Merge Similar Pixels
Problem Statement
In image segmentation, adjacent pixels with similar intensities are often merged into the same region.
Given a list of pixel values and a list of edges (i, j) indicating which pixels are adjacent, merge pixels into groups where the absolute intensity difference is at most threshold.
Return the number of distinct groups after merging.
Applications
- Region-based segmentation
- Color quantization
- Superpixel generation
Constraints
- 1≤len(pixels)≤1000
- 0≤pixels[i]≤255
- 0≤threshold≤255
Example:
pixels = [10, 12, 50, 52], edges = [[0,1], [1,2], [2,3]], threshold = 5
2
Pixels 0,1 merge (diff=2). Pixels 2,3 merge (diff=2). But 1,2 don't merge (diff=38>5).
Merge Similar Pixels: Comprehensive Background and Solution Guide
1. Background Knowledge
This problem models image segmentation as union-find clustering on a graph where pixels are nodes and edges represent adjacency. Key concepts:
- Graph Representation: Pixels as vertices, adjacency list as edges
- Similarity Metric: ∣pixels[i]−pixels[j]∣≤threshold defines mergeable regions
- Connected Components: Final count of distinct groups after transitive merging (if A↔B and B↔C, then A,B,C form one group)
- Applications: Region growing, superpixel generation, color quantization
Prerequisites: Graph theory, Union-Find (Disjoint Set Union - DSU) data structure.
2. Algorithm Approach
Union-Find (DSU) with Path Compression and Union by Rank is optimal:
Core Operations:
1. find(x): Return root parent of x (with path compression)
2. union(x, y): Merge sets containing x and y (by rank for balance)
3. count_components(): Number of unique roots
Why DSU? Handles transitive closure efficiently (A connects B, B connects C → A,B,C connected automatically).
Alternatives (less efficient here):
- DFS/BFS traversal: O(V+E) but requires explicit graph traversal
- Kruskal's MST: Similar but more complex
3. Step-by-Step Strategy
class UnionFind:
def __init__(self, n):
self.parent = list(range(n))
self.rank = * n
self.components = n
def find(self, x):
if self.parent[x] != x:
self.parent[x] = self.find(self.parent[x]) # Path compression
return self.parent[x]
Continue the full explanation
You're reading the free preview. Unlock the complete walkthrough, the code editor, test runner and reference solution with Premium.
Editor locked
The code editor is locked for Pro problems. It is only available for free problems. Please upgrade to gain access to the code editor for all problems.