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.

Rust Solution

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);
}

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