> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Find Critical and Pseudo-Critical Edges in

> Tested Python solution for LeetCode 1489 with 12 pytest cases. Generate a practice environment with lcpy.

LeetCode 1489, Hard. Topics: Union-Find, Graph Theory, Sorting, Minimum Spanning Tree, Strongly Connected Component. [View on LeetCode](https://leetcode.com/problems/find-critical-and-pseudo-critical-edges-in-minimum-spanning-tree/description/).

Generate this problem as a practice environment: tested reference solution, 12 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 1489   # by problem number
lcpy gen -s find_critical_and_pseudo_critical_edges_in_minimum_spanning_tree   # by problem name
```

## Problem

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.

### Examples

![Example 1](https://assets.leetcode.com/uploads/2020/06/04/ex1.png)

```
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 two edges 0 and 1 appear in all MSTs, therefore they are critical edges.
The edges 2, 3, 4, and 5 are only part of some MSTs, therefore they are considered pseudo-critical edges.
```

![All possible MSTs](https://assets.leetcode.com/uploads/2020/06/04/msts.png)

![Example 2](https://assets.leetcode.com/uploads/2020/06/04/ex2.png)

```
Input: n = 4, edges = [[0,1,1],[1,2,1],[2,3,1],[0,3,1]]
Output: [[],[0,1,2,3]]
Explanation: 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**

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/find_critical_and_pseudo_critical_edges_in_minimum_spanning_tree/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/find_critical_and_pseudo_critical_edges_in_minimum_spanning_tree/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    # Time: O(m^2 * alpha(n)) where m = edges; runs Kruskal once + 2m times
    # Space: O(n + m)
    def find_critical_and_pseudo_critical_edges(
        self, n: int, edges: list[list[int]]
    ) -> list[list[int]]:
        m = len(edges)
        # Sort edges by (weight, original index) so tie-breaks are deterministic.
        order = sorted(range(m), key=lambda i: (edges[i][2], i))
        inf = 10**12

        def kruskal(skip: int = -1, force: int = -1) -> int:
            parent = list(range(n))

            def find(x: int) -> int:
                while parent[x] != x:
                    parent[x] = parent[parent[x]]
                    x = parent[x]
                return x

            weight = 0
            components = n

            if force != -1:
                u, v, w = edges[force]
                parent[find(u)] = find(v)
                weight += w
                components -= 1

            for i in order:
                if i in (skip, force):
                    continue
                u, v, w = edges[i]
                ru, rv = find(u), find(v)
                if ru != rv:
                    parent[ru] = rv
                    weight += w
                    components -= 1
                    if components == 1:
                        break

            return weight if components == 1 else inf

        base = kruskal()

        critical: list[int] = []
        pseudo: list[int] = []
        for i in range(m):
            # Excluding edge i: if the MST gets heavier (or impossible), it is critical.
            if kruskal(skip=i) > base:
                critical.append(i)
            # Forcing edge i into the MST: if weight is unchanged, it is in some MST.
            elif kruskal(force=i) == base:
                pseudo.append(i)

        return [critical, pseudo]
```

## Complexity

| Time                                                             | Space    |
| ---------------------------------------------------------------- | -------- |
| O(m^2 \* alpha(n)) where m = edges; runs Kruskal once + 2m times | O(n + m) |

## Tags

[NeetCode 250](/catalog/neetcode-250), [NeetCode All](/catalog/neetcode).