<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.rank = [0 for i in range(n+1)]
        self.size = [1 for i in range(n+1)]
        self.parent = [i for i in range(n+1)]
        
    def findUPar(self,u):
        if u == self.parent[u]:
            return u 
        self.parent[u] = self.findUPar(self.parent[u])
        return self.parent[u]
        
    def unionByRank(self,u,v):
        parentA = self.findUPar(u)
        parentB = self.findUPar(v)
        
        if parentA == parentB:
            return
        
        if self.rank[parentA] < self.rank[parentB]:
            self.parent[parentA] = parentB
        elif self.rank[parentB] < self.rank[parentA]:
            self.parent[parentB] = parentA
        else:
            self.parent[parentB] = parentA
            self.rank[parentA] += 1

    def unionBySize(self,u,v):
        parentA = self.findUPar(u)
        parentB = self.findUPar(v)

        if parentA == parentB:
            return
        
        if self.size[parentA] < self.size[parentB]:
            self.parent[parentA] = parentB
            self.size[parentB] += self.size[parentA]
        else:
            self.parent[parentB] = parentA
            self.size[parentA] += self.size[parentB]
       
class Solution:
    def removeStones(self, stones: List[List[int]]) -> int:
        maxR, maxC = 0,0
        for stone in stones:
            maxR = max(maxR, stone[0])
            maxC = max(maxC, stone[1])
        
        ds = DisJointSet(maxR + maxC + 2)
        totalStones = 0
        stoneNodes = dict()

        for stone in stones:
            a,b = stone
            ds.unionBySize(a, b + maxR + 1)
            totalStones += 1
            stoneNodes[a] = 1
            stoneNodes[b + maxR + 1] = 1

        component = 0
        print(stoneNodes)
        print(ds.parent)
        for i in stoneNodes.keys():
            if ds.parent[i] == i:
                component += 1        

        return totalStones - component

# 0 1 2 3, 4 5 6 7
# 0 1 2 3
18400