Problem Description

The problem asks to find critical and pseudo-critical edges in a Minimum Spanning Tree (MST) of a given graph. A critical edge is an edge in the MST whose removal increases the MST's total weight. A pseudo-critical edge is an edge that can be part of some MSTs but not all.

Solution Explanation

The solution employs a Union-Find data structure to efficiently determine MSTs and their weights. The algorithm iterates through each edge, checking if it's critical or pseudo-critical. The core idea revolves around calculating the MST weight with and without the edge in question.

Steps:

1. Preprocessing: Each edge is augmented with its original index. Edges are then sorted by weight for efficient MST construction.

2. Finding the MST Weight: A Union-Find structure determines the MST weight (v) by iteratively adding edges (sorted by weight) to the MST, only including those that don't create cycles.

3. Critical Edge Check: For each edge, the algorithm creates an MST excluding the edge. If the resulting MST weight is greater than v (original MST weight), the edge is critical. This means its removal forces a higher-weight edge into the MST.

4. Pseudo-Critical Edge Check: For each edge, the algorithm creates an MST including the edge (even if it creates a cycle initially). If the resulting MST weight equals v (original MST weight), the edge is pseudo-critical. This implies it can be part of an MST, but not necessarily all MSTs.

Data Structures:

• UnionFind: This custom class provides union (to merge sets) and find (to check set membership) operations, crucial for efficiently detecting cycles during MST construction.

Time Complexity Analysis:

The algorithm has a time complexity of O(E * (E + α(V))), where:

• E is the number of edges.
• V is the number of vertices.
• α(V) is the inverse Ackermann function (practically a constant for all realistic input sizes).

The O(E) factor comes from iterating through each edge. The O(E + α(V)) factor represents the time complexity of building the MST using Union-Find, which is dominated by the union and find operations. The sort operation takes O(E log E) time. As E < V^2, this is still O(E log E)

Space Complexity Analysis:

The space complexity is O(V + E) due to the Union-Find data structure and the storage of edges and results.

Code Explanation (Python)

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]]:
        # Add index to each edge
        for i, e in enumerate(edges):
            e.append(i)
        
        # Sort edges by weight
        edges.sort(key=lambda x: x[2])
        
        # Find MST weight
        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:
            # Check for critical 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
 
            #Check for pseudo-critical edges
            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

The code in other languages (Java, C++, Go) follows the same logic, with minor syntactic differences to accommodate the respective language features. The core algorithm remains consistent across all implementations.