684. Redundant Connection¶

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684. Redundant Connection

Medium

In this problem, a tree is an undirected graph that is connected and has no cycles.

You are given a graph that started as a tree with n nodes labeled from 1 to n, with one additional edge added. The added edge has two different vertices chosen from 1 to n, and was not an edge that already existed. The graph is represented as an array edges of length n where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the graph.

Return an edge that can be removed so that the resulting graph is a tree of n nodes. If there are multiple answers, return the answer that occurs last in the input.

Example 1:

Input: edges = [[1,2],[1,3],[2,3]]
Output: [2,3]

Example 2:

Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
Output: [1,4]

Constraints:

• n == edges.length
• 3 <= n <= 1000
• edges[i].length == 2
• 1 <= ai < bi <= edges.length
• ai != bi
• There are no repeated edges.
• The given graph is connected.

Solution¶

class Solution {
    private ArrayList<ArrayList<Integer>> adj;
    public int[] findRedundantConnection(int[][] edges) {
        int n = edges.length;
        int ans[] = new int[2];
        adj = new ArrayList<>();
        for(int i = 0; i <= n + 1; i++) adj.add(new ArrayList<>());
        int vis[] = new int[n + 1];
        for(int i = 0; i < edges.length; i++) {
            adj.get(edges[i][0]).add(edges[i][1]);
            adj.get(edges[i][1]).add(edges[i][0]);
            Arrays.fill(vis,0);
            if(cycle(edges[i][0], -1, vis) == true) return new int[]{edges[i][0], edges[i][1]};
         }
         return new int[]{0, 0};
    }
    private boolean cycle(int u, int par, int vis[]) {
        vis[u] = 1;
        boolean temp = false;
        for (int child : adj.get(u)) {
            if (vis[child] == 1 && child == par) continue;
            if (vis[child] == 1) return true;
            temp |= cycle(child, u, vis);
        }
        return temp;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here