The Question

Dynamic Cycle Parity Maintenance

You are given $n$ nodes labeled from $0$ to $n-1$ and a sequence of undirected edges. Each edge is given as a triplet $[u_i, v_i, w_i]$, where $u_i$ and $v_i$ are nodes and $w_i \in \{0, 1\}$ is the weight. Initially, the graph has no edges. You must process the edges in the given order. An edge is added to the graph if and only if its addition does not create any cycle whose sum of edge weights is odd. Return the total number of edges successfully added to the graph. Constraints: - $1 \le n \le 10^5$ - $1 \le \text{edges.length} \le 10^5$ - $0 \le u_i, v_i < n$ - $w_i \in \{0, 1\}$

Java

DSU

Questions & Insights

Clarifying Questions

What are the constraints on $n$ and the number of edges?

Assumption:

n

and the number of edges can be up to

10^5

, requiring an

O(E \alpha(n))

O(E \log n)

solution.

Can the graph contain self-loops or multiple edges between the same nodes?

Assumption: Yes. A self-loop

(u, u, w)

creates a cycle of length 1. If

w=1

, the cycle sum is odd, so the edge should be rejected. If

w=0

, it's allowed. For multiple edges, each must be processed sequentially based on the current state of the graph.

Does "every cycle" include cycles formed by adding the new edge that consist of previously added edges?

Assumption: Yes. Adding an edge

(u, v, w)

is only permitted if all cycles created by this addition (and all cycles existing previously) have an even weight sum. Since we only add edges that maintain this property, we only need to check if the new cycle formed by the edge

(u, v, w)

and the existing path between

u

and

v

has an even sum.

Is the graph potentially disconnected?

Assumption: Yes, the graph starts with no edges and nodes may remain in separate components. The "even cycle" rule applies within each component.

Thinking Process

Parity Property: A graph where every cycle has an even sum of weights is analogous to a bipartite graph. In a bipartite graph, all cycles have even length. Here, if we treat weight 1 as "different" and weight 0 as "same", we are checking if the graph is "consistent" with a global parity assignment

P(u) \in \{0, 1\}

such that for every edge

(u, v, w)

P(u) \oplus P(v) = w

Cycle Consistency: When adding an edge

(u, v, w)

, if

u

and

v

are already in the same connected component, there exists a unique XOR-path sum

dist(u, v)

between them (since all existing cycles are even). The new edge forms a cycle with weight

dist(u, v) \oplus w

. For the cycle to be even, we must have

dist(u, v) \oplus w = 0

, or

dist(u, v) = w

Disjoint Set Union (DSU) with Weights: To efficiently track components and XOR-distances to a root, use DSU where each node stores its parent and the XOR distance to its parent. During find operations (with path compression), we calculate the XOR distance from the node to the root.

Union Logic: If

u

and

v

are in different components, adding the edge will not create a cycle. We merge the components and set the XOR distance between the roots such that the path through the new edge satisfies

P(u) \oplus P(v) = w

Implementation Breakdown

Problem Set

Functional Requirements:

Maintain a dynamic graph.

For each edge, check if it maintains the "even cycle sum" property.

Return the total count of edges added.

Constraints:

n

is a positive integer.

w_i \in \{0, 1\}

Time complexity should be near-linear relative to the number of edges.

Approach

Algorithm: Disjoint Set Union (DSU) with Path Compression and Union by Rank (or Size).

Data Structure: Weighted DSU (specifically, XOR-weighted DSU).

Complexity:

Time:

O(E \cdot \alpha(n))

, where

E

is the number of edges and

\alpha

is the inverse Ackermann function.

Space:

O(n)

to store parents, weights, and ranks.

Implementation

Wrap Up

Advanced Topics

Graph Theory Context: This problem is equivalent to maintaining a 2-colorable graph where the colors are constrained by edge weights. This is a specific case of the 2-SAT problem or T-join in graph theory, though DSU is the most efficient way to handle it dynamically.

Alternative - Dynamic Connectivity: If edges could be deleted, we would need a more complex structure like a Link-Cut Tree or Euler Tour Tree to maintain XOR sums along paths in

O(\log n)

Parallelization: Processing edges in order is strictly sequential. However, if the query was "are these edges valid in a static graph", we could use parallel BFS/DFS. In the dynamic version, we can only parallelize if edges are independent (different components), but determining independence requires sequential processing anyway.

Readability vs Performance: While recursive find is elegant, for extremely deep trees (though unlikely with rank and compression), an iterative find with a two-pass approach could prevent StackOverflowError.