Leetcode daily question (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.

In fact, this is a graph problem , Find isolated subgraphs from the graph , At the end of each subgraph, there will be a point that cannot be eliminated , So the number of points that can be eliminated is the number of all points - Number of orphan subgraphs .

struct Solution;

use std::collections::{
    HashMap, HashSet};

impl Solution {
    
    fn dfs(
        current: (usize, usize),
        i: usize,
        xs: &Vec<Vec<usize>>,
        ys: &Vec<Vec<usize>>,
        parents: &mut HashMap<(usize, usize), usize>,
        visited: &mut HashSet<(usize, usize)>,
    ) {
    
        if visited.contains(&current) {
    
            return;
        }
        visited.insert(current);
        for x in &xs[current.1] {
    
            Solution::dfs((*x, current.1), i, xs, ys, parents, visited);
        }
        for y in &ys[current.0] {
    
            Solution::dfs((current.0, *y), i, xs, ys, parents, visited);
        }
        parents.insert(current, i);
    }
    pub fn remove_stones(stones: Vec<Vec<i32>>) -> i32 {
    
        let mut xs = vec![Vec::new(); 10usize.pow(4) + 1];
        let mut ys = vec![Vec::new(); 10usize.pow(4) + 1];
        for stone in &stones {
    
            ys[stone[0] as usize].push(stone[1] as usize);
            xs[stone[1] as usize].push(stone[0] as usize);
        }
        let mut parents = HashMap::new();
        for (i, stone) in stones.iter().enumerate() {
    
            Solution::dfs(
                (stone[0] as usize, stone[1] as usize),
                i,
                &xs,
                &ys,
                &mut parents,
                &mut HashSet::new(),
            );
        }
        let iso: HashSet<usize> = parents.values().map(|v| *v).collect();
        (stones.len() - iso.len()) as i32
    }
}