grid47 Exploring patterns and algorithms
Aug 30, 2024
4 min read
Solution to LeetCode 684: Redundant Connection Problem
You are given a graph that started as a tree with n nodes. One additional edge has been added, creating a cycle. Your task is to find and return the redundant edge that, when removed, would turn the graph back into a tree.
Problem
Approach
Steps
Complexity
Input: The input is a list of edges in a graph, where each edge connects two nodes. The graph has one extra edge added, causing a cycle.
Example: [[1,2],[1,3],[2,3]]
Constraints:
• 3 <= n <= 1000
• edges[i].length == 2
• 1 <= ai < bi <= edges.length
• ai != bi
• There are no repeated edges.
• The graph is connected.
Output: The output should be the redundant edge that can be removed to make the graph a valid tree.
Example: [2, 3]
Constraints:
• The output will be a pair of nodes representing the redundant edge.
Goal: The goal is to find the redundant edge that, when removed, removes the cycle and returns the graph to a tree structure.
Steps:
• 1. Traverse the list of edges.
• 2. For each edge, check if adding it forms a cycle.
• 3. If a cycle is formed, return the current edge as the redundant one.
Goal: The problem has constraints that ensure the graph remains manageable.
Steps:
• The graph contains at least three nodes and at most 1000 nodes.
• Each edge connects two distinct nodes, and there are no repeated edges.
Assumptions:
• The graph is initially a tree, and only one extra edge is added to form a cycle.
• Input: [[1,2],[1,3],[2,3]]
• Explanation: In this example, adding the edge [2,3] forms a cycle, and removing this edge restores the graph to a tree.
• Input: [[1,2],[2,3],[3,4],[1,4],[1,5]]
• Explanation: Here, the edge [1,4] is redundant because its addition creates a cycle between nodes 1, 2, 3, and 4.
Approach: We can use the Union-Find algorithm to detect cycles and identify the redundant edge.
Observations:
• We need to detect if adding an edge creates a cycle, which is the point at which we find the redundant edge.
• The Union-Find algorithm helps efficiently determine if two nodes are already connected, which helps us find the redundant edge.
Steps:
• 1. Create a Union-Find data structure to keep track of connected components.
• 2. Iterate through each edge, attempting to union the nodes it connects.
• 3. If the union operation fails (i.e., the nodes are already connected), this is the redundant edge.
Empty Inputs:
• There will always be at least three nodes and one extra edge.
Large Inputs:
• The algorithm needs to handle up to 1000 edges efficiently.
Special Values:
• A graph with the minimum number of nodes (3) will still have one extra edge causing a cycle.
Constraints:
• The Union-Find approach works efficiently under the given constraints.
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
int n = edges.size();
UF* uf =new UF(n);
for(int i =0; i < n; i++)
if(!uf->uni(edges[i][0], edges[i][1])) {
return edges[i];
}
return edges[0];
}
