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Question
Solution

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.

Approach #1: DFS [Accepted]
Approach #2: Union-Find [Accepted]

Approach #1: DFS [Accepted]

Intuition and Algorithm

For each edge (u, v), traverse the graph with a depth-first search to see if we can connect u to v. If we can, then it must be the duplicate edge.

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Complexity Analysis

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Time Complexity: $O(N^2)$ where $N$ is the number of vertices (and also the number of edges) in the graph. In the worst case, for every edge we include, we have to search every previously-occurring edge of the graph.
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Space Complexity: $O(N)$ . The current construction of the graph has at most $N$ nodes.
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Approach #2: Union-Find [Accepted]

Intuition and Algorithm

If we are familiar with a Disjoint Set Union (DSU) data structure, we can use this in a straightforward manner to solve the problem: we simply find the first edge occurring in the graph that is already connected. The rest of this explanation will focus on the details of implementing DSU.

A DSU data structure can be used to maintain knowledge of the connected components of a graph, and query for them quickly. In particular, we would like to support two operations:

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dsu.find(node x), which outputs a unique id so that two nodes have the same id if and only if they are in the same connected component, and:
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dsu.union(node x, node y), which draws an edge (x, y) in the graph, connecting the components with id find(x) and find(y) together.
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To achieve this, we keep track of parent, which remembers the id of a smaller node in the same connected component. If the node is it\'s own parent, we call this the leader of that connected component.

A naive implementation of a DSU structure would look something like this:

Psuedocode

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We use two techniques to improve the run-time complexity: path compression, and union-by-rank.

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Path compression involves changing the x = parent[x] in the find function to parent[x] = find(parent[x]). Basically, as we compute the correct leader for x, we should remember our calculation.
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Union-by-rank involves distributing the workload of find across leaders evenly. Whenever we dsu.union(x, y), we have two leaders xr, yr and we have to choose whether we want parent[x] = yr or parent[y] = xr. We choose the leader that has a higher following to pick up a new follower.
\nSpecifically, the meaning of rank is that there are less than 2 ^ rank[x] followers of x. This strategy can be shown to give us better bounds for how long the recursive loop in dsu.find could run for.
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Alternate Implementation of DSU without Union-By-Rank\n

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Complexity Analysis

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Time Complexity: $O(N\\alpha(N)) \\approx O(N)$ , where $N$ is the number of vertices (and also the number of edges) in the graph, and $\\alpha$ is the Inverse-Ackermann function. We make up to $N$ queries of dsu.union, which takes (amortized) $O(\\alpha(N))$ time. Outside the scope of this article, it can be shown why dsu.union has $O(\\alpha(N))$ complexity, what the Inverse-Ackermann function is, and why $O(\\alpha(N))$ is approximately $O(1)$ .
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Space Complexity: $O(N)$ . The current construction of the graph (embedded in our dsu structure) has at most $N$ nodes.
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Analysis written by: @awice

684. Redundant Connection

Approach #1: DFS [Accepted]

Approach #2: Union-Find [Accepted]