🚀Day 18. Minimum Weight Cycle 🧠

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

💡 Problem Description:

Given an undirected, weighted graph with V vertices numbered from 0 to V-1, and E edges, find the minimum weight cycle in the graph.

Each edge is represented by edges[i] = [u, v, w] meaning there is an edge between nodes u and v with weight w.

If no cycle exists, return -1.

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

🔍 Example Walkthrough:

Example 1

Input:

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

Output:

6

Explanation:

Minimum-weighted cycle is 0 → 1 → 4 → 0 with total weight = 2 + 1 + 3 = 6.

Example 2

Input:

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

Output:

5

Explanation:

Minimum-weighted cycle is 1 → 3 → 4 → 1 with total weight = 1 + 2 + 2 = 5.

Constraints

• $(1 \leq V \leq 100)$

• $(1 \leq E = \text{edges.length} \leq 10^3)$

• $(1 \leq \text{edges[i][j]} \leq 100)$

🎯 My Approach:

Dijkstra-Based (Per Node Check)

Algorithm Steps:

1. Build an adjacency list from the edge list.

2. For every node i from 0 to V-1:

   • Run Dijkstra's algorithm from node i to compute the shortest paths to all other nodes.

   • While processing neighbors, if you find a back-edge to an already visited node, and it doesn’t form a direct parent-child, it could form a cycle.

   • Update the minimum cycle weight if a better cycle is found.

3. After checking from all nodes, return the minimum cycle length found, or -1 if no cycle exists.

✅ Why It Works

• Each shortest path computed from i guarantees minimal distance to v, so combining two shortest paths and an extra edge efficiently checks all minimal cycles passing through that edge.

• The parent check ensures we’re not just tracing the same edge forward and backward.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(V × E × log V), since we run Dijkstra's algorithm (O(E log V)) for each vertex.

• Expected Auxiliary Space Complexity: O(V + E), to store the adjacency list, distance array, parent tracking, and priority queue per run.

📝 Solution Code

Code (C++)

class Solution {
  public:
    int findMinCycle(int V, vector<vector<int>>& edges) {
        vector<vector<pair<int,int>>> adj(V);
        for (auto& e : edges)
            adj[e[0]].push_back({e[1], e[2]});
        int res = INT_MAX;
        for (int i = 0; i < V; ++i) {
            vector<int> dist(V, 1e9), par(V, -1);
            dist[i] = 0;
            priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> pq;
            pq.push({0, i});
            while (!pq.empty()) {
                pair<int,int> top = pq.top(); pq.pop();
                int d = top.first, u = top.second;
                for (int j = 0; j < adj[u].size(); ++j) {
                    int v = adj[u][j].first, w = adj[u][j].second;
                    if (dist[v] > d + w) {
                        dist[v] = d + w;
                        par[v] = u;
                        pq.push({dist[v], v});
                    } else if (par[u] != v && par[v] != u)
                        res = min(res, dist[u] + dist[v] + w);
                }
            }
        }
        return res == INT_MAX ? -1 : res;
    }
};

Code (Java)

class Solution {
    public int findMinCycle(int V, int[][] E) {
        List<int[]>[] A = new ArrayList[V];
        for (int i = 0; i < V; i++) A[i] = new ArrayList<>();
        for (int[] e : E) A[e[0]].add(new int[]{e[1], e[2]});
        int r = Integer.MAX_VALUE;
        for (int i = 0; i < V; i++) {
            int[] D = new int[V], P = new int[V];
            Arrays.fill(D, (int)1e9);
            Arrays.fill(P, -1);
            D[i] = 0;
            PriorityQueue<int[]> Q = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
            Q.add(new int[]{0, i});
            while (!Q.isEmpty()) {
                int[] t = Q.poll(); int d = t[0], u = t[1];
                for (int[] a : A[u]) {
                    int v = a[0], w = a[1];
                    if (D[v] > d + w) {
                        D[v] = d + w; P[v] = u; Q.add(new int[]{D[v], v});
                    } else if (P[u] != v && P[v] != u)
                        r = Math.min(r, D[u] + D[v] + w);
                }
            }
        }
        return r == Integer.MAX_VALUE ? -1 : r;
    }
}

Code (Python)

class Solution:
    def findMinCycle(self, V, edges):
        from heapq import heappush, heappop
        A = [[] for _ in range(V)]
        for u, v, w in edges:
            A[u].append((v, w))
        r = float('inf')
        for i in range(V):
            D = [int(1e9)] * V
            P = [-1] * V
            D[i] = 0
            Q = [(0, i)]
            while Q:
                d, u = heappop(Q)
                for v, w in A[u]:
                    if D[v] > d + w:
                        D[v] = d + w
                        P[v] = u
                        heappush(Q, (D[v], v))
                    elif P[u] != v and P[v] != u:
                        r = min(r, D[u] + D[v] + w)
        return -1 if r == float('inf') else r

🎯 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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