Redundant Connection — LC Medium

684. Redundant Connection

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.

class Solution:
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        self.root = [i for i in range(len(edges) + 1)]
        
        for vertex1, vertex2 in edges:
            if not self.union(vertex1, vertex2):
                return [vertex1, vertex2]
            
        
    def findRoot(self, target: int) -> int:
        while target != self.root[target]:
            target = self.root[target]
        return target
    
    def union(self, val1: int, val2: int) -> bool:
        root1 = self.findRoot(val1)
        root2 = self.findRoot(val2)
        
        if root1 == root2:
            return False
        else:
            self.root[root2] = root1
            return True