684. Redundant Connection

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

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:
    def findRedundantConnection(self, edges: list[list[int]]) -> list[int]:
        N = len(edges)
        par = [i for i in range(N + 1)]
        rank = [1] * (N + 1)

        def find(n):
            if n != par[n]:
                par[n] = find(par[n])
            return par[n]

        def union(n1, n2):
            p1, p2 = find(n1), find(n2)
            if p1 == p2:
                return False

            if rank[p1] > rank[p2]:
                par[p2] = p1
                rank[p1] += rank[p2]
            else:
                par[p1] = p2
                rank[p2] += rank[p1]

            return True

        for n1, n2 in edges:
            if not union(n1, n2):
                return [n1, n2]