You are given a weighted undirected connected graph with n vertices numbered from 0 to n - 1, and an array edges where edges[i] = [a[i], b[i], weight[i]] represents a bidirectional and weighted edge between nodes a[i] and b[i]. 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 = 4, edges = [[0,3,2],[0,2,5],[1,2,4]]
Output: [[0,2,1],[]]Example 2:
Input: n = 5, edges = [[0,3,2],[0,4,2],[1,3,2],[3,4,2],[2,3,1],[1,2,3],[0,1,1]]
Output: [[4,6],[0,1,2,3]]Constraints:
2 <= n <= 1001 <= edges.length <= min(200, n * (n - 1) / 2)edges[i].length == 30 <= a[i] < b[i] < n1 <= weight[i] <= 1000(a[i], b[i]) are distinct.Before attempting this problem, you should be comfortable with:
An edge is critical if removing it increases the MST weight or disconnects the graph. An edge is pseudo-critical if it can appear in some MST but is not mandatory. We test each edge by building the MST without it (to check criticality) and by forcing it into the MST first (to check if it can be part of a valid MST without increasing weight).
mstWeight.mstWeight, the edge is pseudo-critical.class UnionFind:
def __init__(self, n):
self.par = [i for i in range(n)]
self.rank = [1] * n
def find(self, v1):
while v1 != self.par[v1]:
self.par[v1] = self.par[self.par[v1]]
v1 = self.par[v1]
return v1
def union(self, v1, v2):
p1, p2 = self.find(v1), self.find(v2)
if p1 == p2:
return False
if self.rank[p1] > self.rank[p2]:
self.par[p2] = p1
self.rank[p1] += self.rank[p2]
else:
self.par[p1] = p2
self.rank[p2] += self.rank[p1]
return True
class Solution:
def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
for i, e in enumerate(edges):
e.append(i) # [v1, v2, weight, original_index]
edges.sort(key=lambda e: e[2])
mst_weight = 0
uf = UnionFind(n)
for v1, v2, w, i in edges:
if uf.union(v1, v2):
mst_weight += w
critical, pseudo = [], []
for n1, n2, e_weight, i in edges:
# Try without curr edge
weight = 0
uf = UnionFind(n)
for v1, v2, w, j in edges:
if i != j and uf.union(v1, v2):
weight += w
if max(uf.rank) != n or weight > mst_weight:
critical.append(i)
continue
# Try with curr edge
uf = UnionFind(n)
uf.union(n1, n2)
weight = e_weight
for v1, v2, w, j in edges:
if uf.union(v1, v2):
weight += w
if weight == mst_weight:
pseudo.append(i)
return [critical, pseudo]Where is the number of vertices and is the number of edges.
This is a cleaner implementation of the same approach. We use a helper function findMST that optionally skips one edge or forces one edge to be included first. By comparing results against the baseline MST weight, we classify each edge as critical or pseudo-critical.
mstWeight by calling findMST(-1, false) (no exclusions or inclusions).i:findMST(i, false) returns a weight greater than mstWeight, the edge is critical.findMST(i, true) equals mstWeight, the edge is pseudo-critical.class UnionFind:
def __init__(self, n):
self.n = n
self.Parent = list(range(n + 1))
self.Size = [1] * (n + 1)
def find(self, node):
if self.Parent[node] != node:
self.Parent[node] = self.find(self.Parent[node])
return self.Parent[node]
def union(self, u, v):
pu = self.find(u)
pv = self.find(v)
if pu == pv:
return False
self.n -= 1
if self.Size[pu] < self.Size[pv]:
pu, pv = pv, pu
self.Size[pu] += self.Size[pv]
self.Parent[pv] = pu
return True
def isConnected(self):
return self.n == 1
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 e : e[2])
def findMST(index, include):
uf = UnionFind(n)
wgt = 0
if include:
wgt += edges[index][2]
uf.union(edges[index][0], edges[index][1])
for i, e in enumerate(edges):
if i == index:
continue
if uf.union(e[0], e[1]):
wgt += e[2]
return wgt if uf.isConnected() else float("inf")
mst_wgt = findMST(-1, False)
critical, pseudo = [], []
for i, e in enumerate(edges):
if mst_wgt < findMST(i, False):
critical.append(e[3])
elif mst_wgt == findMST(i, True):
pseudo.append(e[3])
return [critical, pseudo]Where is the number of vertices and is the number of edges.
