Understanding the Problem
We are given a set of stones on a 2D grid. Each stone is placed at a unique coordinate (x, y)
. We are allowed to remove a stone if there is at least one other stone in the same row or column.
The goal is to find the maximum number of stones we can remove, following the given rule.
At first glance, this may not seem like a graph problem—but it is. Think of each stone as a node in a graph. If two stones share the same row or column, draw an edge between them. The stones form connected components through these edges.
From each connected component, we can remove all the stones except one. Therefore, the number of stones we can remove is:
total number of stones - number of connected components
Step-by-Step Solution with Example
step 1: Represent the Problem as a Graph
Each stone is a node. If two stones have the same row or the same column, draw an edge between them.
For example, suppose the stones are placed at the following coordinates:
[[0, 0], [0, 1], [1, 0], [1, 2], [2, 2], [2, 3]]
We connect stones sharing rows or columns:
- [0, 0] is connected to [0, 1] and [1, 0]
- [1, 2] is connected to [2, 2]
- [2, 2] is connected to [2, 3]
We get two connected components: one with [0, 0], [0, 1], [1, 0] and another with [1, 2], [2, 2], [2, 3].
step 2: Count Connected Components
We can use Depth-First Search (DFS) to visit all stones in a connected component.
Maintain a visited set. For each unvisited stone, do a DFS and mark all stones in that component as visited. Count how many times you initiate DFS—this gives the number of connected components.
step 3: Calculate the Answer
Once we know the number of connected components, subtract that from the total number of stones.
In our example:
Total stones = 6
Connected components = 2
Removable stones = 6 - 2 = 4
Edge Cases
- No stones: If the input is empty, the answer is 0.
- All stones in different rows and columns: No stone shares a row or column, so no stone can be removed. Each stone is its own component.
- All stones in the same row or column: All stones belong to one component, so we can remove all except one.
- Duplicate coordinates: Problem assumes all coordinates are unique, but validation might be necessary in real input.
Finally
This problem is a classic example of reducing a 2D grid problem into a graph traversal problem. By modeling the connections between stones based on rows and columns, we were able to use DFS to find connected components and calculate the solution efficiently.
Always remember: when you’re allowed to remove things based on indirect connections, it often signals that a graph approach will help.
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