Most Stones Removed with Same Row or Column
A medium Graph interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Graph
- Sheets
- 1
- Core for
- 12 roles
The problem
Given a list of stones where each stone is at a coordinate [row, col] on a 2D plane, return the maximum number of stones that can be removed. A stone can be removed if another stone remains in the same row or column.
Example 1
- Input
- stones = [[0,1],[1,0],[1,2],[2,1],[3,4]]
- Output
- 2
- Why
- Three groups share rows or columns: the column-1 pair, the row-1 pair, and the lone stone. 5 stones minus 3 groups leaves 2 removable.
Example 2
- Input
- stones = [[0,0],[1,2],[2,1],[3,3],[4,2]]
- Output
- 1
- Why
- Four groups remain: (0,0), (2,1), (3,3) and the column-2 pair (1,2) with (4,2). Five stones minus four groups leaves 1 removable.
Constraints
- 1 <= stones.length <= 1000
- 0 <= stones[i][0], stones[i][1] <= 10^4
How to think about it
Updated 2026-09-29Stones that share a row or column belong to a single connected component. Within any connected component of size k, deleting leaves of a spanning tree removes k - 1 stones one at a time until a single pivot stone remains. Thus the answer is always `total stones - number of connected components`. Representing rows and columns as bipartite nodes makes component detection immediate.
Approaches, worst first
Pairwise stone graph traversal
time O(N^2) · space O(N^2)
Construct an explicit adjacency list by comparing every pair of stones and creating an edge whenever they share a row or column. Count connected components via DFS and subtract that count from total stones. Pairwise edge evaluation requires quadratic comparisons across stone coordinates.
DSU bipartite coordinate unionWrite this one
time O(N * α(N)) · space O(N)
Create a DSU keyed on coordinates. For each stone `[r, c]`, union row index `r` with bit-inverted column index `~c` to prevent collisions. Record each visited coordinate in a set. The answer is `stones.length - uniqueComponentRoots`.
Where people lose marks · 3
- Index collisions between row 0 and column 0; using bitwise negation `~c` or adding a safe constant offset `c + 10001` keeps row and column node namespaces disjoint.
- Counting parent entries for coordinates that have no stones present; only count unique roots for coordinates referenced by actual stones.
- A single stone (length 1) must return 0 because it cannot be removed without leaving another stone in its row or column.
The theory behind it
Graph — the ground this problem stands on. All Graph problems
What Graph is
A graph is a network of individual points, called vertices or nodes, connected by lines called edges. Think of a subway transit map, an electrical circuit, or a web of social friends. Unlike a tree, a graph has no designated top node and no parent-child hierarchy. Connections can run one-way or both ways, and paths can loop back on themselves to form closed cycles.
When to reach for it
Reach for graph algorithms when inputs describe relationships, networks, flights between cities, course prerequisites, or clone networks. Signals include finding the shortest route across unweighted connections, ordering tasks that depend on earlier tasks, counting isolated clusters, or checking whether a path contains an infinite loop. Whenever problems present pairs of related entities and ask for reachability, distances, or dependencies, graph representations apply.
How the pattern works
First convert edge lists into an adjacency list, mapping each node to an array of its neighbors. Choose your exploration strategy based on the goal: use a queue and breadth-first search to find the shortest path in unweighted networks, or use recursion and depth-first search to explore full paths and detect cycles. Because graphs can have loops, always track visited nodes in a set or boolean array. Add nodes to the visited set at the moment they enter the queue so they are never visited twice.
More Graph problems
Target Roles
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