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:
def DFS(self, node, graph, visited, src, dest):
visited.add(node)
for i in graph.get(node, []):
if i not in visited:
if (node, i) != (src, dest) and (i, node) != (dest, src):
self.DFS(i, graph, visited, src, dest)
def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
n = len(edges)
graph = {}
numnodes = 0
for a, b in edges:
numnodes = max(numnodes, a, b)
graph[a] = graph.get(a, set()).union({b})
graph[b] = graph.get(b, set()).union({a})
for k in range(n - 1, -1, -1):
a, b = edges[k]
visited = set()
self.DFS(a, graph, visited, a, b)
if len(visited) == numnodes:
return edges[k]
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