1489. Find Critical and Pseudo-Critical Edges in Minimum Spanning Tree

Description

Given a weighted undirected connected graph with n vertices numbered from 0 to n - 1, and an array edges where edges[i] = [ai, bi, weighti] represents a bidirectional and weighted edge between nodes ai and bi. A minimum spanning tree (MST) is a subset of the graph's edges that connects all vertices without cycles and with the minimum possible total edge weight.

Find all the critical and pseudo-critical edges in the given graph's minimum spanning tree (MST). An MST edge whose deletion from the graph would cause the MST weight to increase is called a critical edge. On the other hand, a pseudo-critical edge is that which can appear in some MSTs but not all.

Note that you can return the indices of the edges in any order.

 

Example 1:

Input: n = 5, edges = [[0,1,1],[1,2,1],[2,3,2],[0,3,2],[0,4,3],[3,4,3],[1,4,6]]
Output: [[0,1],[2,3,4,5]]
Explanation: The figure above describes the graph.
The following figure shows all the possible MSTs:

Notice that the two edges 0 and 1 appear in all MSTs, therefore they are critical edges, so we return them in the first list of the output.
The edges 2, 3, 4, and 5 are only part of some MSTs, therefore they are considered pseudo-critical edges. We add them to the second list of the output.

Example 2:

Input: n = 4, edges = [[0,1,1],[1,2,1],[2,3,1],[0,3,1]]
Output: [[],[0,1,2,3]]
Explanation: We can observe that since all 4 edges have equal weight, choosing any 3 edges from the given 4 will yield an MST. Therefore all 4 edges are pseudo-critical.

 

Constraints:

• 2 <= n <= 100
• 1 <= edges.length <= min(200, n * (n - 1) / 2)
• edges[i].length == 3
• 0 <= ai < bi < n
• 1 <= weighti <= 1000
• All pairs (ai, bi) are distinct.

Solutions

Solution 1

Python3JavaC++Go

class UnionFind:
    def __init__(self, n):
        self.p = list(range(n))
        self.n = n

    def union(self, a, b):
        if self.find(a) == self.find(b):
            return False
        self.p[self.find(a)] = self.find(b)
        self.n -= 1
        return True

    def find(self, x):
        if self.p[x] != x:
            self.p[x] = self.find(self.p[x])
        return self.p[x]


class Solution:
    def findCriticalAndPseudoCriticalEdges(
        self, n: int, edges: List[List[int]]
    ) -> List[List[int]]:
        for i, e in enumerate(edges):
            e.append(i)
        edges.sort(key=lambda x: x[2])
        uf = UnionFind(n)
        v = sum(w for f, t, w, _ in edges if uf.union(f, t))
        ans = [[], []]
        for f, t, w, i in edges:
            uf = UnionFind(n)
            k = sum(z for x, y, z, j in edges if j != i and uf.union(x, y))
            if uf.n > 1 or (uf.n == 1 and k > v):
                ans[0].append(i)
                continue

            uf = UnionFind(n)
            uf.union(f, t)
            k = w + sum(z for x, y, z, j in edges if j != i and uf.union(x, y))
            if k == v:
                ans[1].append(i)
        return ans

class Solution {
    public List<List<Integer>> findCriticalAndPseudoCriticalEdges(int n, int[][] edges) {
        for (int i = 0; i < edges.length; ++i) {
            int[] e = edges[i];
            int[] t = new int[4];
            System.arraycopy(e, 0, t, 0, 3);
            t[3] = i;
            edges[i] = t;
        }
        Arrays.sort(edges, Comparator.comparingInt(a -> a[2]));
        int v = 0;
        UnionFind uf = new UnionFind(n);
        for (int[] e : edges) {
            int f = e[0], t = e[1], w = e[2];
            if (uf.union(f, t)) {
                v += w;
            }
        }
        List<List<Integer>> ans = new ArrayList<>();
        for (int i = 0; i < 2; ++i) {