graph.shortestDistSoFar[currNodeId] = 0;
    // A doesn't have the parent node
    graph.parentNodes[currNodeId] = 0;

    while (!graph.nodes[currNodeId].processed)
        graph.nodes[currNodeId].processed = true;
        // take a look at adjacent nodes
            [&graph, currNodeId](const Edge &destNodeEdge)
                // an adjacent node from current node being processed
                int destNodeId = destNodeEdge.destNodeId;
                // weight of the edge
                int weight = destNodeEdge.weight;
                // if this is shorter than the record, update it
                if (graph.shortestDistSoFar[currNodeId] + weight < graph.shortestDistSoFar[destNodeId])
                    graph.parentNodes[destNodeId] = currNodeId;
                    graph.shortestDistSoFar[destNodeId] = graph.shortestDistSoFar[currNodeId] + weight;
        // find next node to process
        // need to process shortest distance first
        int minDistSoFar = numeric_limits<int>::max();
            [&currNodeId, &graph, &minDistSoFar](const Node &node)
                if (!node.processed && graph.shortestDistSoFar[node.nodeId] < minDistSoFar)
                    minDistSoFar = graph.shortestDistSoFar[node.nodeId];
                    currNodeId = node.nodeId;
    vector<Node> shortestPath;
    // need to find shortest path by recursively tracking parent node
    currNodeId = 4;
    while (graph.parentNodes[currNodeId] != currNodeId)
        currNodeId = graph.parentNodes[currNodeId];
    // make sure A is in the path
    reverse(shortestPath.begin(), shortestPath.end());
    return shortestPath;

int main()
    Graph graph;
    // dijkstra from A of which node id is 0
    vector<Node> shortestPath = dijkstra(graph, 0);
        [](const Node &node)
            switch (node.nodeId)
            case 0:
                cout << "A -> ";
            case 1:
                cout << "B -> ";
            case 2:
                cout << "C -> ";
            case 3:
                cout << "D -> ";
            case 4:
                cout << "E ";
    cout << endl;
    return 0;


Time complexity: O(n^2)
It’s because the algorithm needs to visit each node to find the shortest path.
(You can simply find out there is nested loop)


The algorithm works correctly as long as there aren’t any negative weight edges.
The reason is that the huge negative edge may change the shortest path record to certain edge which was already processed.
But I assume there won’t be many cases that graph will have negative edges.
Thank you for reading the post and please let me know if you have any questions/suggestions.

Intro to Topological Sort

Topological sort is an important algorithm in DAG – directed acyclic graphs.

It orders each vertex in line from left to right such that left most vertex could be considered as the highest level or root or entrance of the graph and rightmost as the lowest of the graph.

In the end, topological sort tells you the order of vertices you need to go through from one node to the other without missing any middle vertices.

There could be multiple topological sorts in the same graph.

Let’s take a look at an example for a better understanding

Course Prerequisite Example

topological sort

The course prerequisite is a great example of this kind of problem.

Let’s say there are 4 courses A, B, C, D and here are some prerequisites for each course.

  1. Course C requires A, B, D
  2. Course B requires A
  3. CourseD requires A, B
  4. A is a fundamental course you need to complete before any courses

With that in mind, in order to reach from A to C, we need to follow these steps.

  1. We see a path to node C directly from A but we also see B and D
  2. Visit B
  3. From B we see a path to C and D
  4. Visit D
  5. Now the order of node visit is A, B, D, C which is the correct order for course completion

However, the above steps are not complete yet for topological sort algorithm since some intuition plays a role there.
Here is the correct algorithm.

