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[11/14] 947. Most Stones Removed with Same Row or Column

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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.
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Runtime: 356 ms, faster than 64.53% of Python3 online submissions for Most Stones Removed with Same Row or Column.
Memory Usage: 18.2 MB, less than 5.27% of Python3 online submissions for Most Stones Removed with Same Row or Column.
from collections import defaultdict
class Solution:
    def removeStones(self, stones: List[List[int]]) -> int:
        rowdic = defaultdict(list)
        coldic = defaultdict(list)
        
        for row, col in stones:
            rowdic[row].append((row,col))
            coldic[col].append((row,col))
            
        visited = set()
        
        def dfs(r,c):
            visited.add((r,c))
            
            for row, col in rowdic[r]:
                if (row,col) not in visited:
                    dfs(row,col)
            
            for row, col in coldic[c]:
                if (row,col) not in visited:
                    dfs(row,col)
                    
        
        island = 0
        
        for row, col in stones:
            if (row,col) not in visited:
                dfs(row,col)
                island += 1
            
        return len(stones) - island
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