**Problem Statement:** Given a weighted, directed and connected graph of V vertices and E edges, Find the shortest distance of all the vertices from the source vertex S.

**Note:** If the Graph contains a negative cycle then return an array consisting of only -1.

**Example 1**:

Input Format:V = 6, E = [[3, 2, 6], [5, 3, 1], [0, 1, 5], [1, 5, -3], [1, 2, -2], [3, 4, -2], [2, 4, 3]], S = 0Result: 0 5 3 3 1 2Explanation:Shortest distance of all nodes from the source node is returned.

**Example 2**:

Input Format:V = 2, E = [[0,1,9]], S = 0Result:0 9Explanation:Shortest distance of all nodes from the source node is returned.

**Solution**

** Disclaimer**:

*Don’t jump directly to the solution, try it out yourself first.*

*Problem link*

*.*

**Solution:**

** The bellman-Ford algorithm** helps to find the shortest distance from the source node to all other nodes. But, we have already learned

**Dijkstra’s algorithm article link**

*Dijkstra’s algorithm (***to fulfill the same purpose. Now, the question is**

*)***.**

*how this algorithm is different from Dijkstra’s algorithm*While learning Dijkstra’s algorithm, we came across the following two situations, where Dijkstra’s algorithm failed:

*If the graph contains negative edges.**If the graph has a negative cycle (In this case Dijkstra’s algorithm fails to minimize the distance, keeps on running, and goes into an infinite loop. As a result it gives TLE error).*

*Negative Cycle: **A cycle is called a negative cycle if the sum of all its weights becomes negative. The following illustration is an example of a negative cycle:*

** Bellman-Ford’s algorithm **successfully solves these problems.

**as well as**

*It works fine with negative edges***. But this algorithm is only applicable for**

*it is able to detect if the graph contains a negative cycle***. In order to apply this algorithm to an undirected graph, we just need to convert the undirected edges into directed edges like the following:**

*directed graphs***Explanation:** An undirected edge between nodes u and v necessarily means that there are two opposite-directed edges, one towards node u and the other towards node v. So the above conversion is valid.

After converting the undirected graph into a directed graph following the above method, we can use the Bellman-Ford algorithm as it is.

**Intuition:**

*In this algorithm, the edges can be given in any order. The intuition is to relax all the edges for N-1( N = no. of nodes) times sequentially. After N-1 iterations, we should have minimized the distance to every node.*

*Let’s understand what the relaxation of edges means using an example.*

Let’s consider the above graph with dist[u], dist[v], and wt. Here, wt is the weight of the edge and dist[u] signifies the shortest distance to reach node u found until now. Similarly, dist[v](maybe infinite) signifies the shortest distance to reach node v found until now. If the distance to reach v through u(i.e. dist[u] + wt) is smaller than dist[v], we will update the value of dist[v] with (dist[u] + wt). This process of updating the distance is called the relaxation of edges.

We will apply the above process(i.e. minimizing the distance to reach every node) N-1 times in the Bellman-Ford algorithm.

** Two follow-up questions about the algorithm: Why do we need exact N-1 iterations?**Let’s try to first understand this using an example:

- In the above graph, the algorithm will minimize the distance of the i
^{th}node in the i^{th}iteration like dist[1] will be updated in the 1st iteration, dist[2] will be updated in the 2nd iteration, and so on. So we will need a total of 4 iterations(i.e. N-1 iterations) to minimize all the distances as dist[0] is already set to 0.**Note:***Points to remember since, in a graph of N nodes we will take at most N-1 edges to reach from the first to the last node, we need exact N-1 iterations. It is impossible to draw a graph that takes more than N-1 edges to reach any node.* *How to detect a negative cycle in the graph?*- We know if we keep on rotating inside a negative cycle, the path weight will be decreased in every iteration. But according to our intuition, we should have minimized all the distances within N-1 iterations(that means, after N-1 iterations no relaxation of edges is possible).
- In order to check the existence of a negative cycle, we will relax the edges one more time after the completion of N-1 iterations. And if in that N
^{th}iteration, it is found that further relaxation of any edge is possible, we can conclude that the graph has a negative cycle. Thus, the Bellman-Ford algorithm detects negative cycles.

**Approach**:

**Initial Configuration:**

**distance array(dist[ ]): **The dist[] array will be initialized with infinity, except for the source node as dist[src] will be initialized to 0.

The algorithm steps will be the following:

- First, we will initialize the source node in the distance array to 0 and the rest of the nodes to infinity.
- Then we will run a loop for N-1 times.
- Inside that loop, we will try to relax every given edge.

