<p>On a 2D plane, we place <code>n</code> stones at some integer coordinate points. Each coordinate point may have at most one stone.</p>

<p>A stone can be removed if it shares either <strong>the same row or the same column</strong> as another stone that has not been removed.</p>

<p>Given an array <code>stones</code> of length <code>n</code> where <code>stones[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> represents the location of the <code>i<sup>th</sup></code> stone, return <em>the largest possible number of stones that can be removed</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
<strong>Output:</strong> 5
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> stones = [[0,0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> [0,0] is the only stone on the plane, so you cannot remove it.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= stones.length &lt;= 1000</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
	<li>No two stones are at the same coordinate point.</li>
</ul>

class DisjointSet:
    def __init__(self,n):
        self.parent = [i for i in range(n+1)]
        self.size = [1 for i in range(n+1)]
        self.rank = [0 for i in range(n+1)]

    def FindUltiParent(self,node):
        if self.parent[node] == node:
            return node
        self.parent[node] = self.FindUltiParent(self.parent[node])
        return self.parent[node]

    def UnionBySize(self,u,v):
        ultip_u = self.FindUltiParent(u)
        ultip_v = self.FindUltiParent(v)
        if ultip_u == ultip_v:
            return
        if self.size[ultip_u] > self.size[ultip_v]:
            self.parent[ultip_v] = ultip_u
            self.size[ultip_u] += self.size[ultip_v]
        else:
            self.parent[ultip_u] = ultip_v
            self.size[ultip_v] += self.size[ultip_u]

    def UnionByRank(self,u,v):
        ultip_u = self.FindUltiParent(u)
        ultip_v = self.FindUltiParent(v)
        if ultip_u == ultip_v:
            return
        if self.rank[ultip_u] < self.rank[ultip_v]:
            self.parent[ultip_u] = ultip_v
        elif self.rank[ultip_v] < self.rank[ultip_u]:
            self.parent[ultip_v] = ultip_u
        else:
            self.parent[ultip_v] = ultip_u
            self.rank[ultip_u] += 1


class Solution:
    def removeStones(self, stones: List[List[int]]) -> int:
        maxrow = 0
        maxcol = 0
        n = len(stones)
        for i in stones:
            maxrow = max(maxrow,i[0])
            maxcol = max(maxcol,i[1])
        
        ds = DisjointSet(maxrow + maxcol + 1)
        d = {}

        for i in stones:
            noderow = i[0]
            nodecol = i[1] + maxrow + 1
            ds.UnionBySize(noderow,nodecol)
            d[noderow] = 1
            d[nodecol] = 1

        ct = 0
        for i in d:
            if ds.FindUltiParent(i) == i:
                ct += 1
        
        return n-ct
            
17800