684. Redundant Connection

Description

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.

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Solutions

Solution 1

Intuition

UNION FIND

Code

class Solution {
 public:
  const static int N = 1003;
  int p[N];
  int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }
  vector<int> findRedundantConnection(vector<vector<int>> &edges) {
    vector<int> ans;
    for (int i = 1; i <= edges.size(); i++) {
      p[i] = i;
    }
    for (auto &edge : edges) {
      // already in same graph(== has sampe parent), so this edge is redundent
      if (find(edge[0]) == find(edge[1])) {
        ans.push_back(edge[0]);
        ans.push_back(edge[1]);
      } else {
        p[find(edge[0])] = find(edge[1]);
      }
    }

    return ans;
  }
};

Complexity

• Time: O(N)
• Space: O(N)

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