# 947 Most Stones Removed with Same Row or Column

On a 2D plane, we place `n` stones at some integer coordinate points. Each coordinate point may have at most one stone.

A stone can be removed if it shares either the same row or the same column as another stone that has not been removed.

Given an array `stones` of length `n` where `stones[i] = [xi, yi]` represents the location of the `ith` stone, return the largest possible number of stones that can be removed.

Example 1:

``````Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
Output: 5
Explanation: One way to remove 5 stones is as follows:
1. Remove stone [2,2] because it shares the same row as [2,1].
2. Remove stone [2,1] because it shares the same column as [0,1].
3. Remove stone [1,2] because it shares the same row as [1,0].
4. Remove stone [1,0] because it shares the same column as [0,0].
5. Remove stone [0,1] because it shares the same row as [0,0].
Stone [0,0] cannot be removed since it does not share a row/column with another
stone still on the plane.
``````

Example 2:

``````Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
Output: 3
Explanation: One way to make 3 moves is as follows:
1. Remove stone [2,2] because it shares the same row as [2,0].
2. Remove stone [2,0] because it shares the same column as [0,0].
3. Remove stone [0,2] because it shares the same row as [0,0].
Stones [0,0] and [1,1] cannot be removed since they do not share a row/column
with another stone still on the plane.
``````

Example 3:

``````Input: stones = [[0,0]]
Output: 0
Explanation: [0,0] is the only stone on the plane, so you cannot remove it.
``````

Constraints:

• `1 <= stones.length <= 1000`
• `0 <= xi, yi <= 104`
• No two stones are at the same coordinate point.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 `````` ``````class Solution: def removeStones(self, stones: List[List[int]]) -> int: UF = {} def find(x: int) -> int: if x != UF[x]: UF[x] = find(UF[x]) return UF[x] def union(x: int, y: int) -> None: UF.setdefault(x, x) UF.setdefault(y, y) UF[find(x)] = find(y) for i, j in stones: union(i, ~j) return len(stones) - len({find(x) for x in UF})``````