Most Stones Removed with Same Row or Column

Problem

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 <= 10^4
No two stones are at the same coordinate point.

Solution (Java)

class Solution {
    private final int[] roots = new int[20002];

    public int removeStones(int[][] stones) {
        for (int[] stone : stones) {
            init(stone[0] + 1, roots);
            init(stone[1] + 10000, roots);
            union(stone[0] + 1, stone[1] + 10000);
        }

        HashSet<Integer> set = new HashSet<>();
        for (int n : roots) {
            if (n == 0) {
                continue;
            }
            set.add(find(n));
        }

        return stones.length - set.size();
    }

    private void init(int i, int[] roots) {
        if (roots[i] != 0) {
            return;
        }
        roots[i] = i;
    }

    private void union(int i, int j) {
        int ri = find(i);
        int rj = find(j);
        if (ri == rj) {
            return;
        }
        roots[ri] = rj;
    }

    private int find(int i) {
        int cur = i;
        while (cur != roots[cur]) {
            cur = roots[roots[cur]];
        }
        return cur;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).