For each edge connecting nodes u and v with weight w, we ask: is there an alternate path from u to v using only edges with weight at most w? We use a modified Dijkstra that finds the minimax path (minimizing the maximum edge weight along the path). If the edge's weight is strictly less than the minimax without it, the edge is critical. If equal to the minimax, it is pseudo-critical.
(u, v, w, idx):minimax(u, v, idx): the minimum possible maximum edge weight on any path from u to v, excluding edge idx.w < minimax(u, v, idx), the edge is critical.minimax(u, v, -1) (no exclusion). If w == minimax(u, v, -1), the edge is pseudo-critical.class Solution:
def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
for i, edge in enumerate(edges):
edge.append(i)
adj = defaultdict(list)
for u, v, w, idx in edges:
adj[u].append((v, w, idx))
adj[v].append((u, w, idx))
def minimax(src, dst, exclude_idx):
dist = [float('inf')] * n
dist[src] = 0
pq = [(0, src)]
while pq:
max_w, u = heappop(pq)
if u == dst:
return max_w
for v, weight, edge_idx in adj[u]:
if edge_idx == exclude_idx:
continue
new_w = max(max_w, weight)
if new_w < dist[v]:
dist[v] = new_w
heappush(pq, (new_w, v))
return float('inf')
critical, pseudo = [], []
for i, (u, v, w, idx) in enumerate(edges):
if w < minimax(u, v, idx):
critical.append(idx)
elif w == minimax(u, v, -1):
pseudo.append(idx)
return [critical, pseudo]Where is the number of vertices and is the number of edges.
After building one MST, edges not in the MST create cycles when added. We use DFS to find the path in the MST between the endpoints of each non-MST edge. If any edge on this path has the same weight as the non-MST edge, both edges are pseudo-critical (they can be swapped). MST edges not identified as pseudo-critical are critical.
DFS to find the unique path in the MST between its endpoints.class UnionFind:
def __init__(self, n):
self.Parent = list(range(n + 1))
self.Size = [1] * (n + 1)
def find(self, node):
if self.Parent[node] != node:
self.Parent[node] = self.find(self.Parent[node])
return self.Parent[node]
def union(self, u, v):
pu = self.find(u)
pv = self.find(v)
if pu == pv:
return False
if self.Size[pu] < self.Size[pv]:
pu, pv = pv, pu
self.Size[pu] += self.Size[pv]
self.Parent[pv] = pu
return True
class Solution:
def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
mst = [[] for _ in range(n)]
mstEdge = []
edge_list = [(w, u, v, i) for i, (u, v, w) in enumerate(edges)]
edge_list.sort()
uf = UnionFind(n)
for w, u, v, i in edge_list:
if uf.union(u, v):
mst[u].append((v, i))
mst[v].append((u, i))
mstEdge.append(i)
def dfs(node):
for next, ind in mst[node]:
if path and ind == path[-1]:
continue
path.append(ind)
if next == dst or dfs(next):
return True
path.pop()
return False
pseudo, mstEdge = set(), set(mstEdge)
for ind in range(len(edges)):
if ind in mstEdge:
continue
path, dst = [], edges[ind][1]
dfs(edges[ind][0])
for i in path:
if edges[i][2] == edges[ind][2]:
pseudo.add(i)
pseudo.add(ind)
return [list(mstEdge - pseudo), list(pseudo)]Where is the number of vertices and is the number of edges.
When sorting edges by weight for Kruskal's algorithm, the original indices get lost. You must store the original index with each edge before sorting, as the problem requires returning indices from the original edge list, not the sorted order.
When testing if an edge is critical by removing it, simply checking if the MST weight increases is insufficient. The graph may become disconnected entirely, making it impossible to form any spanning tree. Always verify that all nodes remain connected (e.g., by checking if the Union-Find structure has exactly one component).
A critical edge must be in every MST, while a pseudo-critical edge can be in some MST but not all. The testing order matters: first check if removing the edge breaks the MST (critical), then check if forcing the edge into the MST still yields minimum weight (pseudo-critical). An edge cannot be both.
Each edge must be tested with a fresh Union-Find structure. Reusing the same Union-Find instance across multiple edge tests will carry over previous union operations, leading to incorrect connectivity states and wrong classifications.
When multiple edges have the same weight, they may be interchangeable in forming an MST. This is precisely what creates pseudo-critical edges. Failing to account for weight ties can cause edges to be incorrectly classified as critical when they are actually pseudo-critical.