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] = [x`

represents the location of the _{i}, y_{i}]`i`

stone, return ^{th}*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:5Explanation: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:3Explanation: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:0Explanation:[0,0] is the only stone on the plane, so you cannot remove it.

**Constraints:**

`1 <= stones.length <= 1000`

`0 <= x`

_{i}, y_{i}<= 10^{4}- No two stones are at the same coordinate point.

```
struct Solution;
use std::collections::HashMap;
impl Solution {
fn remove_stones(stones: Vec<Vec<i32>>) -> i32 {
let n = stones.len();
let mut row: HashMap<i32, Vec<usize>> = HashMap::new();
let mut col: HashMap<i32, Vec<usize>> = HashMap::new();
for i in 0..n {
let r = stones[i][0];
let c = stones[i][1];
row.entry(r).or_default().push(i);
col.entry(c).or_default().push(i);
}
let mut visited: Vec<bool> = vec![false; n];
let mut island = 0;
for i in 0..n {
if !visited[i] {
visited[i] = true;
Self::dfs(i, &mut visited, &stones, &row, &col);
island += 1;
}
}
(n - island) as i32
}
fn dfs(
u: usize,
visited: &mut [bool],
stones: &[Vec<i32>],
row: &HashMap<i32, Vec<usize>>,
col: &HashMap<i32, Vec<usize>>,
) {
let r = stones[u][0];
let c = stones[u][1];
for &v in &row[&r] {
if !visited[v] {
visited[v] = true;
Self::dfs(v, visited, stones, row, col);
}
}
for &v in &col[&c] {
if !visited[v] {
visited[v] = true;
Self::dfs(v, visited, stones, row, col);
}
}
}
}
#[test]
fn test() {
let stones = vec_vec_i32![[0, 0], [0, 1], [1, 0], [1, 2], [2, 1], [2, 2]];
let res = 5;
assert_eq!(Solution::remove_stones(stones), res);
let stones = vec_vec_i32![[0, 0], [0, 2], [1, 1], [2, 0], [2, 2]];
let res = 3;
assert_eq!(Solution::remove_stones(stones), res);
let stones = vec_vec_i32![[0, 0]];
let res = 0;
assert_eq!(Solution::remove_stones(stones), res);
}
```