Approach

Algorithm

In this problem, it is clearly stated that a stone can be removed if it shares either the same row or the same column as another stone that has not been removed. So, we can assume that these types of stones, sharing either the same row or column, are connected and belong to the same group. If we consider example 2:

We can easily spot two different groups in this example. The first group includes the stones [0,0], [0,2], [3,2], and [3,1], and the second one includes [1,3] and [4,3].
If we carefully observe, for each group we can remove all the stones leaving one stone intact. Thus, we can conclude that at most we can remove (size of the group - 1) number of stones from a group as we need to leave one stone untouched for each group.

In other words we can also say that, since from every group one stone remains intact whereas other are removed, the total number of stones that remains after removing maximum number of stones is equal to number of groups of stones.

Consider each stone as a node in a graph.
Create edges between any two stones that are in the same row or same column.
To efficiently group connected stones, use Disjoint Set Union (Union-Find) to unite nodes (stones) that are in the same row or column.
While building DSU, treat each row and each column as a unique node
For each stone, perform union operation between its row index and its column index.
Track the number of unique connected components using DSU parent roots.
The number of stones that can be removed is: total stones - number of connected components.