🚀Day 16. Bellman-Ford 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given a weighted graph with V vertices (numbered from 0 to V-1) and a list of E edges, where each edge is represented as a triplet [u, v, w] denoting a directed edge from u to v with weight w.

Also given a source vertex src. Your task is to compute the shortest distance from the source to all other vertices using the Bellman-Ford algorithm.

• If a vertex is unreachable from the source, mark its distance as 1e8.

• If the graph contains a negative weight cycle, return [-1].

📝 Note: Sorry for uploading late, my Final Sem exam is going on.

🔍 Example Walkthrough:

Example 1:

Input:

V = 5
edges = [[1, 3, 2], [4, 3, -1], [2, 4, 1], [1, 2, 1], [0, 1, 5]]
src = 0

Output:

[0, 5, 6, 6, 7]

Explanation:

• 0 → 1: 5

• 0 → 1 → 2: 6

• 0 → 1 → 2 → 4 → 3: 6

• 0 → 1 → 2 → 4: 7

Example 2:

Input:

V = 4
edges = [[0, 1, 4], [1, 2, -6], [2, 3, 5], [3, 1, -2]]
src = 0

Output:

[-1]

Explanation:

• The cycle: 1 → 2 → 3 → 1 has a negative weight → Negative Cycle Detected

Constraints

• $( 1 \leq V \leq 100 )$

• $( 1 \leq E = \text{edges.size()} \leq V \times (V-1) )$

• $( -1000 \leq w \leq 1000 )$

• $( 0 \leq \text{src} < V )$

🎯 My Approach:

Standard Bellman-Ford Algorithm

The Bellman-Ford algorithm is designed to find the shortest paths from a single source node to all other nodes, even when negative edge weights exist.

Algorithm Steps:

1. Initialize a distance array dist[] of size V with all elements as 1e8, and set dist[src] = 0.

2. Relax all edges V - 1 times.

3. Check for negative weight cycles by attempting one more relaxation:

   • If any edge still updates, it means a negative cycle exists → return [-1].

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: $( O(V \times E) )$, as we perform edge relaxation V-1 times over all E edges.

• Expected Auxiliary Space Complexity: $( O(V) )$, to store the distance array of size V.

📝 Solution Code

Code (C++)

// ✅ Standard Bellman-Ford Algorithm
class Solution {
  public:
    vector<int> bellmanFord(int V, vector<vector<int>>& edges, int src) {
        vector<int> dist(V, 1e8);
        dist[src] = 0;
        for (int i = 0; i < V - 1; ++i)
            for (auto& e : edges)
                if (dist[e[0]] != 1e8 && dist[e[0]] + e[2] < dist[e[1]])
                    dist[e[1]] = dist[e[0]] + e[2];
        for (auto& e : edges)
            if (dist[e[0]] != 1e8 && dist[e[0]] + e[2] < dist[e[1]])
                return {-1};
        return dist;
    }
};

⚡ Alternative Approaches

📊 2️⃣ Bellman-Ford with Early Termination Optimization

Algorithm Steps:

1. Initialize a distance array with a large value (here, 1e8) and set the source distance to 0.

2. For each of the V - 1 iterations, update the distances of the vertices.

   • If no update is made in an iteration, break out early to optimize runtime.

3. Check once more for any possible relaxation to detect negative cycles.

4. Return {-1} if a negative cycle is detected, else return the distance array.

class Solution {
  public:
    vector<int> bellmanFord(int V, vector<vector<int>>& edges, int src) {
        vector<int> dist(V, 1e8);
        dist[src] = 0;

        for (int i = 0; i < V - 1; ++i) {
            bool updated = false;
            for (auto& e : edges) {
                int u = e[0], v = e[1], w = e[2];
                if (dist[u] != 1e8 && dist[u] + w < dist[v]) {
                    dist[v] = dist[u] + w;
                    updated = true;
                }
            }
            if (!updated) break;  // Optimization
        }

        for (auto& e : edges)
            if (dist[e[0]] != 1e8 && dist[e[0]] + e[2] < dist[e[1]])
                return {-1};

        return dist;
    }
};

📝 Complexity Analysis

• Time Complexity: O(V * E) in worst-case, better with early break.

• Space Complexity: O(V)

✅ Why This?

This can significantly reduce computation when the graph stabilizes early. A practical tweak that saves time in sparse or nearly optimal graphs.

🆚 Comparison of Approaches

Approach	⏱️ Time Complexity	🗂️ Space Complexity	✅ Pros	⚠️ Cons
Standard Bellman-Ford	🟢 O(V × E)	🟢 O(V)	Simple, classic, handles all cases	May do redundant passes
Optimized with Early Termination	🟡 Best: O(E), Worst: O(V × E)	🟢 O(V)	Faster on converging graphs, fewer passes	Still worst-case O(V × E)

• Use Bellman-Ford when negative weights or cycles are possible.

• Use Early Termination for performance optimization.

Code (Java)

class Solution {
    public int[] bellmanFord(int V, int[][] edges, int src) {
        int[] dist = new int[V];
        Arrays.fill(dist, (int)1e8);
        dist[src] = 0;
        for (int i = 0; i < V - 1; i++)
            for (int[] e : edges)
                if (dist[e[0]] < 1e8 && dist[e[0]] + e[2] < dist[e[1]])
                    dist[e[1]] = dist[e[0]] + e[2];
        for (int[] e : edges)
            if (dist[e[0]] < 1e8 && dist[e[0]] + e[2] < dist[e[1]])
                return new int[]{-1};
        return dist;
    }
}

Code (Python)

class Solution:
    def bellmanFord(self, V, edges, src):
        dist = [int(1e8)] * V
        dist[src] = 0
        for _ in range(V - 1):
            for u, v, w in edges:
                if dist[u] < 1e8 and dist[u] + w < dist[v]:
                    dist[v] = dist[u] + w
        for u, v, w in edges:
            if dist[u] < 1e8 and dist[u] + w < dist[v]:
                return [-1]
        return dist

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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