Dijkstra Algorithm

Problem Statement

Given a weighted, undirected and connected graph of V vertices and an adjacency list adj where adj[i] is a list of lists containing two integers where the first integer of each list j denotes there is edge between i and j , second integers corresponds to the weight of that edge . You are given the source vertex S and You to Find the shortest distance of all the vertex's from the source vertex S. You have to return a list of integers denoting shortest distance between each node and Source vertex S.

Note: The Graph doesn't contain any negative weight cycle.

Example 1:

Input:
V = 2
adj [] = {{{1, 9}}, {{0, 9}}}
S = 0
Output:
0 9
Explanation:

The source vertex is 0. Hence, the shortest 
distance of node 0 is 0 and the shortest 
distance from node 1 is 9.

Example 2:

Input:
V = 3, E = 3
adj = {{{1, 1}, {2, 6}}, {{2, 3}, {0, 1}}, {{1, 3}, {0, 6}}}
S = 2
Output:
4 3 0
Explanation:

For nodes 2 to 0, we can follow the path-
2-1-0. This has a distance of 1+3 = 4,
whereas the path 2-0 has a distance of 6. So,
the Shortest path from 2 to 0 is 4.
The shortest distance from 0 to 1 is 1 .

Your Task: You don't need to read input or print anything. Your task is to complete the function dijkstra() which takes the number of vertices V and an adjacency list adj as input parameters and Source vertex S returns a list of integers, where ith integer denotes the shortest distance of the ith node from the Source node. Here adj[i] contains a list of lists containing two integers where the first integer j denotes that there is an edge between i and j and the second integer w denotes that the weight between edge i and j is w.

Intuition

Take All edges and Relax

Time Complexity = ElogV

Video Links

Approach 1:

C++

 vector <int> dijkstra(int V, vector<vector<int>> adj[], int s){
        vector<int> distance(V,INT_MAX);
        // Distance, node
        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> pq; 
        pq.push({0,s});
        distance[s] = 0;

        while (!pq.empty()) {
            int node = pq.top().second;

            for(auto &it: adj[node]){
                if(distance[node] + it[1] < distance[it[0]]){
                    int new_dis = distance[node] + it[1];
                    distance[it[0]] = new_dis;
                    pq.push({new_dis, it[0]});
                }
            }
            pq.pop();
        }
        
        for(int i = 0; i < V; i++) {
            if(distance[i] == INT_MAX)
                distance[i] = -1;
        }
        return distance;
    }

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++