For example, one of the given edge information is like (u, v, wt), where u = starting node of the edge, v = ending node, and wt = edge weight. For all edges like this we will be checking if node u is reachable and if the distance to reach v through u is less than the distance to v found until now(i.e.**dist[u] and dist[u]+ wt < dist[v]**). - After repeating the 3rd step for N-1 times, we will apply the same step one more time to check if the negative cycle exists. If we found further relaxation is possible, we will conclude the graph has a negative cycle and from this step, we will return a distance array of -1(i.e. minimization of distances is not possible).
- Otherwise, we will return the distance array which contains all the minimized distances.

**Note**: *If you wish to see the dry run of the above approach, you can watch the video attached to this article.***Code**:

## C++ Code

```
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
/* Function to implement Bellman Ford
* edges: vector of vectors which represents the graph
* S: source vertex to start traversing graph with
* V: number of vertices
*/
vector<int> bellman_ford(int V, vector<vector<int>>& edges, int S) {
vector<int> dist(V, 1e8);
dist[S] = 0;
for (int i = 0; i < V - 1; i++) {
for (auto it : edges) {
int u = it[0];
int v = it[1];
int wt = it[2];
if (dist[u] != 1e8 && dist[u] + wt < dist[v]) {
dist[v] = dist[u] + wt;
}
}
}
// Nth relaxation to check negative cycle
for (auto it : edges) {
int u = it[0];
int v = it[1];
int wt = it[2];
if (dist[u] != 1e8 && dist[u] + wt < dist[v]) {
return { -1};
}
}
return dist;
}
};
int main() {
int V = 6;
vector<vector<int>> edges(7, vector<int>(3));
edges[0] = {3, 2, 6};
edges[1] = {5, 3, 1};
edges[2] = {0, 1, 5};
edges[3] = {1, 5, -3};
edges[4] = {1, 2, -2};
edges[5] = {3, 4, -2};
edges[6] = {2, 4, 3};
int S = 0;
Solution obj;
vector<int> dist = obj.bellman_ford(V, edges, S);
for (auto d : dist) {
cout << d << " ";
}
cout << endl;
return 0;
}
```

**Output: 0 5 3 3 1 2**

**Time Complexity: **O(V*E), where V = no. of vertices and E = no. of Edges.

**Space Complexity: **O(V) for the distance array which stores the minimized distances.

## Java Code

```
import java.util.*;
/*
* edges: vector of vectors which represents the graph
* S: source vertex to start traversing graph with
* V: number of vertices
*/
class Solution {
static int[] bellman_ford(int V,
ArrayList<ArrayList<Integer>> edges, int S) {
int[] dist = new int[V];
for (int i = 0; i < V; i++) dist[i] = (int)(1e8);
dist[S] = 0;
// V x E
for (int i = 0; i < V - 1; i++) {
for (ArrayList<Integer> it : edges) {
int u = it.get(0);
int v = it.get(1);
int wt = it.get(2);
if (dist[u] != 1e8 && dist[u] + wt < dist[v]) {
dist[v] = dist[u] + wt;
}
}
}
// Nth relaxation to check negative cycle
for (ArrayList<Integer> it : edges) {
int u = it.get(0);
int v = it.get(1);
int wt = it.get(2);
if (dist[u] != 1e8 && dist[u] + wt < dist[v]) {
int temp[] = new int[1];
temp[0] = -1;
return temp;
}
}
return dist;
}
}
public class tUf {
public static void main(String[] args) {
int V = 6;
int S = 0;
ArrayList<ArrayList<Integer>> edges = new ArrayList<>() {
{
add(new ArrayList<Integer>(Arrays.asList(3, 2, 6)));
add(new ArrayList<Integer>(Arrays.asList(5, 3, 1)));
add(new ArrayList<Integer>(Arrays.asList(0, 1, 5)));
add(new ArrayList<Integer>(Arrays.asList(1, 5, -3)));
add(new ArrayList<Integer>(Arrays.asList(1, 2, -2)));
add(new ArrayList<Integer>(Arrays.asList(3, 4, -2)));
add(new ArrayList<Integer>(Arrays.asList(2, 4, 3)));
}
};
int[] dist = Solution.bellman_ford(V, edges, S);
for (int i = 0; i < V; i++) {
System.out.print(dist[i] + " ");
}
System.out.println("");
}
}
```

**Output: 0 5 3 3 1 2**

**Time Complexity: **O(V*E), where V = no. of vertices and E = no. of Edges.

**Space Complexity: **O(V) for the distance array which stores the minimized distances.

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