Topological Sort Algorithm

  1. Visit a node and mark it visited (but not processed yet)
  2. Search any nodes connected to the node from #1 and visit only if it’s not visited yet
  3. If you see any visited but not processed node then there is a cycle and you cannot find topological sort. (It will be an infinite loop!)
  4. Mark the node processed and push it into stack
  5. After all the nodes are processed pop each value from stack and pop order is the correct order for topological sort

Code Example

enum Status

struct Node
    Status      status;
    vector<int> destNodeIds;

// return false if there is a cycle
bool findOrderHelper(int currNodeId, unordered_map<int, Node> &nodes, vector<int> &order)
    auto it = nodes.find(currNodeId);
    it->second.status = DISCOVERED;

    bool canPublish = true;

        [&nodes, &canPublish, &order](int destNodeId)
            auto it = nodes.find(destNodeId);
            if (it->second.status == UNDISCOVERED)
                canPublish &= findOrderHelper(destNodeId, nodes, order);
            else if (it->second.status == DISCOVERED)
                canPublish = false;

    it->second.status = PROCESSED;
    return canPublish;

vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) 
    vector<int> order;

    // nodes is a graph constructed from prerequisites vector
    unordered_map<int, Node> nodes;
        [&nodes](const vector<int> &prerequisite)
            auto it = nodes.insert(make_pair(prerequisite.back(), Node()));

            nodes.insert(make_pair(prerequisite.front(), Node()));

    // now iterate nodes (graph) to find topological sort
    bool canPublish = true;
        [&nodes, &order, &canPublish](const pair<int, Node> &nodeInfo)
            if (nodeInfo.second.status == UNDISCOVERED)
                canPublish &= findOrderHelper(nodeInfo.first, nodes, order);

    if (!canPublish)
    else if (order.size() != numCourses)
        // courses don't have many prerequisites
        for (int i = 0; i < numCourses; ++i)
            auto it = find(order.begin(), order.end(), i);
            if (it == order.end())

    reverse(order.begin(), order.end());

    return order;

Intro to Graphs

A graph is an abstract representation of a complex system such as networks, human relationships, and any organization.
A graph is very important because it can represent and simplify any complex relationships in a visual format.
It is pretty amazing that messy applied problems often could be described with simple and classical graph properties.

In this post, I am going to discuss some popular graphs and brief examples.

Graph Examples

A graph G can be represented as a set of vertices and edges which can be said as G = (V, E)
Now, let’s take a look at various graphs and their properties.

Undirected vs. Directed Graph

undirected vs directed graph

A graph G = (V, E) is undirected if an edge between two vertices is equally applicable.
In other words, if you can travel from one vertex to another vertex in both directions with one edge it’s an undirected graph.
If not, it is a directed graph.
In an undirected graph, you can visit A from B, C, D whereas there isn’t any vertex that can visit A in a directed graph.

Weighted vs. Unweighted Graph

weighted vs unweighted graph

If each edge in a graph is assigned with numerical value then it is considered as a weighted graph.
Each edge may have different weights and therefore the costs of visiting node may be different even if the number of edges are the same.
If each edge in a graph is not assigned with numerical value then it is an unweighted graph. (All edges are assumed to be of equal weight)

As you can see above graph, the shortest path from A to C in the unweighted graph is edge A, C.
However, the story is different in weighted graph.
Since it costs 10 to use direct path it is cheaper to visit B and then C which is 5 (4 + 1)

Cyclic vs. Acyclic Graph

An acyclic graph does not have any cycles.
There is a cycle in a graph if you can visit a vertex and come back to the same vertex while traversing.

In the above cyclic graph example you can see that starting from vertex A you can visit C, B, and can come back to A which is a cycle.
But there isn’t any cycle in the acyclic graph.

Code Example

Now, let’s discuss what kind of data structure we can use in order to represent a graph.
First, we need information about all the vertices and edges associated with each vertex.
In that sense, it would be reasonable to define a user-defined type for each edge.
And each vertex will know which edges are associated with it.

Here is a sample code representing data structure for a graph.
There could be many other ways as this is just one way to represent it.

struct Edge
    // destination vertex id for this edge
    // it could be any other data structure to represent destination
    int vertexId;
    // weight of the edge if application. 0 means unweighted.
    int weight;

struct Vertex
    // all the edges associated with this vertex
    vector<Edge> edges;    

struct Graph
    // here I use index of each vertex as vertex id for simplicity
    // but you can use your own way to represent it differently.
    vector<Vertex> vertices;
    // is this directed graph?
    bool directed;


There are many other types such as simple vs non-simple, sparse vs dense and etc which are not discussed here.
I just wanted to show very basic concepts of important graphs and their properties.