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LeetCode 684: Redundant Connection Solution

Master LeetCode problem 684 (Redundant Connection), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation:

To solve this problem, we can use the Union-Find (Disjoint Set) data structure. We iterate through the edges and for each edge, we check if the two vertices are already part of the same connected component. If they are, it means adding this edge will create a cycle. In that case, we return this edge as the redundant connection.

Algorithm:

  1. Initialize a parent array to keep track of the parent of each node.
  2. Create a helper function find to find the root parent of a node.
  3. Iterate through the edges:
    • For each edge, find the root parent of both vertices.
    • If the root parents are the same, return this edge as the redundant connection.
    • Otherwise, union the two vertices by setting one as the parent of the other.
  4. If no redundant connection is found in the above steps, return an empty array.

Time Complexity:

The time complexity of this algorithm is O(nα(n)), where α(n) is the inverse Ackermann function, which grows very slowly and can be considered as nearly constant.

Space Complexity:

The space complexity is O(n) for storing the parent array.

:

Solution Code

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
        int n = edges.length;
        int[] parent = new int[n + 1];
        Arrays.fill(parent, -1);

        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            if (find(parent, u) == find(parent, v)) {
                return new int[]{u, v};
            }
            union(parent, u, v);
        }
        return new int[0];
    }

    private int find(int[] parent, int node) {
        if (parent[node] == -1) return node;
        return find(parent, parent[node]);
    }

    private void union(int[] parent, int u, int v) {
        int rootU = find(parent, u);
        int rootV = find(parent, v);
        parent[rootU] = rootV;
    }
}

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