2025-01-29 Daily Challenge

Today I have done leetcode's January LeetCoding Challenge with cpp.

January LeetCoding Challenge 29

Description

Redundant Connection

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 {
  int parent[1000];

  int find(int x) {
    if(x != parent[x]) parent[x] = find(parent[x]);
    return parent[x];
  }
  
  void merge(int x, int y) {
    int px = find(x);
    int py = find(y);
    parent[px] = py;
  }

  void init() {
    for(int i = 0; i < 1000; i++) parent[i] = i;
  }
public:
  vector<int> findRedundantConnection(vector<vector<int>>& edges) {
    init();
    for(auto &edge : edges) {
      int x = edge[0] - 1;
      int y = edge[1] - 1;
      int px = find(x);
      int py = find(y);
      if(px != py) merge(x, y);
      else return edge;
    }
    return {-1, -1};
  }
};

// Accepted
// 39/39 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 99.47 % of cpp submissions (12.5 MB)