684. Redundant Connection

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.

Update (2017-09-26): We have overhauled the problem description + test cases and specified clearly the graph is an undirected graph. For the directed graph follow up please see Redundant Connection II ). We apologize for any inconvenience caused.

# @lc code=start
using LeetCode

function UFS()
    function findRoot(u::Int, farther::Vector{Int})
        return (u == farther[u]) ? (u) : (farther[u] = findRoot(farther[u], farther))
    end

    function merge(u::Int, v::Int, farther::Vector{Int})
        u = findRoot(u, farther)
        v = findRoot(v, farther)
        (u == v) || (farther[u] = v)
        nothing
    end

    function issameRoot(u::Int, v::Int, farther::Vector{Int})
        return findRoot(u, farther) == findRoot(v, farther)
    end

    findRoot, merge, issameRoot
end

function findRedundantConnection(edges::Vector{Vector{Int}})::Vector{Int}
    farther = collect(1:1000)
    findRoot, merge, issameRoot = UFS()
    for edge in edges
        if issameRoot(edge..., farther)
            return edge
        end
        merge(edge..., farther)
    end
end
# @lc code=end

findRedundantConnection (generic function with 1 method)

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