# 947. Most Stones Removed with Same Row or Column

## Description

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.

## Solutions

• class Solution {
private int[] p;

public int removeStones(int[][] stones) {
int n = 10010;
p = new int[n << 1];
for (int i = 0; i < p.length; ++i) {
p[i] = i;
}
for (int[] stone : stones) {
p[find(stone[0])] = find(stone[1] + n);
}
Set<Integer> s = new HashSet<>();
for (int[] stone : stones) {
}
return stones.length - s.size();
}

private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
}

• class Solution {
public:
vector<int> p;

int removeStones(vector<vector<int>>& stones) {
int n = 10010;
p.resize(n << 1);
for (int i = 0; i < p.size(); ++i) p[i] = i;
for (auto& stone : stones) p[find(stone[0])] = find(stone[1] + n);
unordered_set<int> s;
for (auto& stone : stones) s.insert(find(stone[0]));
return stones.size() - s.size();
}

int find(int x) {
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
};

• class Solution:
def removeStones(self, stones: List[List[int]]) -> int:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]

n = 10010
p = list(range(n << 1))
for x, y in stones:
p[find(x)] = find(y + n)

s = {find(x) for x, _ in stones}
return len(stones) - len(s)


• func removeStones(stones [][]int) int {
n := 10010
p := make([]int, n<<1)
for i := range p {
p[i] = i
}
var find func(x int) int
find = func(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
for _, stone := range stones {
p[find(stone[0])] = find(stone[1] + n)
}
s := make(map[int]bool)
for _, stone := range stones {
s[find(stone[0])] = true
}
return len(stones) - len(s)
}

• class UnionFind {
p: number[];
size: number[];
constructor(n: number) {
this.p = Array.from({ length: n }, (_, i) => i);
this.size = Array(n).fill(1);
}

find(x: number): number {
if (this.p[x] !== x) {
this.p[x] = this.find(this.p[x]);
}
return this.p[x];
}

union(a: number, b: number): boolean {
const [pa, pb] = [this.find(a), this.find(b)];
if (pa === pb) {
return false;
}
if (this.size[pa] > this.size[pb]) {
this.p[pb] = pa;
this.size[pa] += this.size[pb];
} else {
this.p[pa] = pb;
this.size[pb] += this.size[pa];
}
return true;
}
}

function removeStones(stones: number[][]): number {
const n = stones.length;
const uf = new UnionFind(n);
let ans = 0;
for (let i = 0; i < n; ++i) {
for (let j = 0; j < i; ++j) {
if (stones[i][0] === stones[j][0] || stones[i][1] === stones[j][1]) {
ans += uf.union(i, j) ? 1 : 0;
}
}
}
return ans;
}