Redundant Connection
Created: March 2, 2020 by [lek-tin]
Last updated: March 2, 2020
In this problem, a tree is an undirected graph that is connected and has no cycles.
The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, ..., 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 resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] with u < v, that represents an undirected edge connecting nodes u and v.
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 given 2D-array. The answer edge [u, v] should be in the same format, with u < v.
Example 1
Input: [[1,2], [1,3], [2,3]]
Output: [2,3]
Explanation: The given undirected graph will be like this:
1
/ \
2 - 3
Example 2
Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
Output: [1,4]
Explanation: The given undirected graph will be like this:
5 - 1 - 2
| |
4 - 3
Note
- The size of the input 2D-array will be between 3 and 1000.
- Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.
Solution (union-find with rank)
class Solution:
root = []
rank = []
def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
# we skip 0 index
N = len(edges)+1
self.root = [i for i in range(N)]
self.rank = [0 for i in range(N)]
for connec in edges:
x, y = connec
if not self.union(x, y):
return connec
return []
def find(self, x):
if self.root[x] != x:
# path compression
self.root[x] = self.find(self.root[x])
return self.root[x]
def union(self, x, y):
root_x = self.find(x)
root_y = self.find(y)
if root_x == root_y:
return False
if self.rank[root_x] > self.rank[root_y]:
self.root[root_y] = root_x
elif self.rank[root_x] < self.rank[root_y]:
self.root[root_x] = root_y
else:
self.root[root_x] = root_y
self.rank[root_y] += 1
return True
alternatively: we can implement find iteratively
def find(x, roots):
while root[x] != x:
roots[x] = roots[roots[x]]
x = roots[x]
return x