LeetCode 1489. Find Critical and Pseudo-Critical Edges in Minimum Spanning Tree Solution in Java, C++, Python & Go | Explanation + Code

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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

PythonJavaC++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(code-box)

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