<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 Solution:
    def removeStones(self, stones: List[List[int]]) -> int:
        #line sweep algorithm (hmm if this idea doesnt work out think connected compo)
        #line sweep in x direction
        #line sweep in y direction
        #in particular sweep in x direction, but you must leave one
        #make sure the one you leave is the one that has stones in the x direction
        #if possible
        #well turns out thats a good strategy to consider
        #but the issue is at a corner L, we can have vertices/stones in one row
        #affecting stones in another row
        #this makes it difficult to do a sweep since block x in sweep may be affected
        #by an earlier block y removal of elements
        #this may cause use to think we may need to keep alternatively trying
        #x sweep -> y sweep -> x sweep --> y sweep
        #in that case recognize that the x and y cannot be treated independently
        #this essentially is showing that some stonnes/vertices are essentially
        #connected to another
        #we can like see this a rubber band and reduce the component down to 1 point
        
        #(0,0) (0,1)
        #(1,0)         (1,2)
        #      (2,1)   (2,2)

        #(0,0) (0,1)
        #(1,0)         (1,2)
        #      (2,1)

        #(0,0) (0,1)
        #(1,0)         (1,2)
        #     

        #(0,0) (0,1)
        #           (1,2)
        #  

        #connected components can be solved in two general ways
        #dfs & union find data structures
        #first we construct edges from nodes given
        #since every edge can be between any pair of nodes/vertices
        #we iterate over every pair of edges
        #we form a complete graph. We essentialy don't care about the tireless
        #work of try to form the exact graph representation with direct/indirectly
        #reachable nodes. If it can be reached, we go straight there.
        #This will require O(V^2) time.
        #We can actually construct the actual graph based on a set of actual direct
        #edges in O(V) time by separating work between x and y coordinates
        #and applying a sweep line algorithm to check if two nodes share an edge
        #For simplicity, we'll do the easy one (create all direct edges) with dfs
        #Essentially creating a complete graph. (V^2) edges
        #we can also use a row map/array and index map/array but will inevitably
        #have O(V^2) time complexity due to dfs run

        def dfs(stone):
            if stone in visited:
                return
            visited.add(stone)
            for v2 in g[stone]:
                dfs(v2)
            
        
        N=len(stones)
        g=dict()
        for i in range(N):
            stone1=tuple(stones[i])
            for j in range(i+1):
                stone2=tuple(stones[j])
                if stone1[0]==stone2[0] or stone1[1]==stone2[1]:
                    if stone1 not in g:
                        g[stone1]=[]
                    if stone2 not in g:
                        g[stone2]=[]
                    g[stone1].append(stone2)
                    g[stone2].append(stone1)

        visited=set()
        count=0
        for stone in g:
            if stone in visited:
                continue
            count+=1
            dfs(stone)
        
        return N-count

        #connected components can be improved with Union Find Data structure
        #Do for homework



        

    

        
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