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)